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How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Can you notate a function as an argument of another function?. <p>So I'm writing a research paper. It's a computer science paper and I have written the following expression for calculating the average similarity of a set of images:
<span class="math-container">\begin{equation}
\alpha(X) = \sum_{\{x_1,x_2\} \in [X]^2} \frac{f(x_1,x_2)}{\vert[X]^2\vert}
\end{equation}</span></p>
<p>where <span class="math-container">$\alpha(X)$</span> iterates through all pairs within set of images <span class="math-container">$X$</span>, and <span class="math-container">$f(x_1,x_2)$</span> is a measure of similarity between the pairs.</p>
<p>I'm trying to demonstrate that <span class="math-container">$f(x_1,x_2)$</span> can be either a computational measure of similarity (using some computer vision technique for example), or a human inputted one.</p>
<p>Would it be appropriate to demonstrate this with the following notation
<span class="math-container">\begin{align}
\alpha(X,f) &= \sum_{\{x_1,x_2\} \in [X]^2} \frac{f(x_1,x_2)}{\vert[X]^2\vert} \\
\end{align}</span></p>
<p>so that when I refer to the computational or human version I can write <span class="math-container">$\alpha(X,f^\mathit{comp})$</span> or <span class="math-container">$\alpha(X,f^\mathit{hum})$</span> respectively?</p>
| 0non-cybersec
| Stackexchange |
[WP] To the best of your knowledge, you have been the last person alive for the past 6 years. Your iPhone, however, has somehow regularly received updates this entire time. You have dedicated the rest of your life to finding out the cause and reason of these updates.. | 0non-cybersec
| Reddit |
Meddler on XP and leveling to 30 - "What we're going to do instead, probably in the next patch, is lower the total amount of XP needed to get to 30 by around half...". | 0non-cybersec
| Reddit |
Evaluate: $ \int_0^1 \int_y^1 e^{x^2 }dxdy$. <blockquote>
<p>Evaluate: $$ \int_0^1 \int_y^1 e^{x^2 }dxdy=?$$</p>
</blockquote>
<p>MyTry :</p>
<p>$$\begin{align} \int_0^1 \big(\int_y^1 e^{x^2 }dx\big)dy=?\end{align}$$</p>
<p>Now what ? </p>
<p>$$I=\big(\int_y^1 e^{x^2 }dx\big)=?$$</p>
| 0non-cybersec
| Stackexchange |
Trigonometric Ratios Of Multiple Angles. <p>If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$</p>
<p>I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.</p>
| 0non-cybersec
| Stackexchange |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Probability based on partially ordered sets. <p>Is there any alternative study of "probability" where random variables are defined as maps between a sample space and a <a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="noreferrer">partially ordered set</a>, rather than a sample space and the real interval $[0,1]$? If it exists, what is this theory called and what are some good references?</p>
<p>In such a setting, one could talk about one event being more likely than another, but perhaps not about the absolute probabilty of an event occurring.</p>
<p>I googled around, but what I found were different concepts involving partially ordered sets and probability (e.g., defining stochastic processes on partially ordered sets to generalize the concept of "time"), but not what I'm talking about.</p>
| 0non-cybersec
| Stackexchange |
Is it possible to mount a partition only for some users?. <p>This thread comes from <a href="https://android.stackexchange.com/questions/123615/external-ext4-card-mounted-only-for-root-on-android-5-1-64-bits">here</a>. Even when it is an Android question,
I think the theoretical command line part should better be asked here.</p>
<p>In short: I have achieved to mount a Ext4 file system, but this mount only <strong>exists for root</strong> user.</p>
<p><strong>Details</strong>:</p>
<pre><code>root@unknown:/ # mount | grep "sdcard" -i
/dev/block/mmcblk1p1 /storage/sdcard1 ext4 rw,seclabel,relatime,data=ordered 0 0
root@unknown:/ # exit
u0_a98@unknown:/ $ mount | grep "sdcard"
u0_a98@unknown:/ $ mount | grep "mmcblk"
</code></pre>
<p>As can be seen, the <strong>normal user can not see</strong> the device as mounted. This is, obviously, very different from having <em>no permissions</em> to access it.<br>
This could be some sort of bug, by the way.</p>
<p>Or <strong>is it possible</strong> to do this on Linux?</p>
| 0non-cybersec
| Stackexchange |
Looks like fun..... | 0non-cybersec
| Reddit |
My main hard drive on ubuntu has no free space left. <p>I have a hard drive that keeps filling up with data, but I'm unable to figure out what is causing it.
I tried moving files, but the system just replaces that data with more useless data. This made it tough for me to install anything on my drive. </p>
<p>As soon as I restart my computer after removing files, the info box says:</p>
<blockquote>
<p>Free space: 0 bytes</p>
</blockquote>
<p>I've even freed a gig of data on my hard drive but it still goes back to reducing my free space down to zero. I've looked all over the place, but I couldn't find anything useful that I could use to help me.
Disks is telling me that File Partition 6 74 GB Ext 4 is completely full. File Partition 6 Ext 4 is my Root and Home partition. Unfortunately I'm unable to send a screen shot of the program because I can't open my web browser at the moment, so I'm using my Windows laptop to report the problem.
My current Linux system is Xubuntu 16.04 LTS.</p>
| 0non-cybersec
| Stackexchange |
[Singapore] Red light running van hits little car. | 0non-cybersec
| Reddit |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Are these two functors same (isomorphic)?. <p>Let <span class="math-container">$\mathscr{C}$</span> be the category of schemes (that is the category of schemes over <span class="math-container">$\mathrm{Spec}(\mathbb{Z})$</span>), <span class="math-container">$X,Y$</span> be two schemes. Then we can have two functors: <span class="math-container">$\mathscr{C}^{op}\to \{\mathrm{Sets}\}$</span>, one is <span class="math-container">$h_{X\coprod Y}$</span> (<span class="math-container">$X\coprod Y$</span> is the disjoint union (or coproduct) of <span class="math-container">$X,Y$</span>), where <span class="math-container">$h_{X\coprod Y}(T)$</span> is the <span class="math-container">$Mor(T, X\coprod Y)$</span> for all scheme <span class="math-container">$T$</span>. The other is <span class="math-container">$h_{X}\coprod h_{Y}$</span> defined by <span class="math-container">$(h_{X}\coprod h_{Y})(T)=Mor(T,X)\coprod Mor(T,Y)$</span>. I think these two functors are not isomorphic.</p>
<p>For example, if <span class="math-container">$T=X\coprod Y$</span>, then we have <span class="math-container">$\mathrm{id}: T\to X\coprod Y$</span> which belongs to <span class="math-container">$h_{X\coprod Y}(T)$</span>,but we could not find an element in <span class="math-container">$(h_{X}\coprod h_{Y})(T)$</span> which corresponds it. Hence there is no canonical isomorphism between <span class="math-container">$h_{X\coprod Y}(T)$</span> and <span class="math-container">$(h_{X}\coprod h_{Y})(T)$</span>. Thus these two functors are not isomorphic. Am I right?</p>
| 0non-cybersec
| Stackexchange |
IBM MobileFirst certificate pinning best practices. <p>We are developing an IBM MobileFirst 7.1 Hybrid mobile application, and planning to use the certificate pinning feature.</p>
<p>We can find information on IBM website about the SSL certificate pinning on IBM MobileFirst website : <a href="https://www.ibm.com/support/knowledgecenter/SSHS8R_7.1.0/com.ibm.worklight.dev.doc/monitor/c_cert_pinning_intro.html" rel="noreferrer">Here is a post in the IBM Knowledge Center</a> and <a href="https://mobilefirstplatform.ibmcloud.com/blog/2015/08/14/certificate-pinning-in-ibm-mobilefirst-platform-foundation-7-1/" rel="noreferrer">Here is a tutorial</a> and <a href="https://github.com/IdanAdar/CertificatePinningProject" rel="noreferrer">its sample project/code</a></p>
<p>All these resources are great, but I have some questions :</p>
<ul>
<li>Are there any advantages to use native certificate pinning implementation for Android and iOS when my application is a hybrid application ? </li>
<li>If I use the hybrid environment code, will the SSL certificate be included in the wlapp, and then can be updated using direct update ?</li>
<li>Does the hybrid implementation of certificate pinning work on WindowsPhone 8 (Hybrid app for Silverlight environment) ?</li>
<li>In the tutorial video, I have seen that when SSL pinning is ON, and we call a URL outside our server (google for example) it will fail. Does this mean google maps will fail loading if I enable certificate pinning ? </li>
<li>What happen when the SSL certificate is revoked ?</li>
<li>What happen when the SSL certificate is expired ?</li>
<li>What is the best SSL renewal strategy to keep certificate pinning working and the application up while updating our server SSL certificate ?</li>
</ul>
<p>Please advice</p>
| 0non-cybersec
| Stackexchange |
Vector bundle with flat connection over simply connected manifold is trivial. <p>I am trying to finish a proof of the statement in the title. Let <span class="math-container">$M$</span> be a simply connected smooth manifold and <span class="math-container">$E \twoheadrightarrow M$</span> be a vector bundle with a flat connection. Since the bundle is flat we can choose a local trivialization at every point such that the connection is trivial in all trivializations (with null 1-form matrix). Because of this, and because of the transition formula between the connection matrices of 1-forms, the transition functions between such trivializations are constant. I would like to use this fact, together with the simple connectedness of <span class="math-container">$M$</span> to either reduce the structure group to the trivial group or to build a basis of global sections, but I am a bit rusty with these proofs and I cannot come up with an intelligent idea to do this. Any hint?</p>
| 0non-cybersec
| Stackexchange |
How to anonymize the programs from the terminal?. <p>How to anonymize the programs from your terminal (such as <code>wget</code>, <code>curl</code>, etc.)?</p>
| 0non-cybersec
| Stackexchange |
New York files suit against Trump, alleging his charity engaged in ‘illegal conduct’. | 0non-cybersec
| Reddit |
Koran art exhibit pulled in France after near riot - The
Local. | 0non-cybersec
| Reddit |
Laravel / Eloquent Model Attribute Visibility. <p>Previously ORM's i've used have mapped database columns directly to class properties which allowed you to specific property visibility, just as you normally would to restrict access to certain properties e.g. passwords.</p>
<p>With Eloquent i can't seem to replicate this because database columns are mapped to the internal attributes array which contain no visibility.</p>
<p>My desire is to restrict the scope of access to a user password to only the object i.e. private.</p>
<p>Setting a class property with visibility doesn't work because this property is then outside the scope of the Eloquent model attributes and thus the property is not mapped to the column.</p>
<p>Eloquent $hidden and $guarded properties do not work as these deal with mass output (toArray, toJSON) and mass assignment rather than direct assignment.</p>
<p>I have attempted to use the accessors/mutators (getters/setters) in order to achieve this with mixed results.</p>
<p>Specifying the visibility on the accessor doesn't work because the accessor method called (e.g. getPasswordAttribute) is called from Eloquent\Model->getAttribute method and as such public/protected will always work and private will always fail regardless of where the attribute it accessed from.</p>
<p>What does work however is to stop the Eloquent accessor returning the attribute altogether so any request to $user->password or $user->getAttribute ('password') fails, and then having a separate method with visibility defined in order to return the attribute directly from the Eloquent attributes array only in the scope allowed e.g.</p>
<pre><code>/**
* Return password string only for private scope
* @return string
*/
private function getPassword ()
{
return $this->attributes['password'];
}
/**
* Don't return password with accessor
* @param string $password Password
* @return void
* @throws Exception
*/
public function getPasswordAttribute ($password)
{
throw new Exception ('Password access denied');
}
</code></pre>
<p>This same approach also works for mutators (setters) for anyone wanting setter method visibility.</p>
<p>Does this seem correct or is there a better "Laravel-Approved" way of dealing with this? :)</p>
| 0non-cybersec
| Stackexchange |
How to easily scan a website for destructive malware.. | 0non-cybersec
| Reddit |
Misty day near the peak of El Yunque rainforest in Puerto Rico, incredibly beautiful. | 0non-cybersec
| Reddit |
[Season 1/AGoT Spoilers] Joffery gets internet access.. | 0non-cybersec
| Reddit |
mocking a singleton class. <p>I recently read that making a class singleton makes it impossible to mock the objects of the class, which makes it difficult to test its clients. I could not immediately understand the underlying reason. Can someone please explain what makes it impossible to mock a singleton class? Also, are there any more problems associated with making a class singleton?</p>
| 0non-cybersec
| Stackexchange |
Concerning: presentations of rational numbers into sums. <p>Problem: Prove that all positive rational numbers can be expressed as the finite sum of different numbers $\displaystyle \frac {1} {n}$ ($n$ is a natural number).</p>
<p>Example: $\displaystyle \frac {19}{16}=1+ \frac {1}{8} + \frac {1}{16}.$</p>
<p>*We cant sum numbers as $\displaystyle \frac {3}{16}$ (denominator > 1) but we can sum $\displaystyle \frac {1}{8}+ \frac {1}{16}.$</p>
<p>Any solutions? Suggestions?</p>
| 0non-cybersec
| Stackexchange |
[WDYWT] My outfit to go to the grocery store and pick up oat milk. | 0non-cybersec
| Reddit |
Artwork i had commissioned of my puppy Oni and her namesake Onyxia.. | 0non-cybersec
| Reddit |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Finding the Sum of a series $\frac{1}{1!} + \frac{1+2}{2!} +\frac{1+2+3}{3!}+...$. <p>I need to find the sum of this series $$\dfrac{1}{1!} + \dfrac{1+2}{2!} + \dfrac{1+2+3}{3!}+\ldots$$ But somehow I am not even convinced this converges. I tried writing it as $\sum \dfrac{n(n+1)}{2(n!)}$. I also tried to rewrite it as a telescoping series. But I seem to be stuck. Some hints on where I am going wrong would help. I am supposed to get the answer as $\dfrac{3e}{2}$.</p>
| 0non-cybersec
| Stackexchange |
Is it well-defined behavior to return a vector of const char*, filled inside a function. <p>I am currently learning vulkan. In one of the tutorials I saw a function which roughly does the following:</p>
<pre><code>#define SOMESTRING "Hello World!"
std::vector<const char*> buildVector() {
std::vector<const char*> vec;
vec.push_back(SOMESTRING);
return vec;
}
</code></pre>
<p>When I saw this, I was wondering: is this defined behavior? Isn't the content of the string <code>"Hello World!"</code> located on the stack and thus is invalid once the function returns? If this is undefined behavior, what would be the correct way to do this? Unfortunately, using <code>std::string</code> is not an option because of the vulkan API.</p>
| 0non-cybersec
| Stackexchange |
On a certain sphere-packing problem. <p>Suppose we have a closed ball <span class="math-container">$S$</span> of radius <span class="math-container">$r$</span> centered at the origin <span class="math-container">$O$</span> of <span class="math-container">$\mathbb{R}^3$</span>. Suppose further that we are given a point <span class="math-container">$v$</span> on the ball's boundary (i.e., <span class="math-container">$\|v\|=r$</span>), and a point <span class="math-container">$u$</span> in the ball's interior (i.e., <span class="math-container">$\|u\|<r$</span>). We assume that <span class="math-container">$v$</span> and <span class="math-container">$u$</span> are not too far from each other, so <span class="math-container">$\|v-u\| \leq r$</span>. Let <span class="math-container">$\theta$</span> denote the angle between the vectors <span class="math-container">$\overrightarrow{vO}$</span> and <span class="math-container">$\overrightarrow{vu}$</span>. </p>
<blockquote>
<p>Is it true that the ball <span class="math-container">$S'$</span> of radius <span class="math-container">$\|v-u\| \cos \theta$</span> centered at <span class="math-container">$u$</span> is contained in <span class="math-container">$S$</span>? </p>
</blockquote>
<p>And if not, is there any (interesting) radius of the form <span class="math-container">$\|v-u\|f(\theta)$</span> which works?</p>
<p>This arose in the context of an (elementary) differential geometry problem. We're asked to prove that the osculating circles of a curve <span class="math-container">$\mathbf{r}(t)$</span> (with certain hypotheses on the curve) on some sphere <span class="math-container">$S$</span> centered at the origin must be contained within sphere. The above claim is one avenue of attack I'm attempting, but it also makes for an interesting geometric problem in and of itself. </p>
<p>The intuition seems fairly clear. If the angle <span class="math-container">$\theta$</span> is large, there is much little wiggle room for the ball containing <span class="math-container">$u$</span>, and <span class="math-container">$\cos \theta$</span> comes off as a natural factor. Also, the extreme cases <span class="math-container">$\theta=0$</span> or <span class="math-container">$\theta=\frac{\pi}{2}$</span> seem to be okay.</p>
| 0non-cybersec
| Stackexchange |
Parallels Virtuozzo under Windows. <p>Here the logic chain:</p>
<ol>
<li>Virtuozzo is a commercial version of OpenVZ</li>
<li>OpenVZ works by modifying the core of hosting OS, so it doesn't
support Windows</li>
<li>Virtuozzo supports Windows</li>
</ol>
<p>I do not understand how we can come to the fact that Virtuozzo supports Windows (but it really does). Have they modified Windows core some way? Or may be there are some other way?</p>
| 0non-cybersec
| Stackexchange |
Not sure how I stumbled on this but WHAT THE *UCK?. | 0non-cybersec
| Reddit |
My Ciri from The Witcher 3 cosplay. | 0non-cybersec
| Reddit |
I want to Automate saving all characters of a font in Photoshop. <p>So I have a square Photoshop document, containing 1-character text from a font I like, and with special effects (drop shadow, emboss, etc...).</p>
<p>The manual way is I type a character, then save a PNG. I have to repeat this for the other characters.</p>
<p>How can I automate saving the letters A-Z, a-z, and 0-9 using actions?</p>
| 0non-cybersec
| Stackexchange |
Find the amplitude of the steady state solution. <p>Heave motion of a point energy absorber in the form of a buoy due t incident wave of frequency $w$ and amplitude $a$ is described by</p>
<p>$(m+m_a)y'' + (c_h + c_e)y' +ky = F_i(t)$</p>
<p>$F_i(t)=aF(w)\cos(wt)$ is the hydrodynamic force acting on the buoy due to incident wave.</p>
<p>When the waves first arrive at the buoy there are transient parts to the solution which die down as time progresses until a steady state oscillation is established.
The steady state oscillation of the buoy is described</p>
<p>$y(t)=Y\cos(wt+d)$</p>
<p>(i)Find the amplitude Y of the steady state solution.</p>
<p>So i have differentiated $y(t)=Y\cos(wt+d)$ to find $y''$ and $y'$ and I have substituted this into the given equation but I am not really sure what I need to do next nothing seems to cancel. Is this right? Do I need to substitute $F_i(t)$ for $aF(w)\cos(wt)$?</p>
<p>Thanks</p>
| 0non-cybersec
| Stackexchange |
Help with graph search problem. <p>For the next problem I can not think of how to find a solution with graph searches (I thought it was backtracking but my college professor told me that I should use graph searches, which I do not know much about, to solve it).</p>
<p>As input we obtain a matrix (3x4) of connections ("|" or "-") which we must decode in the least amount of movements, to avoid that the bomb explodes. As an output inform one of the possible solutions.</p>
<p>The decoding consists in that the matrix must contain only characters "-". To achieve this they tell us that if we move an element of the matrix, both the values of the column and row of the matrix change. For example if I move the element (0,0) of the matrix in this case the "-" change to "|" and vice versa in the column 0 and row 0. In the image (where (1) represents the input matrix and (2) the decoded matrix) a solution to inform would be:</p>
<p><code>move (0,3) and then (2,1)</code></p>
<p><a href="https://i.stack.imgur.com/sxQEZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sxQEZ.png" alt="enter image description here"></a></p>
<p>How to design an algorithm that solves this problem without using backtracking directly?</p>
| 0non-cybersec
| Stackexchange |
Invoke-command specify scriptblock on the same line. <p>I was wondering is it possible to pass a tiny script into the ScriptBlock parameter but do it all in one line?</p>
<p>For instance, I want to run the following 2 commands:</p>
<p>import-module lync
get-csuser</p>
<p>I can do this if I have a powershell script file and call that file explicitly. The contents of the script look like this</p>
<pre><code>invoke-command -ComputerName mycomputer.mylab.com -ScriptBlock {
import-module lync
get-csuser
}
</code></pre>
<p>I want to be able to do the above without putting this into a temporary script file and do it on one lime. Is this possible?</p>
<p>Thanks</p>
| 0non-cybersec
| Stackexchange |
Implementing GraphQL (GraphQL for .Net) in ASP.NET Core, why is my Query:ObjectGraphType class not being registered by dependency injection?. <p>I've setup a basic Web Api demonstrate the use of GraphQL using ASP.Net Core. I've Followed a tutorial, what feels like exactly but am getting an error I don't understand.</p>
<p>I'm using GraphQL for .NET v2.4.0</p>
<p>This is the error:</p>
<pre><code>System.InvalidOperationException: No service for type 'Land.GraphQL.Queries.LandQuery' has been registered.
at at Microsoft.Extensions.DependencyInjection.ServiceProviderServiceExtensions.GetRequiredService(IServiceProvider provider, Type serviceType)
at at GraphQL.FuncDependencyResolver.Resolve(Type type)
at at GraphQL.FuncDependencyResolver.Resolve[T]()
at Land.GraphQL.LandSchema..ctor(IDependencyResolver resolver) in .../LandSchema.cs:11
</code></pre>
<p>I'd be grateful for any light shed :)</p>
<p>Here's the code:</p>
<p>I created a LandType:ObjectGraphType to define the Type:</p>
<pre><code> public class LandType : ObjectGraphType<Entities.Land>
{
public LandType(ILandDataAccess landDataAccess)
{
Name = "Land";
Field(land => land.Id, type: typeof(IdGraphType)).Description("Land Id in LandApi context");
Field(land => land.Apn)
.Description(
"Assessor's Parcel Number (APN) is a unique number that is assigned to each tract of land in a county by the Tax Assessor.");
Field(land => land.Address).Description("");
Field(land => land.ZipCode).Description("");
Field(land => land.City).Description("");
Field(land => land.County).Description("");
Field(land => land.State).Description("");
Field(land => land.Country).Description("");
Field(land => land.GisNumber).Description("");
Field(land => land.AssessedValue).Description("");
Field(land => land.LegalDescription).Description("");
Field(land => land.Acreage, type: typeof(FloatGraphType)).Description("Acreage of Land");
}
}
</code></pre>
<p>I created a LandQuery:ObjectGraphType to define the Query:</p>
<pre><code>public class LandQuery : ObjectGraphType
{
public LandQuery(ILandDataAccess dataAccess)
{
Field<ListGraphType<LandType>>(
"Land",
resolve: context => dataAccess.GetLandsAsync());
}
}
</code></pre>
<p>I created a LandSchema:Schema to define the Schema:</p>
<pre><code>public class LandSchema : Schema
{
public LandSchema(IDependencyResolver resolver) : base(resolver)
{
Query = resolver.Resolve<LandQuery>();
}
}
</code></pre>
<p>I added the service and middleware to the Startup file:</p>
<pre><code>public void ConfigureServices(IServiceCollection services)
{
services.AddScoped<IDependencyResolver>(s => new FuncDependencyResolver(
s.GetRequiredService));
services.AddScoped<LandSchema>();
services.AddGraphQL(o => { o.ExposeExceptions = true; })
.AddGraphTypes(ServiceLifetime.Scoped);
}
public void Configure(IApplicationBuilder app, IHostingEnvironment env)
{
app.UseGraphQL<LandSchema>();
}
</code></pre>
<hr>
<p>Edit:</p>
<p>Thanks to the commenters @NateBarbettini and @TonyNgo inspired me to find the answer. Turns out <code>.AddGraphTypes</code> only searches the calling assembly. My GraphTypes are stored in a referenced Assembly. passing the referenced assembly fixed the problem: <code>.AddGraphTypes(typeof(LandSchema).Assembly, ServiceLifetime.Scoped);</code></p>
| 0non-cybersec
| Stackexchange |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Ramleh - Bite The Bolster [Power Electronics]. | 0non-cybersec
| Reddit |
Female Air France crew can opt out of Iran flights over headscarf, official says. | 0non-cybersec
| Reddit |
Green riding the original, single horned, Charizard! One of my favorites by Ken Sugimori!. | 0non-cybersec
| Reddit |
Metro station. | 0non-cybersec
| Reddit |
Batman walks into a room which alfred is ìn, late at night.. "Alfred could you fill up the bathtub please" batman said after entering the room.
Alfred replied with, "what's a htub sir?" | 0non-cybersec
| Reddit |
Floor. | 0non-cybersec
| Reddit |
ELI5: Why do Americans write the date differently to everyone else?. While I'm not certain it's the rest of the world, in Europe the date is written day/month/year, which - to me - seems entirely logical. So how did it come to be that Americans write the date month/day/year? | 0non-cybersec
| Reddit |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Sharing remote folder in subdir. <p>I want to use a remote shared folder use in a subdirectory.
For example the share \somePC\folder => x:\folder\somePCsRemoteFolder </p>
<p>is there an easy way to do this ?</p>
<p>endo</p>
| 0non-cybersec
| Stackexchange |
Mobile Augmented Reality Latest Trend. | 0non-cybersec
| Reddit |
End-extension in Gödel's constructible universe. <p>Given two ordinals <span class="math-container">$\alpha < \beta$</span>, considering the subsets of Gödel's constructible universe, one say that <span class="math-container">$L_\beta$</span> is a <span class="math-container">$\Sigma_n$</span> end-extension of <span class="math-container">$L_\alpha$</span> (and <span class="math-container">$L_\alpha$</span> is an <span class="math-container">$\Sigma_n$</span> elementary submodel of <span class="math-container">$L_\beta$</span>), written <span class="math-container">$L_\alpha \prec_{\Sigma_n} L_\beta$</span>, when for all <span class="math-container">$\Sigma_n$</span> formulas <span class="math-container">$\varphi$</span>,
<span class="math-container">\begin{align}
L_\beta \models \varphi \iff L_\alpha \models \varphi
\end{align}</span></p>
<p>Given such sets <span class="math-container">$L_\beta$</span> there exists results, showed by Jensen and others using Skolem functions, that helps us constructing elementary submodels <span class="math-container">$L_\alpha \subset L_\beta$</span>. However I have had a hard time finding results that lets us do the converse, that is constructing end-extensions <span class="math-container">$L_\delta \supset L_\beta$</span>. </p>
<blockquote>
<p>Does there exists such general results about end-extension of Gödel's constructible sets ?</p>
</blockquote>
<p>And more particulary, let (<span class="math-container">$\zeta$</span>, <span class="math-container">$\Sigma$</span>) be the least ordered-pair such that <span class="math-container">$L_\zeta \prec_{\Sigma_2} L_\Sigma$</span>, as characterized by Welch using infinite time Turing machines or by Harrington using <span class="math-container">$\Sigma_2$</span> truth sets :</p>
<blockquote>
<p>Can we find an ordinal <span class="math-container">$\kappa > \Sigma$</span>
such that <span class="math-container">$L_\Sigma \prec_{\Sigma_2} L_\kappa$</span> ? And if so, can we continue this sequence of end-extensions <span class="math-container">$\zeta, \Sigma, \kappa \ldots$</span> ?</p>
</blockquote>
| 0non-cybersec
| Stackexchange |
I'm a DJ, and I play a knock-off version of 'Everybody' by a group called Bachelor Party, instead of the original Backstreet Boys version.. Nobody has ever noticed. | 0non-cybersec
| Reddit |
So Sorry - Tolerate or Die EP [Saturday Night Original progressive Alt rock/punk]. | 0non-cybersec
| Reddit |
How to use spot instance with amazon elastic beanstalk?. <p>I have one infra that use amazon elastic beanstalk to deploy my application.
I need to scale my app adding some spot instances that EB do not support.</p>
<p>So I create a second autoscaling from a launch configuration with spot instances.
The autoscaling use the same load balancer created by beanstalk.</p>
<p>To up instances with the last version of my app, I copy the user data from the original launch configuration (created with beanstalk) to the launch configuration with spot instances (created by me).</p>
<p>This work fine, but:</p>
<ol>
<li><p>how to update spot instances that have come up from the second autoscaling when the beanstalk update instances managed by him with a new version of the app?</p>
</li>
<li><p>is there another way so easy as, and elegant, to use spot instances and enjoy the benefits of beanstalk?</p>
</li>
</ol>
<p><strong>UPDATE</strong></p>
<p>Elastic Beanstalk add support to spot instance since 2019... see:
<a href="https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html" rel="nofollow noreferrer">https://docs.aws.amazon.com/elasticbeanstalk/latest/relnotes/release-2019-11-25-spot.html</a></p>
| 0non-cybersec
| Stackexchange |
Fresh strawberries with orange, lemongrass and Thai-basil dressing. | 0non-cybersec
| Reddit |
Why not have many edges instead of one edge connecting to many nodes (in hypergraphs). <p>As a follow-up to <a href="https://mathoverflow.net/questions/13750/what-are-the-applications-of-hypergraphs">What are the Applications of Hypergraphs</a>, the linked article:</p>
<p><a href="https://www.microsoft.com/en-us/research/publication/learning-hypergraphs-clustering-classification-embedding/?from=http%3A%2F%2Fresearch.microsoft.com%2Fen-us%2Fum%2Fpeople%2Fdenzho%2Fpapers%2Fhyper.pdf" rel="nofollow noreferrer">Learning with Hypergraphs: Clustering, Classification, and Embedding</a></p>
<p>Mentions this:</p>
<blockquote>
<p>Let us consider a problem of grouping a collection of articles into different topics. Given an article, assume the only information that we have is who wrote this article. One may construct an undirected graph in which <strong>two vertices are joined together by an edge if there is at least one common author of their corresponding articles</strong>...</p>
<p>A natural way of remedying the information loss issue occurring in the above methodology is to represent the data as a hypergraph instead. A hypergraph is a graph in which an edge can connect more than two vertices [2]. In other words, <strong>an edge is a subset of vertices</strong>. In what follows, we shall unifiedly refer to the usual undirected or directed graphs as simple graphs. Moreover, without special mentioning, the referred simple graphs are undirected. It is obvious that a simple graph is a special kind of hypergraph with each edge containing two vertices only. In the problem of clustering articles stated before, it is quite straightforward to construct a hypergraph with the vertices representing the articles, and the edges the authors (Figure 1). <strong>Each edge contains all articles by its corresponding author</strong>.</p>
</blockquote>
<p>What I'm wondering is, instead of using <em>1 edge with many vertices</em>, why not use <em>many edges</em>. That way it's just like a normal graph, not a hypergraph.</p>
<p>I don't yet understand the value in making 1 edge become essentially like a new type of node, if you could do the same thing by creating many edges. In hypergraphs they even call these new things "tentacles" or "attachments", connecting the edge to many things. Then it's like an edge is just another node. Hoping for some clarification on why this is necessary or what advantage this provides that multiple edges can't accomplish.</p>
| 0non-cybersec
| Stackexchange |
Nier: Automata'a Uplifting Existentialism (Story Discussion) - Super Bunnyhop. | 0non-cybersec
| Reddit |
Puppies being total jerks. | 0non-cybersec
| Reddit |
Something light brown.. | 0non-cybersec
| Reddit |
First network in hyperledger. <p>I am trying to work on my first network in Hyperledger Fabric. Using the following documentation
<a href="http://hyperledger-fabric.readthedocs.io/en/latest/build_network.html" rel="nofollow noreferrer">http://hyperledger-fabric.readthedocs.io/en/latest/build_network.html</a></p>
<p>I have completed the setup till
<a href="http://hyperledger-fabric.readthedocs.io/en/latest/build_network.html#create-join-channel" rel="nofollow noreferrer">http://hyperledger-fabric.readthedocs.io/en/latest/build_network.html#create-join-channel</a></p>
<p>but when I run the </p>
<pre><code>peer channel create -o orderer.example.com:7050 -c $CHANNEL_NAME -f ./channel-artifacts/channel.tx --tls $CORE_PEER_TLS_ENABLED --cafile /opt/gopath/src/github.com/hyperledger/fabric/peer/crypto/ordererOrganizations/example.com/orderers/orderer.example.com/msp/tlscacerts/tlsca.example.com-cert.pem
</code></pre>
<p>I got error as </p>
<blockquote>
<p><strong>Error: Got unexpected status: BAD_REQUEST</strong></p>
</blockquote>
<p>Searching for the issue i came across
<a href="http://hyperledgerdocs.readthedocs.io/en/latest/asset_trouble.html" rel="nofollow noreferrer">http://hyperledgerdocs.readthedocs.io/en/latest/asset_trouble.html</a></p>
<p>Hence I've tried to use a new channel name as given(old channel name=mychannel),
I've tried below cmds</p>
<p><code>CORE_PEER_COMMITTER_LEDGER_ORDERER=orderer:7050 peer channel create -c myc1</code></p>
<p><code>CORE_PEER_COMMITTER_LEDGER_ORDERER=orderer.example.com:7050 peer channel create -c myc1</code></p>
<p><code>CORE_PEER_COMMITTER_LEDGER_ORDERER=172.18.0.2:7050 peer channel create -c myc1</code></p>
<p>For all the three I am getting error</p>
<blockquote>
<p><strong>Error: Ordering service endpoint is not valid or missing
Usage:
peer channel create [flags]</strong></p>
</blockquote>
<p>Also I've tried to skip it assuming the channel is already created.
hence ran</p>
<pre><code>peer channel join -b ./mychannel.block
</code></pre>
<p>But got </p>
<blockquote>
<p><strong>Error: proposal failed (err: rpc error: code = Unknown desc = chaincode error (status: 500, message: Cannot create ledger from genesis block, due to LedgerID already exists))</strong></p>
</blockquote>
<p>My OS is Ubuntu 16.04</p>
<p>docker ps
<a href="https://i.stack.imgur.com/E25Gh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E25Gh.png" alt="Docker ps output"></a>
Kindly help</p>
| 0non-cybersec
| Stackexchange |
Generate graph in Grafana from API. <p>I'm looking for a way to generate an arbitrary graph from the Grafana API, ideally by just feeding it a query.
After looking in the doc I don't see anything to do it directly, so the only way I can see would be to :</p>
<ul>
<li>Generate a dashboard json with just the graph I want</li>
<li>Create the dashboard through the API by sending that json</li>
<li>Export that graph as jpg</li>
<li>Delete this darshboard</li>
</ul>
<p>That seems a bit silly, isn't there a way to just generate a graph from a specific query directly ?
The goal here is to add a graph in our monitoring alerts, that way if we get a high load alert on a server for example I could generate a query to get that server's load graph, and include that in the alert e-mail. Nothing life changing, but it would be a nice feature to have I think.</p>
| 0non-cybersec
| Stackexchange |
Fields of characteristic $p$ and $p$-torsion. <p>Let $X$ be an affine variety over a field $k$ of characteristic $p$. Let $A$ be the set of morphisms $X\to \mathbb A_1$. Let $B\subset A$ be a sub-algebra. Suppose $f\in A$ is such that </p>
<p>$$ \bigcap_{g\in B} \ker g \subset \ker f.$$</p>
<p>Then if $A/B$ has no $p$-torsion, we have that $f\in B$. My proof goes as follow: For any $h\in A$, let $\bar h$ be the image of $h$ in $A/B$. Then,</p>
<p>$$ X = \bigcap_{g\in B} \ker\bar{g} \subset \ker\bar{f}.$$</p>
<p>Thus $X = \ker\bar f$, which implies that $\bar f = 0$, so $f\in B$. I did not use $p$-torsion. Can you please tell me what is wrong with this proof and why $p$-torsion is so important? </p>
| 0non-cybersec
| Stackexchange |
CIFS VFS: Server X has not responded in 120 seconds. Reconnecting. <pre><code>CIFS VFS: Server X has not responded in 120 seconds. Reconnecting...
...
</code></pre>
<p>This message is spamming my <code>dmesg</code> output. The warning is caused by the following mount of a windows share:</p>
<pre><code># CIFS shares
//X/share/folder /mnt/some_folder cifs vers=2.1,credentials,uid=some_user,gid=some_user,file_mode=0764,dir_mode=0765,nosetuids,noserverino 0 0
</code></pre>
<p>The share works. From various posts I consulted, it seems this warning is normal. Any way to suppress it (if it's normal after all), or adjust the parameters so it doesn't show?</p>
| 0non-cybersec
| Stackexchange |
$(BDE) \cap (CDF) = P \not\equiv D$. Prove that $PD$ passes through the excenter of $\triangle ABC$ in $\angle A$.. <blockquote>
<p><span class="math-container">$D$</span>, <span class="math-container">$E$</span> and <span class="math-container">$F$</span> are points respectively on sides <span class="math-container">$BC$</span>, <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> such that <span class="math-container">$AE = CD$</span> and <span class="math-container">$AF = BD$</span>. <span class="math-container">$(BDE) \cap (CDF) = P \not\equiv D$</span>. Prove that <span class="math-container">$PD$</span> passes through the excenter of <span class="math-container">$\triangle ABC$</span> in <span class="math-container">$\angle A$</span>.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/62fhB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/62fhB.png" alt="enter image description here"></a></p>
<p>I have noticed that if <span class="math-container">$K$</span> is the midpoint of arc <span class="math-container">$BC$</span> not containing point <span class="math-container">$A$</span> and <span class="math-container">$EF \cap KD = H$</span> then <span class="math-container">$B, E, H, K$</span> and <span class="math-container">$C, F, H, K$</span> are concyclic.</p>
<p><span class="math-container">$\implies \angle AEF = \angle BKD$</span> and <span class="math-container">$\angle AFE = \angle CKD$</span>.</p>
<p>Not sure what is next though.</p>
<p><a href="https://i.stack.imgur.com/qKKL2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qKKL2.png" alt="enter image description here"></a></p>
| 0non-cybersec
| Stackexchange |
ar
X
iv
:1
90
5.
08
23
4v
2
[
ec
on
.T
H
]
1
3
Ju
l
20
20
Empirical bias of extreme-price auctions: analysis∗
Rodrigo A. Velez† and Alexander L. Brown‡
Department of Economics, Texas A&M University, College Station, TX 77843
July 15, 2020
Abstract
We advance empirical equilibrium analysis (Velez and Brown, 2020b) of the winner-
bid and loser-bid auctions for the dissolution of a partnership. We show, in a complete
information environment, that even though these auctions are essentially equivalent for
the Nash equilibrium prediction, they can be expected to differ in fundamental ways
when they are operated. Besides the direct policy implications, two general conse-
quences follow. First, a mechanism designer who accounts for the empirical plausibility
of equilibria may not be constrained by Maskin invariance. Second, a mechanism de-
signer who does not account for the empirical plausibility of equilibria may inadvertently
design biased mechanisms.
JEL classification: C72, D47, D91.
Keywords : equilibrium refinements; implementation theory; behavioral mechanism
design; empirical equilibrium.
1 Introduction
This paper studies the plausibility of Nash equilibria of two popular auctions for the dis-
solution of a partnership. By using the recently introduced empirical equilibrium analysis
(Velez and Brown, 2020b), we show that even though the Nash equilibria of the complete
information games that ensue when these mechanisms are operated are essentially identical,
these mechanisms can be expected to differ in fundamental ways. Two general conclusions
∗This paper benefited from comments of Dustin Beckett, Ernan Haruvy, Thomas Palfrey, and Anastasia
Zervou as well as audiences at TETC17, U. Maryland, U. Virginia, Caltech, UC Santa Barbara, U. Rochester,
Kellogg (MEDS), 2018 Naples Workshop on Equilibrium Analysis, SAET18, SCW18. The results in this
paper circulated in a paper entitled Empirical Equilibrium, which is now dedicated to study the foundations
of this refinement (Velez and Brown, 2020a). All errors are our own.
†[email protected]; https://sites.google.com/site/rodrigoavelezswebpage/home
‡[email protected]; http://people.tamu.edu/∼alexbrown
1
http://arxiv.org/abs/1905.08234v2
mailto:[email protected]
https://sites.google.com/site/rodrigoavelezswebpage/home
mailto:[email protected]
http://people.tamu.edu/%7Ealexbrown
for the theory of full implementation follow. First, a mechanism designer who accounts
for empirical plausibility of equilibria may not be constrained by usual invariance proper-
ties. Second, a mechanism designer who does not account for the empirical plausibility of
equilibria may design biased mechanisms.
We consider a symmetric partnership dissolution problem in which two agents who
collectively own an object need to decide who receives the object when monetary compen-
sation, chosen out of a finite but fine greed, is possible. In the spirit of the implementation
literature with complete information (see Jackson, 2001, for a survey) we assume that an
arbitrator who makes a recommendation for this division knows that the agents know each
other well but does not know the agents’ preferences on the possible divisions.1 This is a
relevant benchmark for the dissolution of a marriage or a long standing partnership. We
assume that agents are expected utility maximizers with quasi-linear utility indices. For
concreteness, suppose that agents’ values for the object are vl ≤ vh, which for simplicity we
assume to be even numbers. We study two prominent mechanisms that we refer to as the
extreme-price auctions (EPs). These mechanisms operate as follows. First, the arbitrator
asks the agents to bid for the object. Then assigns the object to a higher bidder, breaking
ties uniformly at random. The transfer from the agent who receives the object to the other
agent is determined as follows: the transfer is the winner bid, the winner-bid auction (WB);
the transfer is the loser bid, the loser-bid auction (LB).2
Our first step to evaluate the performance of EPs is to characterize the Nash equilibrium
outcomes of the games that ensue when these mechanisms are operated. We consider
both pure and mixed strategy equilibria. When agents have equal types, both auctions
give, in each Nash equilibrium, equal expected payoff to each agent (Lemma 1). When
valuations are different, the set of efficient Nash equilibrium payoffs of both auctions coincide
(Proposition 1).3 This common set can be placed in a one to one correspondence with the
integers {vl/2, vl/2+1, ...., vh/2}. Each equilibrium has a unique payoff-determinant bid in
this set and for each such integer there is an equilibrium with this payoff-determinant bid
(Proposition 2). We refer to this set as the Nash range. Between two elements of the Nash
1The assumption of complete information in our partnership dissolution problem allows us to contrast
empirical equilibrium analysis with the well-understood restrictions of Nash implementation in this en-
vironment. Our approach to implementation theory, and in general mechanism design, does not impose
restrictions on the information structure that the modeler believes is a reasonable description of reality.
For instance, one can define and study empirical equilibria in general incomplete information environments,
under a Bayesian information structure (Velez and Brown, 2020b).
2EPs belong to the family of α-auctions studied by Cramton et al. (1987) and McAfee (1992). Note that
the partnership dissolution environment differs substantially from a buyer-seller environment in which the
payoff of the loser of a first-price auction or a second-price auction does not depend on the price paid by the
winner.
3The maximal aggregate loss in an inefficient equilibrium is one unit. The analysis of inefficient equilibria
leads to the same conclusions we state here in the introduction for efficient equilibria (Sec. 4.2.1).
2
range, the higher valuation agent prefers the left one (paying less), and the lower valuation
agent prefers the right one (being paid more).
Not all Nash equilibria are plausible to be observed, even approximately, when a game
takes place (c.f. van Damme, 1991). Thus, the equivalence of Nash equilibrium outcomes
for EPs may not be realistic. If one refines the Nash equilibria of EPs by ruling out
weakly dominated behavior, the associated equilibrium payoff-determinant bids collapse to
extreme singleton sets: {vl/2} for WB and {vh/2} for LB. Experiments show that WB
exhibits a bias in favor of the higher valuation agent and LB exhibits a bias in favor of
the lower valuation agent (Brown and Velez, 2020). However, observed biases are not as
extreme as those predicted by undominated equilibrium. Indeed, the best performance of
these auctions, when behavior of both types clearly separates, involves persistent weakly
dominated behavior (Sec. 6, Brown and Velez, 2020). This is consistent with the extensive
literature that shows weakly dominated behavior in dominant strategy games is persistently
observed (see Velez and Brown, 2020b).
Thus, we are left with the question whether the bias observed in laboratory experiments
with these mechanisms should be expected in general. That is, is there any reason why
the weakly dominated behavior that sustains payoff-determinant bids close to vl/2 for WB
is more plausible than the weakly dominated behavior that sustains payoff-determinant
bids on the other extreme of the Nash range? Symmetrically, is there any reason why the
weakly dominated behavior that sustains payoff-determinant bids close to vh/2 for LB is
more plausible than the weakly dominated behavior that sustains payoff-determinant bids
on the other extreme of the Nash range?
To answer this question we advance empirical equilibrium analysis of these auctions.
This theory was introduced by Velez and Brown (2020b) to discriminate among weakly
dominated Nash equilibria of a normal form game based on their empirical plausibility.4
Empirical equilibrium is defined by means of the following thought experiment. Con-
sider a researcher who samples behavior in normal-form games and constructs a theory that
explains this behavior. The researcher determines the plausibility of Nash equilibria based
on the empirical content of the theory by requiring that Nash equilibria be in its closure.
That is, if a Nash equilibrium cannot be approximated to an arbitrary degree by the em-
pirical content of the researcher’s theory, it is identified as implausible or unlikely to be
observed. Empirical equilibrium is the refinement so defined based on the non-parametric
theory that each agent chooses actions with higher probability only when they are better
for her given what the other agents are doing. More precisely, an empirical equilibrium
4Traditional equilibrium refinements, implicitly of explicitly rule out all weakly dominated behavior (van
Damme, 1991). Thus, these theories are silent about the nature of the phenomenon of our interest.
3
is a Nash equilibrium that is the limit of a sequence of behavior satisfying weak payoff
monotonicity, i.e., between two alternative actions for an agent, say a and b, if the agent
plays a with higher frequency than b, it is because given what the other agents are doing, a
has higher expected utility than b. Remarkably, this property is a common factor in some
of the most popular models that have been proposed to account for observed behavior in
experiments.5
One can give empirical equilibrium a static or dynamic interpretation. First, its defini-
tion simply articulates the logical implication of the hypothesis that the researcher’s theory
is well specified, for this hypothesis is refuted by the observation of a Nash equilibrium that
is not an empirical equilibrium. Alternatively, suppose that the researcher hypothesizes that
behavior will converge to a Nash equilibrium through an unmodeled evolutionary process
that produces a path of behavior that is consistent with her theory. Then, the researcher
can also conclude that the only Nash equilibria that will be approximated by behavior are
empirical equilibria.
In our main results, we characterize the set of empirical equilibrium payoffs of EPs
(Theorems 1 and 2). The highlights of these characterizations are the following. With a
single exception among all type profiles, these sets are disjoint. The empirical equilibrium
payoffs of WB belong to the left half of the Nash range. When vl is not too close to the
minimal bid, i.e., at least 3vh/8, the empirical equilibrium payoffs of WB essentially (up
to rounding) are the left fifth of the Nash range. Symmetrically, the empirical equilibrium
payoffs of LB belong to the right half of the Nash range. When vh is not too close to the
maximal bid, the empirical equilibrium payoffs of LB essentially are the right fifth of the
Nash range.
Thus, empirical equilibrium analysis reveals that the bias observed in experiments is
indeed a characteristic of EPs. Under the testable hypotheses that observable behavior
is weakly payoff monotone and that frequencies of play approach mutual best responses,
WB necessarily benefits the higher valuation agent and LB necessarily benefits the lower
valuation agent. This result has two policy relevant implications. First, a mechanism
designer who accounts for the empirical plausibility of equilibria may not be constrained by
Maskin invariance, a well-known necessary condition for implementation in pure strategy
Nash equilibrium. Second, a mechanism designer who does not account for the empirical
plausibility of equilibria may inadvertently design biased mechanisms.
5These models include the exchangeable randomly perturbed payoff models (Harsanyi, 1973; van Damme,
1991), the control cost model (van Damme, 1991), the structural QRE model (McKelvey and Palfrey, 1995),
the regular QRE models (McKelvey and Palfrey, 1996; Goeree et al., 2005). The common characteristic of
these models that make them suitable the purpose of equilibrium refinement is that their parametric forms
are independent of the game in which they are applied, and have been successful in replicating comparative
statics in a diversity of games (Goeree et al., 2005).
4
The remainder of the paper is organized as follows. Sec. 2 places our contribution in the
context of the literature. Sec. 3 introduces the model. Section 4 presents our results, the
intuition behind them, and their general implications for the theory of full implementation.
Sec. 5 discusses the robustness of our results to violations of weak payoff monotonicity.
Sec. 6 concludes. We present all proofs in the Appendix.
2 Related literature
EPs are part of a family of mechanisms proposed for Cramton et al. (1987) for the dissolution
of a partnership. If ownership is symmetric, in an independent private values environment,
these mechanisms are efficient, individually rational, and incentive compatible. In a com-
plete information environment, EPs are strategically equivalent to a mechanism that elicits
agents valuations and then assigns an allocation in which no agent prefers the allotment of
any other agent to her own (see Velez, 2018, for a survey).
Our study shares some features with the literature on Nash implementation (Jackson,
2001), Bayesian implementation (Jackson, 1991), and undominated implementation (Palfrey
and Srivastava, 1991; Jackson, 1992). Similar to these studies, we evaluate the performance
of EPs based on the set of all equilibrium outcomes that cannot be ruled out when the
mechanisms are operated. An obvious difference is that our equilibrium refinement is a
compromise between these approaches. We acknowledge that not all Nash equilibria are
plausible. We do not rule out all weakly dominated equilibria, which is not supported
by data. At a technical level, our approach differs substantially with most of previous
implementation literature because we consider all Nash equilibria including equilibria in
mixed strategies. To our knowledge the only study that, like ours, also treats pure and
mixed strategy equilibria symmetrically in the context of implementation is Mezzetti and
Renou (2012). These authors show that complete information implementation in Nash
equilibrium (in both pure and mixed strategies) is more permissive than implementation in
pure strategy Nash equilibria. Interestingly, while a continuum of mixed strategy equilibria
of EPs pass our equilibrium refinement, most of its pure strategy equilibria are wiped
out by it (see Sec. 4.3). Thus, our work reinforces the view of these authors that mixed
strategy equilibria and pure strategy equilibria should be treated symmetrically by the
mechanism designer. At a technical level, even though we arrive to a similar conclusion by
considering plausibility of equilibria, our results are independent. The reason Mezzetti and
Renou (2012)’s implementation in pure and mixed Nash equilibria is not bound by Maskin
invariance, is that a mixed strategy equilibrium may have multiple outcomes and Maskin
invariance is a requirement centered at a particular outcome. More precisely, starting from
5
a profile of types in which a certain outcome results in a mixed strategy equilibrium, when
this outcome does not drop in the ranking of any agent, the profile of strategies may not
be a Nash equilibrium anymore depending on what happened to the ranking of the other
outcomes in the equilibrium. Thus, the essential insight of Mezzetti and Renou (2012) is
that one can avoid the restrictions of this property by designing mechanisms that obtain
some desirable outcomes for some realizations of types only in mixed strategy equilibria
with multiple outcomes. Our main insight is that a mechanism designer that accounts for
plausibility of equilibria is not bound by Maskin invariance because starting from a profile
of types in which a certain outcome results in an empirical equilibrium, when this outcome
does not drop in the ranking of any agent, the alleged equilibrium may still be a Nash
equilibrium, but may not pass our refinement anymore (see Sec. 4.3). To provide clarity
about the significance of our results, we show that the normative objectives achieved by
the empirical equilibrium correspondence of EPs are not achieved by the Nash equilibrium
correspondence (including equilibria in mixed strategies) of any mechanism (Lemma 2).
Empirical equilibrium refines Nash equilibrium by means of approximation by behavior
that is observationally sub-optimal. In this sense our work can be related with the growing
literature on mechanism design with (as if) boundedly rational agents. These studies gener-
ally endorse a form of bounded rationality and aim at finding institutions that perform well
when operated on agents who exhibit these particular patterns of behavior (c.f., Anderson
et al., 1998, 2001; Eliaz, 2002; Healy, 2006; Cabrales and Serrano, 2012; Tumennasan, 2013;
de Clippel, 2014; de Clippel et al., 2018; Kneeland, 2017). Empirical equilibrium analy-
sis does not give up the regularity induced by expected utility maximization. It puts a
reality check on it by requiring proximity by plausible boundedly rational behavior. Alter-
natively, one can think of our analysis as the study of boundedly rational behavior when
it is disciplined by proximity to a Nash equilibrium. The studies that are closest to ours
consider forms of implementation that like empirical equilibrium are determined by conver-
gence processes: Anderson et al. (1998, 2001) study limit behavior of logistic QRE models
in all-pay auctions and minimum contribution games; Healy (2006); Cabrales and Serrano
(2012) study implementation in the limits of some evolutionary processes; and Tumennasan
(2013) studies implementation in the limits of logistic equilibria for increasingly sophisti-
cated logistic quantal response functions. Anderson et al. (1998, 2001)’s approach is limited
to the parametric logistic form and intensively uses the structure of the games it analyses in
order to characterize logistic equilibria by means of tractable differential equations.6 Healy
(2006), Cabrales and Serrano (2012), and Tumennasan (2013) concentrate only on pure
6The set of empirical equilibria may differ from the set of Nash equilibria that are limits of logistic
equilibria (Velez and Brown, 2020a).
6
strategy equilibria and require strong conditions be satisfied for convergence, which limit
the applicability of their results to general games.
Finally, this paper makes part of a research agenda that reevaluates the applications of
game theory based on the empirical equilibrium refinement. In Velez and Brown (2020a)
we study the foundations of empirical equilibrium. In Velez and Brown (2020b) we advance
empirical equilibrium analysis of strategy-proof mechanisms. In a companion paper, Brown
and Velez (2020), we experimentally evaluate the predictions of empirical equilibrium for
EPs. In experiments with different valuation structures these auctions exhibit the bias
predicted by empirical equilibrium analysis. The data also supports the positive association
of bids and their expected utility.
3 Model
There are two agents N := {1, 2} who collectively own an object (indivisible good) and
need to allocate it to one of them. Monetary compensation is possible. Each agent’s payoff
type is characterized by the value that he or she assigns to the object. We assume these
type spaces are Θ1 = Θ2 := {v, v + 2, ..., v}, where 2 < v ≤ v are even positive integers.
7
The generic type of agent i is vi ∈ Θi. Let Θ := Θ1 ×Θ2 with generic element v := (v1, v2).
The lower and higher values at v are vl and vh respectively. We also assume that agents
are expected utility maximizers. The expected utility index of agent i with type vi is vi− p
if receiving the object and paying p to the other agent; p if being paid this amount by the
other agent and receiving no object. Whenever we make statements in which the identity
of the agents is not relevant, we conveniently use neutral notation i and −i. The set of
possible allocations is that in which an agent receives the object and transfers an amount
p ∈ {0, 1, ..., p} with p ≥ v/2+2, to the other agent. Let A be the space of these allocations.
For an allocation a ∈ A the value of agent i’s utility index at the allotment assigned by a
to this agent is ui(a|vi).
A social choice correspondence (scc) selects a set of allocations for each possible profile
of types. The generic scc is G : Θ ⇒ A.
A mechanism is a pair (M,ϕ) where M := (Mi)i∈N is an unrestricted message space and
ϕ : M → ∆(A) is an outcome function. Our main results concern the finite mechanisms
we introduce next. For simplicity we complete the set up of the model assuming that
mechanisms are finite. Lemma 2, our only result that refers to general mechanisms, can be
generalized for appropriate extensions of strategies and Nash equilibrium.
7Our restrictions on valuations are only for simplicity in the presentation. The conclusions we draw from
the analysis of this restricted model are general.
7
Definition 1. The winner-bid auction is the mechanism in which each agent selects a bid
in the set {0, 1, 2, ..., p}. An agent with the highest bid receives the object. Ties are resolved
uniformly at random. The agent who receives the object pays the winner bid to the other
agent. The loser-bid auction is the mechanism defined similarly where the payment is the
loser bid. We refer to these two auctions as the extreme-price auctions.
Given a profile of types v ∈ Θ, mechanism (M,ϕ) determines a complete information
game Γ := (M,ϕ, v). A (mixed) strategy for agent i is a probability measure on Mi.
Agent i’s generic strategy is σi ∈ ∆(Mi). The profile of strategies is σ := (σi)i∈N . A pure
strategy places probability one in a given action. We associate pure strategies with actions
themselves. The expected utility of agent i with type vi in Γ from selecting action mi when
the other agent selects an action as prescribed by σ−i is
Uϕ(mi|σ−i; vi) :=
∑
m−i∈M−i
ui(ϕ(mi,m−i)|vi)σ−i(m−i).
A profile of strategies σ is a Nash equilibrium of Γ if for each i ∈ N , each mi in the
support of σi, and eachm
′
i ∈ Mi, Uϕ(m
′
i|σ−i; vi) ≤ Uϕ(mi|σ−i; vi). The set of Nash equilibria
of Γ is N(Γ). The set of Nash equilibrium outcomes of Γ are those obtained with positive
probability for some Nash equilibrium of Γ. Agent i’s expected payoff in equilibrium σ is
πi(σ), where for simplicity in the notation we avoid conditioning these payoffs on the agent’s
types, which can be always inferred from context.
Definition 2 (Velez and Brown, 2020a). σ := (σi)i∈N is weakly payoff monotone for
(M,ϕ, v) if for each i ∈ N and each pair {mi, ni} ⊆ Mi, σi(mi) > σi(ni) implies that
Uϕ(mi|σ−i; vi) > Uϕ(ni|σ−i; vi).
Definition 3 (Velez and Brown, 2020a). σ ∈ N(Γ) is an empirical equilibrium of Γ if there
is a sequence of weakly payoff monotone distributions of Γ, {σλ}λ∈N, such that as λ → ∞,
σλ → σ.
4 Results
4.1 Extreme-price auctions and Nash equilibrium
We start by characterizing the Nash equilibrium outcomes of the games induced by the
EPs.
The theory of fair allocation has produced a series of principles that an arbitrator may
want to adhere to when resolving a partnership dissolution dispute (see Thomson, 2010, for
8
a survey). Two of the most popular are the following: efficiency, i.e., a party who values
the object the most should receive it, or equivalently, that expected payoffs add up to vh;
and equity, i.e., no party should prefer the allotment of the other (Foley, 1967). Abstracting
from incentives issues imagine that the arbitrator knows the agents’ valuations, v ∈ Θ. It
is easy to see that if the arbitrator abides by the principles of efficiency and equity, agents’
payoffs should have the form:8 vl/2 + t for l and vh/2 + (|vh/2 − vl/2| − t) for h, where
0 ≤ t ≤ |vh/2 − vl/2|. In other words, if an arbitrator endorses these two principles, the
only that is left for him or her is to determine a division between the agents of the “equity
surplus,” i.e., ES(v) := |vh/2− vl/2| (Tadenuma and Thomson, 1995).
The following lemma states that each extreme-price auction achieves, in each Nash
equilibrium (including those in mixed strategies), the objectives of efficiency and equity
when agents have equal valuations. We omit the straightforward proof.
Lemma 1. Let (M,ϕ) be an extreme-price auction and v ∈ Θ such that v1 = v2. Then
for each σ ∈ N(M,ϕ, v), π1(σ) = π2(σ) = v1/2 = v2/2. Moreover, for each deterministic
allocation a ∈ A in which each agent receives this common payoff, there is a pure-strategy
Nash equilibrium of (M,ϕ, v) whose outcome is this allocation.
The following proposition states that each extreme-price auction essentially achieves, in
each Nash equilibrium (including those in mixed strategies), the objectives of efficiency and
equity for arbitrary valuation profiles.
Proposition 1. Let (M,ϕ) be an extreme-price auction and v ∈ Θ such that vl < vh.
Let σ be a Nash equilibrium of (M,ϕ, v). Then, there is p ∈ {vl/2, ..., vh/2} such that the
support of σl belongs to {0, ..., p} and the support of σh belongs to {p, ..., p}. If σ is efficient,
the higher value agent receives the object and pays p to the other agent. Moreover,
1. If (M,ϕ) is the winner-bid auction, then p is in the support of σh. If σ is inefficient,
i.e., σl(p) > 0, then p = vl/2; πl(σ) + πh(σ) ≥ vh − 1; and πh(σ) ≥ vh/2 +ES(v)− 1.
2. If (M,ϕ) is the loser-bid auction, then p is in the support of σl. If σ is inefficient, i.e.,
σh(p) > 0, then p = vm/2; πl(σ) + πh(σ) ≥ vh − 1; and πl(σ) ≥ vl/2 + ES(v) − 1.
Proposition 1 reveals that the Nash equilibria of the extreme-price auctions have a
simple structure. In all equilibria, the payoff-determinant bid is in the set {vl/2, ..., vh/2}.
Let us refer to this set of bids as the Nash range. In most of these equilibria agents’ bids
are strictly separated. That is, there is a bid p in the Nash range such that one agent
bids weakly on one side of p and the other agent bids strictly on the other side of p. In
8In our environment no-envy implies efficiency for deterministic allocations (Svensson, 1983). Our state-
ment here refers also to random assignments.
9
these equilibria, which are strictly separated, outcomes are efficient and equitable. There
are inefficient equilibria. For the winner-bid auction, it is possible that both agents bid
vl/2. For the loser-bid auction it is possible that both agents bid vh/2. In both cases the
aggregate welfare loss is at most one unit, i.e., the size of the minimal difference between
bids. This means that if the minimal bid increment is one cent, the maximum that these
auctions can lose in aggregate expected utility for any Nash equilibrium is one cent. Thus,
one can say that these auctions essentially implement the principles of efficiency and equity
in Nash equilibria.9
Proposition 1 allows us to characterize Nash equilibrium payoffs for extreme-price auc-
tions. For efficient equilibria, these payoffs are exactly the integer divisions of the equity
surplus. For simplicity in the presentation, in the remainder of the paper we omit the anal-
ysis of inefficient equilibria. We know from Proposition 1 that the corresponding payoffs
are extreme in the equilibrium set. Thus, all conclusions we obtain for efficient empirical
equilibria extend to inefficient empirical equilibria.
Proposition 2. Let (M,ϕ) be an extreme-price auction and v ∈ Θ such that vl < vh. The
set of efficient Nash equilibrium payoffs of (M,ϕ, v), i.e.,
{(πl(σ), πh(σ)) : σ ∈ N(M,ϕ, v), πl(σ) + πh(σ) = vh}, is the set of integer divisions of
the equity-surplus, i.e., {(vl/2 + ES(v) − t, vh/2 + t) : t ∈ {0, 1, ..., ES(v)}}.
4.2 Empirical equilibrium
4.2.1 Empirical equilibrium payoffs
We characterize the set of empirical equilibrium payoffs of the extreme-price auctions. Re-
call that by Lemma 1, when valuations are equal, the payoff of each agent is the same in
each Nash equilibrium of each extreme-price auction. Thus, we only need to character-
ize the payoffs of empirical equilibria when agents’ valuations differ. Since Proposition 2
characterizes Nash equilibrium payoffs for these auctions, it is convenient to describe em-
pirical equilibrium payoffs by the set of conditions for which a Nash equilibrium payoff is
an empirical equilibrium payoff.
Theorem 1. Let (M,ϕ) be the winner-bid auction, v ∈ Θ such that vl < vh, and σ ∈
N(M,ϕ, v). If σ is efficient, π(σ) is the payoff of an empirical equilibrium of (M,ϕ, v) if
and only if
1. πh(σ) = vh/2 + ES(v) when ES(v) ≤ 2;
9Our proof of Proposition 1 also reveals that the probability of an inefficient outcome is bounded above
by the inverse of the equity surplus, measured in the minimal bid increment.
10
2. πh(σ) ≥ vh/2 + ES(v)/2 + vl/4 − 1/2 when ES(v) > 2 and vl ≤ 3vh/8;
3. πh(σ) ≥ vh/2 + 4ES(v)/5− 4/5 when ES(v) > 2, vl > 3vh/8, and vl < 7vh/12− 7/6;
4. πh(σ) > vh/2 + 4ES(v)/5− 4/5 when ES(v) > 2, vl > 3vh/8, and vl ≥ 7vh/12− 7/6.
Theorem 1 reveals a surprising characteristic of the empircal equilibria of the winner-bid
auction. For simplicity fix vh at a certain value. Let vl ≤ vh. When vl is low, i.e., at most
3vh/8, the minimal share of the equity surplus that the higher value agent obtains in an
empirical equilibrium is, essentially, at least 50% of the equity surplus (since we assumed
vl is a positive even number, the exact share depends on rounding, but is never less than
50%). More precisely, for this range of vl, agent h receives a payoff that is at least
vh/2 + ES(v)/2 + vl/4− 1/2 = vh/2 + vh/4− 1/2.
This means that p is the winner bid in an empirical equilibrium of the winner-bid auction
for such valuations if and only if p ≤ vh/4 + 1/2 (Fig. 1). Thus, while the maximal bid
in an empirical equilibrium is the same for all valuations when vl ≤ 3vh/8, the minimal
percentage of the equity surplus that is assigned to the higher value agent increases from
essentially 50% when vl = 2 to essentially 80% when vl = 3vh/8. For higher values of vl,
i.e., 3vh/8 < vl < vh, the higher value agent receives, essentially, at least 80% of the equity
surplus (Fig. 1).
In summary, Theorem 1 states that the minimal share of the equity surplus that the
higher value agent obtains in an empirical equilibrium of the winner-bid auction depends
on the number of possible bids that are to the left of the Nash range. In the extreme case in
which there is only one bid to the left of the Nash range, the higher value agent essentially
obtains at least 50% of the equity surplus. As the number of bids to the left of the Nash
range increases, the minimal share of the equity surplus that is obtained by the higher
value agent in an empirical equilibrium increases until it reaches essentially 80% when the
number of possible bids to the left of the Nash range is 60% of the number of bids in the
Nash range (equivalently, vl ≤ 3vh/8). When the number of possible bids to the left of
the Nash range is higher than 60% of the number of bids in the Nash range (equivalently,
vl > 3vh/8), the minimal share of the equity surplus that is obtained by the higher value
agent in an empirical equilibrium remains essentially 80%. (For low values of the equity
surplus, rounding has a significant effect; see Fig. 1).
Theorem 2. Let (M,ϕ) be the loser-bid auction, v ∈ Θ such that vl < vh, and σ ∈
N(M,ϕ, v). If σ is efficient, π(σ) is the payoff of an empirical equilibrium of (M,ϕ, v) if
and only if
11
1. πl(σ) ≥ vl/2 + ES(v) if ES(v) ≤ 2;
2. πl(σ) ≥ vl/2 + ES(v)/2 + (p − vh/2)/2 − 1/2 if ES(v) > 2 and vh/2 ≥ vl/2 + 5(p −
vl/2)/8.
3. πl(σ) ≥ vl/2 + 4ES(v)/5 − 4/5 if ES(v) > 2, vh/2 < vl/2 + 5(p − vl/2)/8, and
vh/2 > p− 7(p − vl/2)/24 − 7/12;
4. πl(σ) > vl/2 + 4ES(v)/5 − 4/5 if ES(v) > 2, vh/2 < vl/2 + 5(p − vl/2)/8, and
vh/2 ≤ p− 7(p − vl/2)/24 − 7/12.
The empirical equilibrium payoffs of the loser-bid auction are symmetric to those of the
winner-bid auction. The minimal share of the equity surplus that the lower value agent
obtains in an empirical equilibrium of the winner-bid auction depends on the number of
possible bids that are to the right of the Nash range. In the extreme case in which there is
only one bid to the right of the Nash range, the lower value agent essentially obtains at least
50% of the equity surplus. As the number of bids to the right of the Nash range increases,
the minimal share of the equity surplus that is obtained by the lower value agent in an
empirical equilibrium increases until it reaches essentially 80% when the number of possible
bids to the right of the Nash range is 60% of the number of bids in the Nash range. When
the number of possible bids to the right of the Nash range is higher than 60% of the number
of bids in the Nash range, the minimal share of the equity surplus that is obtained by the
lower value agent in an empirical equilibrium remains essentially 80%. (For low values of
the equity surplus, rounding has a significant effect; see Fig. 1).
4.2.2 The intuition
The characterization of empirical equilibrium payoffs of the extreme-price auctions reveals
that empirical equilibrium makes a delicate selection of Nash equilibria, which is sensitive
to the global structure of the game. An empirical equilibrium may involve an agent playing
a weakly dominated strategy. However, not all Nash equilibria, in particular not all Nash
equilibria in which an agent plays a weakly dominated strategy, are empirical equilibria.
Thus, empirical equilibrium somehow discriminates among actions by assessing how likely
they can be played based on how they compare with the other actions. A discussion of the
proof of Theorem 1 informs us about this feature of empirical equilibrium.
Let v ∈ Θ be such that vl < vh. Let (M,ϕ) be the winner-bid auction and consider
σ ∈ N(M,ϕ, v). By Proposition 1 this equilibrium is characterized by a winner bid p ∈
{vl/2, ..., vh/2}. Suppose that p > vl/2 and let t := p−1−vl/2. Suppose for simplicity that
t ≥ 1. One can easily see that agent l always regrets wining the auction when bidding p−1.
12
vh/2
vl/2
vh/4
1
| | |
(3/8)vh/2
p
ut ut ut ut ut ut rs rs rs rs rs rsb b b b b b b b b b b b
ut ut ut ut ut rs rs rs rs rsb b b b b b b b b b b
ut ut ut ut rs rs rs rsb b b b b b b b b b
ut ut ut rs rs rsb b b b b b b b b
ut ut ut rs rs rsb b b b b b b b
ES(v)/5 + 4/5 ut ut ut rs rs rsb b b b b b b
ut ut rs rsb b b b b b
ut ut rs rsb b b b b
ut ut rs rsb b b b
ut rsb b b
ut rsb b
b Nash range
ut winner-bid
rs loser-bid
qp p
qp
qp
qp
qp
qp
qp
qp
qp
qp
qp
qp
Figure 1: Payoff determinant bids in an empirical equilibrium of the extreme-price auctions.
For a given v ∈ Θ such that vl < vh, the Nash range is the set {vl/2, ..., vh/2}. In this graph we show, for
a fixed vh, the Nash range for each 2 < vl < vh and the bids that are winning bids for empirical equilibria
of the winner-bid and loser-bid auction. In the vertical line we measure vl/2 and in the horizontal line we
measure bids. For vl = 2, the Nash range is the set {1, 2, ..., vh/2}. In the graph this set is represented
by the points (1, 1), (2, 1),...,(vh/2, 1) (black dots). In general, for vl, we represent the Nash range, i.e.,
{vl/2, ..., vh/2} by the set of points (vl/2, vl/2),...,(vh/2, vl/2). We represent the winner bids in an empirical
equilibrium of the winner-bid auction for a given v ∈ Θ by triangles. We represent the winner bids in an
empirical equilibrium of the loser’s-bid auction for a given v ∈ Θ such that vl/2 = p − vh/2 by squares.
Notice that this choice of variable p as vl increases guarantees symmetry, i.e., there is the same number of
possible bids on each side of the Nash range, and is chosen for easier interpretation of the graph.
Indeed, bidding p−1 is weakly dominated for agent l by each bid in {vl/2−1, vl/2, ..., p−2}.
Let σλ be any monotone quantal response equilibrium of (M,ϕ, v). It follows that for each
b ∈ {vl/2 − 1, vl/2, ..., p − 2}, Uϕ(b|σ
λ
h ; vl) ≥ Uϕ(p− 1|σ
λ
h ; vl). This means that there are
always at least t+ 2 bids that are at least as good as p− 1 for agent l. This imposes a cap
on the probability that this agent can place on p− 1. More precisely,
σλl (p− 1) ≤ 1/(t + 2). (1)
Thus, if σ is an empirical equilibrium, by (1), σl(p− 1) needs to be the limit of a sequence
of probabilities bounded above by 1/(t+ 2) and thus,
σl(p− 1) ≤ 1/(t+ 2). (2)
Now, in order for σ to be a Nash equilibrium it has to be the case that agent h has no
13
incentive to bid one unit less than p. This is simply,
vh − p ≥ σl(p− 1)
[
1
2
(vh − (p− 1)) +
1
2
(p− 1)
]
+ (1− σl(p− 1))(vh − (p − 1)).
Equivalently, σl(p− 1) ≥ 1/(vh/2− p+ 1). Together with (2) this implies that p ≤ (vl/2 +
vh/2)/2, or equivalently
πh(σ) ≥ vh/2 + ES(v)/2.
Thus, the reason why empirical equilibrium predicts agent h gets at least half of the equity
surplus for any possible valuation is that it is not plausible that agent l will consistently
bid above half of the Nash range. These high bids are worse than too many bids to their
left for agent l. If this agent’s actions are monotone with respect to utility, the maximum
probability that he or she will end up placing in a bid on the right half of the Nash range
will never be enough to contain the propensity of agent h to lower his or her bid.
When there is more than one bid to the left of the Nash range, the analysis becomes
subtler. Let y := vl/2−3t. The key to complete this analysis is to prove that in any quantal
response equilibrium of (M,ϕ, v) that is close to σ, all bids b ∈ {max{0, y}, ..., p − 1} are
weakly better than p − 1 for agent l. The subtlety here lies in that this is not implied
directly by a weak domination relation as in our analysis above. In order to uncover this we
need to recursively obtain estimates of agent h’s distribution of play, which in turn depend
on agent l’s distribution of play.
Remarkably, we prove that the set of restrictions that we uncover by means of our anal-
ysis are the only ones that need to be satisfied by an empirical equilibrium payoff. Goeree
et al. (2005) characterize the set of regular QRE of an asymmetric matching pennies game.
At a conceptual level, this exercise is similar to the construction of empirical equilibria.
However, the techniques developed by Goeree et al. (2005) are useful only in two-by-two
games, where an agent’s distribution of play is described by a single real number. Thus,
there is virtually no precedent in the construction of empirical equilibria that are not strict
Nash equilibria for games with non-trivial action spaces. We do so as follows. First, we
identify for each target payoff an appropriate Nash equilibrium that produces it, say σ.
Then we take a convex combination of a perturbation of σ and a logistic quantal response.
This defines a continuous operator whose fixed points are the basis of our construction. In
a two-limit process we first get close enough to σ by placing a weight on it that is high
enough so the fixed points of the convex combination operators inherit some key properties
of σ. Then we allow the logistic response to converge to a best response and along this path
of convergence we obtain interior distributions that are close to σ and are payoff-monotone.
14
4.3 Empirical equilibrium analysis and full implementation theory
Full implementation theory is the application of game theory in which one designs mech-
anisms based on their worst-case scenario performance. More precisely, given an scc and
a prediction for the interaction of agents in strategic situations, the designer looks for a
mechanism that obtains outcomes selected by the scc for the true profile of types for all the
predicted outcomes of the game that ensues when the mechanism is operated. When this
exercise is done based on the Nash equilibrium prediction, the mechanism designer finds
that many a desirable social objective is beyond her reach (Maskin, 1999; Jackson, 1991).
By contrast, most of the constraints of the mechanism designer are lifted if the worst-case
scenario discards all Nash equilibria that involve weakly dominated behavior (Palfrey and
Srivastava, 1991; Jackson, 1992).
Since there are Nash equilibria that are intuitively implausible, designing for a worst-
case scenario accounting for all possible Nash equilibria is unnecessarily pessimistic. On the
other extreme, ruling out all weakly dominated behavior, is also unrealistically optimistic
(see Velez and Brown, 2020b). Our partnership dissolution problem epitomizes this conflict
and clearly shows how empirical equilibrium analysis resolves it.
One can envision an arbitrator having a deliberate choice on the division of the equity
surplus in a partnership dissolution problem. For instance the arbitrator may want to
guarantee that the lower valuation agent always receive at least half of the equity surplus.
Proposition 2 implies that this is not achieved by the extreme-price auctions if one evaluates
them with the Nash equilibrium prediction. The following lemma states that this is a
feature not only of these auctions but also of any mechanism with at least one equitable
Nash equilibrium for each profile of types.
Lemma 2. Let (M,ϕ) be a mechanism. Suppose that for each v ∈ Θ, there is a Nash
equilibrium of (M,ϕ, v) (potentially in mixed strategies) that obtains with positive prob-
ability only efficient and equitable allocations for v. Then, for each v ∈ Θ such that
vl < vh, and each {t, t+1} ⊆ {0, 1, ..., ES(v)} there is an efficient σ ∈ N(M,ϕ, v) such that
vl/2 + t ≤ πl(σ) ≤ vl/2 + t+ 1.
Lemma 2 states that if an arbitrator selects a mechanism that obtains efficiency and
equity in at least one Nash equilibrium (partial implementation), then the range of divisions
of the equity surplus that are necessarily obtained when valuations are different is essentially
the whole Nash range. More precisely, for any two consecutive integer divisions of the equity
surplus, there is a Nash equilibrium that obtains a division of the equity surplus between
these two divisions.
Theorems 1 and 2 and Lemma 2 allow us to conclude that a social planner who accounts
15
for the empirical plausibility of equilibria, realizes that some social goals, which would
be ruled out impossible by the full Nash implementation analysis (independently whether
implementation in mixed strategy equilibria as in Mezzetti and Renou (2012) is allowed
or not), are within his or her reach. For instance, using LB could be sensible for a social
planner who is able to exercise some level of affirmative action and chooses to benefit a
segment of the population who are likely to have lower valuations for the objects to be
assigned.
Finally, it is worth noting that our results suggest that it is not without loss of generality
to restrict our attention to pure-strategy equilibria when a mechanism is operated.
Remark 1. Let v ∈ Θ such that vl < vh. The only pure-strategy empirical equilibrium
of the winner-bid auction for v is σl = δvl/2−1 and σh := δvl/2. The only pure-strategy
empirical equilibrium of the loser-bid auction for v is σl = δvh/2 and σh := δvh/2+1.
Jackson (1992) constructed examples in which including arguably plausible mixed-
strategy equilibria in a worst-case scenario analysis would reverse the conclusions one ob-
tains by only analyzing pure-strategy equilibria. Empirical equilibrium analysis goes beyond
these observations and provides a clear framework in which plausibility is built into the pre-
diction of agents’ behavior. It is fair to say then that while empirical mechanism design
opens new possibilities in the design of economic institutions, it also sets the standards
of analysis high by forcing us to consider mixed-strategy equilibria. In this context, our
complete characterization of empirical equilibria of extreme-price auctions in Theorems 1
and 2, and the characterization results in Velez and Brown (2020b), show that the technical
challenges can be resolved in policy relevant environments.
5 Discussion
Our analysis of mechanism based on empirical equilibrium has some challenges that are
shared by the analysis of mechanisms based on any solution concept that is not single-
valued for each game. If a Nash equilibrium is not empirical, we learn that it is implausible
(under our hypothesis of weak payoff monotonicity). By contrast, if a Nash equilibrium is
empirical, we do not know whether the equilibrium will be actually relevant in a particular
environment.
In some applications this may not be a problem. For instance, the main conclusion from
our analysis of empirical equilibria of EPs survives even if some empirical equilibria end up
not being relevant for these auctions. More precisely, the bias predicted for these mecha-
nisms will be present even when not all empirical equilibria end up being relevant when they
are operated. In general, many a social objective is guaranteed to be achieved whenever all
16
empirical equilibrium outcomes that result when a mechanism is operated satisfy it. Think,
for instance, of Pareto efficiency. Suppose that a mechanism guarantees this property for
each empirical equilibrium. This means that as long as behavior is payoff monotone and
approximates mutual best responses, this behavior will obtain Pareto efficiency with high
probability. Again, notice that in order to conclude this, it is not necessary to know which
empirical equilibria will actually be relevant when the mechanism is operated.
Designing based on empirical equilibrium also implicitly assumes that behavior when
the mechanism is operated will approximate a Nash equilibrium. On the one hand, though
not universally observed, there is evidence that frequencies of play in many games move
towards best responses. For instance, it is common that logistic QRE parameter estimates
increase toward best responses when estimated in experiments that are run in multiple
periods without repeated game effects (McKelvey and Palfrey, 1995). On the other hand,
one can also see our analysis of mechanisms based on empirical equilibrium as a departure
from the Nash equilibrium based analysis that makes it more realistic without losing the
regularity and power provided by the assumption that individual incentives are eventually
in equilibrium.
It is worth noting that not all empirical evidence supports the assumption of weak
payoff monotonicity. Velez and Brown (2020b) documented violations of this property in
the second price auction experiments of Andreoni et al. (2007) and the pivotal mechanism
experiment of Cason et al. (2006). In our partnership dissolution environment there is
a positive association between the empirical expected payoff of an action and the prob-
ability with which it is chosen (Brown and Velez, 2020). Rounding bids to multiples of
five, a behavioral regularity observed in these games, can easily induce violations of payoff
monotonicity, however.
Thus, we should see weak payoff monotonicity as a regular expression of the positive
association between choice and expected utility in a game. We should not see this property
as a bullet proof characterization of behavior in games. Consistently, we should verify
whether the conclusions we obtained from our analysis survive for continuous variations of
it for which there is more empirical support. The following parametric form of the axiom
allows us to do this.
Definition 4. Let m ∈ [0, 1]. A profile of strategies σ := (σi)i∈N is m-weakly pay-
off monotone for Γ if for each i ∈ N and each pair of actions {ai, âi} ⊆ Ai such that
Ui(σ−i, ai) ≥ Ui(σ−i, âi), we have that σi(ai) ≥ mσi(âi).
For m = 1 the property exactly corresponds to weak payoff monotonicity. For m > n,
m-weak payoff monotonicity implies n-weak payoff monotonicity. For m = 0 the property
imposes no restrictions in data.
17
Definition 5. Let m > 0. An m-empirical equilibrium of Γ is a Nash equilibrium of Γ that
is the limit of a sequence of m-weakly payoff monotone strategies for Γ.
One can prove that for any m > 0, the m-empirical equilibrium outcome correspondence
violates invariance under Maskin monotonic transformations. Moreover, for values of m
close enough to one, the m-empirical equilibria of WB and LB are separated on the left half
and the right half of the Nash range. It is an interesting open question to determine the
lowest m for which this is so.
Finally, we conducted empirical equilibrium analysis under the assumption of risk neu-
trality. This can be seen as the choice of a model in which we can constructively arrive at
the main conclusion from our study, i.e., that mechanisms that are essentially equivalent
with respect to the Nash equilibrium prediction may be biased towards certain populations.
The interpretation of empirical equilibrium analysis as pointing to particular mechanisms
to implement certain normative objectives (Sec. 4.3), does assume that the social planner
only entertains agents have this type of risk preferences. It would be interesting, but outside
of the scope of this paper, to identify mechanisms that implement in empirical equilibria
ordinal social choice correspondences independently of risk preferences (c.f., Mezzetti and
Renou, 2012).
6 Conclussion
We advanced empirical equilibrium analysis of EPs. Our results reveal that as long as
empirical distributions of play in the games induced by these mechanisms are weakly payoff
monotone, they can approximate mutual best responses only if they exhibit a form of
bias. WB favors the higher valuation agent and LB favors the lower valuation agent. This
implies that an arbitrator who uses an EPA within a legal system in which this type of
affirmative action is forbidden, can be subject to a legitimate challenge supported by theory
and empirical data. Indeed, our results produce a series of comparative statics that are
supported by experimental evidence (see our companion paper, Brown and Velez, 2020).
Our analysis also brings news to the abstract mechanism design paradigm. We learned
that an arbitrator can abide by a principle of equity and at the same time exercise a form
of affirmative action that guarantees a special treatment for either low or high value agents.
This shows that a mechanism designer who accounts for empirical plausibility of equilibria
is not constrained by typical invariance properties.
18
7 Appendix
For x ∈ R, ⌊x⌋ denotes the floor of x, i.e., the greatest integer that is less than or equal to
x; ⌈x⌉ denotes the ceiling of x, i.e., the smallest integer that is greater than or equal to x.
Definition 6. σ := (σi)i∈N is payoff monotone for (M,ϕ, v) if for each i ∈ N and each pair
{mi, ni} ⊆ Mi, Uϕ(mi|σ−i; vi) ≥ Uϕ(ni|σ−i; vi) if and only if σi(mi) ≥ σi(ni).
The following theorem is useful in our characterization of empirical equilibria of EPs.
Theorem 3 (Velez and Brown, 2020a). Let (M,ϕ) be a EPA and v ∈ Θ. Then, σ is an
empirical equilibrium of (M,ϕ, v) if and only if it is a Nash equilibrium of (M,ϕ, v) and it
is the limit of interior payoff monotone distributions for (M,ϕ, v).
The logistic quantal response function with parameter λ ≥ 0, denoted by lλ, assigns to
each mi ∈ Mi and each x ∈ R
Mi the value,
lλimi(x) :=
eλxmi
∑
ni∈Mi
eλxni
. (3)
It can be easily checked that for each λ > 0, the corresponding logistic quantal response
function is continuous and monotone, i.e., for each x ∈ RMi and each pair {m, t} ⊆ Mi,
xm ≥ xt if and only if l
λ
im(x) ≥ l
λ
it(x).
Proof of Proposition 1. In any of the auctions, by bidding ch := vh/2 the high-value
agent guarantees a payoff at least vh − ch = ch. By bidding cl := vl/2 the low-value agent
guarantees a payoff at least cl = vl − cl. Thus, in a Nash equilibrium the high-value agent’s
payoff is at least ch and the low-value agent’s payoff is at least cl.
We prove the proposition for the winner-bid auction, which we denote by (M,ϕ). The
proof for the loser-bid auction is symmetric. Let γ be the probability with which an agent
with the high value gets the object when there is a tie. We prove our proposition for the
slightly more general mechanism in which γ ≥ 1/2. The alternative tie breaker γ = 1 may
be relevant in experimental settings.
Let σ ∈ N(M,ϕ, v) with vl < vh. Let p be in the support of σh. We claim that p ≤ ch.
Suppose without loss of generality that p is the maximal element in the support of σh.
Suppose by contradiction that p ≥ ch + 1. Since p > cl and γ > 0, the expected payoff of
any bid b > p for the low-valuation agent is strictly lower than the expected payoff of p.
19
Thus, there is no b > p in the support of σl. Since p− 1 ≥ ch,
Uϕ(p− 1|σl; vh)− Uϕ(p|σl; vh) ≥ (1− σl(p)− σl(p− 1))
+σl(p− 1)(γ + (1− γ)((p − 1)− (vh − p)))
+σl(p)(p − (vh − p))
= (1− σl(p)− σl(p− 1))
+σl(p− 1)(2γ − 1 + 2(1− γ)(p − ch))
+2σl(p)(p − ch) > 0,
where the last inequality holds because σl is a probability distribution, γ ≥ 1/2, and
p− ch ≥ 1. This contradicts p is in the support of σh.
We claim now that p ≥ cl. Suppose without loss of generality that p is the minimal
element in the support of σh. Suppose by contradiction that p ≤ cl−1. We claim that there
is no b < p in the support of σl. Suppose by contradiction there is b < p in the support of
σl. Since cl − p ≥ 1 and b < p,
Uϕ(p+ 1|σh; vl)− Uϕ(b|σh; vl) ≥ σh(p)(vl − (p+ 1)− p) = σh(p)(2(cl − p)− 1) > 0.
This contradicts b is in the support of σl. We claim that σl(p) = 0. Suppose by contradiction
that σl(p) > 0. Then,
Uϕ(p+ 1|σh; vl)− Uϕ(p|σh; vl) ≥ σh(p) (γ(vl − (p+ 1)− p) + (1− γ)(−1))
= σh(p) (γ(2(cl − p)− 1)− (1− γ))
≥ σh(p)(γ − 1 + γ) ≥ 0.
If the inequality above holds strictly, there is a contradiction to σl(p) > 0. Since
σh(p) > 0, γ ≥ 1/2, and cl − p ≥ 1, the expression above is equal to zero only when
γ = 1/2 and p = cl − 1. Suppose then that γ = 1/2 and p = cl − 1. Since for each b < p,
σl(b) = 0, we have that
Uϕ(p + 1|σl; vh)− Uϕ(p|σl; vh) ≥
1
2
σl(p)(−1) +
1
2
σl(p)(vh − cl − (cl − 1))
= σl(p)(ch − cl) > 0.
This contradicts σh(p) > 0. Thus far we have proved that the support of σl belongs to
(p,+∞). Let b be the minimum element of the support of σl. Thus, p < b. If b < ch,
since γ > 0, agent h would benefit by bidding b instead of p. Thus, b ≥ ch. Recall that the
support of σh belongs to (−∞, ch]. Since the expected payoff of l is at least cl, b ≤ ch, for
otherwise agent l, when bidding b, would receive the object with probability one and pay
the other agent more than ch > cl. Thus, b = ch. Thus, agent l would benefit by bidding
20
b − 1 instead of b, because ch > cl and σh(p) > 0. This contradicts ch is in the support
of σl. Thus, the support of σl is empty. This is a contradiction. Thus, the support of σh
belongs to [cl, ch].
Let p be the minimum element of the support of σh and b an element in the support of
σl. We claim that b ≤ p. Suppose by contradiction that there is b > p in the support of σl.
Since p ≥ cl, b > cl. Thus, since σh(p) > 0, agent l benefits by bidding cl instead of b. This
contradicts b is in the support of σl. Thus, the support of σl belongs to (−∞, p].
Finally, let p be the minimum of the support of σh. If γ = 1/2 and σ is efficient, the
support of σl belongs to {0, ..., p − 1}. Thus, σh = δp, i.e., agent h receives the object and
pays p to the other agent. Suppose now that σ is inefficient, i.e., σl(p) > 0 and γ = 1/2.
We claim that p = cl and σl(p) < 1/(ch − cl). Recall that, p ∈ [cl, ch]. Since σl(p) > 0,
0 ≥ Uϕ(p − 1|σh; vl)− Uϕ(p|σh; vl) = σh(p)[p − [(vl − p)/2 + p/2]]
= σh(p)(p − cl).
Since σh(p) > 0, p = cl. Recall that agent l has guaranteed vl/2 in each Nash equilibrium
of (M,ϕ, v). If the maximum of the support of σl is cl, then agent h gets in equilibrium at
least what he or she would get by bidding cl +1. This bid gives agent h an expected payoff
of vh − (vl/2 + 1). Thus, the aggregate expected payoff is at least vh − 1.
Proof of Proposition 2. We prove the proposition for the winner-bid auction. The proof
for the loser-bid auction is symmetric. Let σ ∈ N(M,ϕ, v). By Proposition 1, there is
p ∈ {vl/2, ..., vh/2} that is in the support of σh such that the support of σl belongs to
{0, ..., p} and the support of σh belongs to {p, ..., p}. Suppose first that σ is efficient. Thus,
the support of σl belongs to {0, ..., p − 1}. Thus, σh = δp. Thus, πl(σ) = vl/2 + t and
πh(σ) = vh/2 + ES(v) − t for some t ∈ {0, 1, ..., ES(v)}. One can easily see that for each
t ∈ {0, 1, ..., ES(v)} the distributions σl = δp−1, σh = δp with p = vl/2 + t is a Nash
equilibrium.
Proof of necessity in Theorem 1. We prove that h’s expected payoff in an empirical
equilibrium of the winner-bid auction is bounded below by the expressions in the theorem.
The proof for the loser-bid auction is symmetric.
Let (M,ϕ) be the winner-bid auction and v ∈ Θ. Let σ ∈ N(M,ϕ, v). For i ∈ N , let
ci := vi/2. By Proposition 1 there is p ∈ {cl, ..., ch} separating the supports of σl and σh,
i.e., such that the support of σl belongs to {0, ..., p}; the support of σh belongs to {p, ..., p};
and p belongs to the support of σh. Suppose that p ≥ cl + 1. Since p is a best response
to σl for h with type vh, h’s expected payoff of p should be at least the expected payoff of
21
p− 1. Thus,
vh − p ≥ σl(p− 1)
[
1
2
(vh − (p− 1)) +
1
2
(p− 1)
]
+ (1− σl(p− 1))(vh − (p − 1)).
Equivalently,
σl(p− 1) ≥ 1/(ch − p+ 1). (4)
Let σ be an efficient empirical equilibrium of (M,ϕ, v) and let p be its associated separating
bid (in the support of σh). Let {σ
λ}λ∈N be a sequence of interior payoff monotone distribu-
tions for (M,ϕ, v) that converges to σ as λ → ∞. This sequence exists by Theorem 3. For
i ∈ N and {r, d} ⊆ {0, 1, ..., p} such that r ≤ d, let ∆i(r, d) be the difference in expected
utility for agent i in the winner-bid auction between the two situations in which agent i
bids strictly to the left of r and bids exactly d, conditional on agent −i bidding r.10 Using
this notation we have that when b < d,
Uϕ(b|σ−i, vi)−Uϕ(d|σ−i, vi) =
∑
r<b
σ−i(r)(d− b)+σ−i(b)(d− ci)+
∑
b<r≤d
σ−i(r)∆i(r, d). (5)
We prove that the expected payoff for h given σ satisfies the lower bound in the statement
of the theorem.
Case 1: ES(v) = 1. Recall that we require valuations to be positive. Thus, cl > 0. Let
λ ∈ N. By (5),
Uϕ(cl − 1|σ
λ
h , vl)− Uϕ(cl|σ
λ
h, vl) =
∑
r<cl−1
σλh(r) + σ
λ
h(cl − 1)(cl − cl) + σ
λ
h(cl − 1)∆l(cl, cl)
=
∑
r<cl−1
σλh(r) + 0 + 0 ≥ 0.
By payoff monotonicity, σλl (cl) ≤ σ
λ
l (cl − 1). By convergence, σl(cl) ≤ σl(cl − 1). Thus,
σl(cl) ≤ 1/2. We claim that p = cl. Suppose by contradiction that p = ch = cl + 1. By
Proposition 1, the support of σl belongs to {0, ..., cl}. Thus,
Uϕ(cl|σl, vh)− Uϕ(ch|σl, vh) = (1− σl(cl))(ch + 1) + σl(cl)ch − ch = 1− σl(cl) > 0.
This contradicts σ is a Nash equilibrium of (M,ϕ, v). Thus, p = cl. Since σ is efficient,
πh = vh/2 + 1.
Case 2: ES(v) > 1. Let t := p− cl − 1 and suppose that t ≥ 1. Let d := p− 1. Then
∆l(d, d) = d− ((vl − d)/2 + d/2) = t.
10Note that conditional on agent −i bidding r, the payoff for agent i when bidding strictly to the left of
r is exactly r. Thus, ∆i(r, d) is well defined.
22
Let r < d and n = d− r. Then,
∆l(r, d) = r − (vl − d) = cl + t− n− (2cl − cl − t) = 2t− n. (6)
We claim that for each max{0, cl − 3t} ≤ b < d there is Λ ∈ N such that for each λ ≥ Λ
Uϕ(b|σ
λ
h , vl)− Uϕ(d|σ
λ
h , vl) ≥ 0. (7)
Note that if this claim is true for b, then by payoff monotonicity and convergence, σl(b) ≥
σl(d).
By (5), for each max{0, cl − 3t} ≤ b < d, we have that
Uϕ(b|σ
λ
h , vl)− Uϕ(d|σ
λ
h , vl) =
∑
r<b σ
λ
h(r)(d− b) + σ
λ
h(b)(d − cl) +
∑
b<r<d σ
λ
h(r)∆l(r, d)
+σλh(d)∆l(d, d)
=
∑
r<b σ
λ
h(r)(d− b) + σ
λ
h(b)t+
∑
b<r<d σ
λ
h(r)∆l(r, d) + σ
λ
h(d)t
≥
∑
b<r<d σ
λ
h(r)∆l(r, d).
(8)
Let cl − t− 1 ≤ b < d. By (6), for each d− 2t ≤ r, ∆l(r, d) ≥ 0. Thus, by (8), for each
λ ∈ N, Uϕ(b|σ
λ
h , vl) − Uϕ(d|σ
λ
h , vl) ≥ 0. We complete the proof of our claim by proving by
induction on η ∈ {1, ..., 2t} that the claim holds for b = cl − t − η. Let η ∈ {2, ..., 2t} and
suppose that for each cl − t− η < g < d, σl(g) ≥ σl(d). Let b := cl − t− η. We claim that
there is Λ ∈ N for which for each λ ≥ Λ,
∑
b<r<d
σλh(r)∆l(r, d) ≥ 0.
Note that
∑
b<r<d σ
λ
h(r)∆l(r, d) =
∑
cl−t−η+1≤r≤cl−t−1
σλh(r)∆l(r, d) + σ
λ
h(c− t)∆l(cl − t, d)
+
∑
cl−t+1≤r≤cl−t+η−1
σλh(r)∆l(r, d) +
∑
cl−t+η≤r<d
σλh(r)∆l(r, d).
Since for each r ≥ cl − t, ∆l(r, d) ≥ 0,
∑
b<r<d σ
λ
h(r)∆l(r, d) ≥
∑
cl−t−η+1≤r≤cl−t−1
σλh(r)∆l(r, d)
+
∑
cl−t+1≤r≤cl−t+η−1
σλh(r)∆l(r, d)
=
∑
1≤r≤η−1 σ
λ
h(cl − t− r))(−r)+
∑
1≤r≤η−1 σ
λ
h(cl − t+ r)r
=
∑
1≤r≤η−1(σ
λ
h(cl − t+ r)− σ
λ
h(cl − t− r))r.
Thus, in order to prove our claim it is enough to show that there is Λ ∈ N such that for
23
each λ ≥ Λ and each r ∈ {1, .., η − 1}, σλh(cl − t+ r) ≥ σ
λ
h(cl − t− r). Let r ∈ {1, .., η − 1}.
Then,
∆h(cl− t+r, cl− t+r) = cl− t+r− ((vh− (cl− t+r))/2+(cl− t+r)/2) = −(ch−cl+ t−r).
Let 1 ≤ s ≤ 2r − 1. Then,
∆h(cl − t+ r− s, cl − t+ r) = cl − t+ r− s− (vh − (cl − t+ r)) = −(2(ch − cl + t− r) + s).
Observe that
Uϕ(cl − t+ r|σl, vh)− Uϕ(cl − t− r|σl, vh) = −
∑
0≤s<cl−t−r
σl(s)2r
+σl(cl − t− r)(vh − (cl − t+ r)− ((vh − (cl − t− r))/2 + (cl − t− r)/2))
−
∑
1≤s≤2r−1 σl(cl − t+ r − s)∆h(cl − t+ r − s, cl − t+ r)
−σl(cl − t+ r)∆h(cl − t+ r, cl − t+ r).
Thus,
Uϕ(cl − t+ r|σl, vh)− Uϕ(cl − t− r|σl, vh) ≥ −(1−
∑
0≤s≤2r σl(cl − t+ r − s))2r
+σl(cl − t− r)(ch − cl + t− r)
+
∑
1≤s≤2r−1 σl(cl − t+ r − s)(2(ch − cl + t− r) + s)
+σl(cl − t+ r)(ch − cl + t− r).
Since cl − t− r ≥ cl − t− η + 1 = b+ 1, for each 0 ≤ s ≤ 2r, b + 1 ≤ cl − t+ r − s. Since
r ≤ η − 1 and η ≤ 2t, cl − t + r ≤ cl + t − 1 < d. By the induction hypothesis, for each
0 ≤ s ≤ 2r, σl(cl − t+ r− s) ≥ σl(d) = σl(p− 1). By (4), σl(cl − t+ r− s) ≥ 1/(ch − p+1).
Thus,
Uϕ(cl − t+ r|σl, vh)− Uϕ(cl − t− r|σl, vh) ≥
−(1− 2r+1
ch−p+1
)2r + 2r
ch−p+1
2(ch − cl + t− r) >
2r
ch−p+1
(2(ch − cl)− (ch − p+ 1)).
Since p ≥ cl and ch > cl + 1, 2(ch − cl) − (ch − p + 1) > 0. Thus, Uϕ(cl − t+ r|σl, vh) −
Uϕ(cl − t− r|σl, vh) > 0. Thus, there is Λ ∈ N such that for each λ ≥ Λ, Uϕ(cl − t+ r|σ
λ
l , vh)−
Uϕ(cl − t− r|σ
λ
l , vh) > 0. By payoff monotonicity, σ
λ
h(cl−t+r) ≥ σ
λ
h(cl−t−r). Thus, there
is Λ ∈ N such that for each λ ≥ Λ and each r ∈ {1, .., η − 1}, σλh(cl − t+ r) ≥ σ
λ
h(cl − t− r).
Thus, for each λ ≥ Λ,
∑
b<r<d σ
λ
h(r)∆l(r, d) ≥ 0.
In summary, we have proved that when p−1 ≥ cl+1, for each max{cl−3t, 0} ≤ b ≤ p−1,
σl(b) ≥ σl(p − 1). (9)
24
There are two sub-cases.
Case 2.1: vl ≤ 3vh/8. We claim that p ≤ ES(v)/2− vl/4+ 1/2 = vh/4+1/2. Suppose
by contradiction that p > vh/4+ 1/2. Since p and cl are integers, p ≥ ⌊vh/4+ 1/2⌋+1 and
cl ≤ ⌈3vh/16⌉ ≤ 3vh/16. Then, p− 1− cl ≥ ⌊vh/4 + 1/2⌋ + 1− ⌈3vh/16⌉. Since vh is even,
⌊vh/4+1/2⌋ ≥ vh/4 and p ≥ vh/4+1. Thus, p−1−cl ≥ vh/16. Thus, p−1−cl ≥ 1 whenever
vh ≥ 16. Since ES(v) > 1, then vh ≥ 6 (recall that we assumed positive even valuations).
Direct calculation determines that ⌊vh/4 + 1/2⌋ + 1 − ⌈3vh/16⌉ ≥ 1 for vh = {6, ..., 14}.
Since, t = p − cl − 1 ≥ vh/4 − cl ≥ vh/16 = ch/8, cl − 3t ≤ cl − 3ch/8 ≤ 0. Thus,
max{cl−3t, 0} = 0. Since t ≥ 1, by (9), σl(p−1) ≤ 1/(cl+t+1) = 1/(cl+p−1−cl+1) = 1/p.
By (4), 1/(ch − p + 1) ≤ 1/p. Thus, p ≤ ch/2 + 1/2 = vh/4 + 1/2. This contradicts
p ≥ vh/4 + 1. Since σ is efficient and σh(p) > 0, the support of σl belongs to {0, ..., p − 1}.
Thus, πh(σ) ≥ vh−p ≥ vh−(vh/4+1/2) = vh/2+vh/4−1/2 = vh/2+ES(v)/2+vl/4−1/2.
Note that if vl ≤ 3vh/8 and ES(v) = 2, then vl = 2. Thus, statements (a) and (b) in the
theorem have no overlap.
Case 2.2: vl > 3vh/8. We claim that p ≤ vl/2 + (vh/2 − vl/2)/5 + 4/5. Suppose by
contradiction that p > q := ⌊vl/2 + (vh/2− vl/2)/5 + 4/5⌋. Since vl/2 and vh/2− vl/2 > 1
are integers, q ≥ vl/2 + 1 = cl + 1. Moreover, q = vl/2 + (vh/2 − vl/2)/5 + ε where
ε ∈ {0, 1/5, 2/5, 3/5, 4/5}. Let n := p−q ≥ 1 and tq := q−cl ≥ 1. Then, tq = vl/2+(vh/2−
vl/2)/5+ε−cl = (ch−cl)/5+ε. Thus, cl−3tq = cl−3(ch−cl)/5−3ε ≥ 8(cl−3ch/8)/5−3ε.
Since vl > 3vh/8, cl > 3ch/8. Since cl and ch are integers, cl − 3ch/8 ≥ 1/8. Thus,
cl − 3tq ≥ 1/5 − 3ε. Thus, if ε ≤ 1/5, cl − 3tq ≥ 0; if 2/5 ≤ ε ≤ 3/5, cl − 3tq ≥ −1; and if
ε = 4/5, cl − 3tq ≥ −2. Suppose that ε ≤ 1/5. Since tq ≥ 1, by (9), σl(p− 1) ≤ 1/(4tq +1).
By (4), 1/(ch − p+ 1) ≤ 1/(4tq + 1). Thus, 1/(ch − q − n+ 1) ≤ 1/(4tq + 1). Equivalently,
ch − 4tq − n ≥ q. Since n ≥ 1 and tq = (ch − cl)/5 + ε, ch − 4(ch − cl)/5 − 4ε − 1 ≥ q.
Equivalently, q−5ε−1 = cl+(ch−cl)/5−5ε−1 ≥ q. This is a contradiction. Suppose that
2/5ε ≤ 3/5. Since tq ≥ 1, by (9), σl(p−1) ≤ 1/(4tq−1+1). By (4), 1/(ch−p+1) ≤ 1/(4tq).
Thus, 1/(ch − q − n + 1) ≤ 1/(4tq). Equivalently, ch − 4tq − n + 1 ≥ q. Since n ≥ 1 and
tq = (ch−cl)/5+ε, ch−4(ch−cl)/5−4ε ≥ q. Equivalently, q−5ε = cl+(ch−cl)/5−5ε−1 ≥ q.
This is a contradiction. Finally, suppose that ε = 4/5. Since tq ≥ 1, by (9), σl(p − 1) ≤
1/(4tq − 2 + 1). By (4), 1/(ch − p + 1) ≤ 1/(4tq − 1). Thus, 1/(ch − q − n + 1) ≤
1/(4tq − 1). Equivalently, ch − 4tq − n + 1 + 1 ≥ q. Since n ≥ 1 and tq = (ch − cl)/5 + ε,
ch − 4(ch − cl)/5− 4ε+ 1 ≥ q. Equivalently, q − 3 = cl + (ch − cl)/5− 5ε+ 1 ≥ q. This is a
contradiction. Since σ is efficient and σh(p) > 0, the support of σl belongs to {0, ..., p − 1}.
Thus, πh(σ) ≥ vh − p ≥ vh − (vl/2 + (vh/2− vl/2)/5 + 4/5) = vh/2 + 4ES(v)/5 − 4/5.
Suppose now that ES(v) = 2. Recall that σ is efficient. Since vl > 3vh/8, vl > 2.
Since πh(σ) ≥ vh/2 + 4ES(v)/5 − 4/5 = vh/2 + 4/5, p ≤ cl + 1. We prove that p = cl.
25
Suppose by contradiction that p = cl+1, i.e., πh(σ) = vh/2+1. By Proposition 1, σh = δp.
Let {σλ}λ∈N be a sequence of interior payoff monotone distributions such that as λ → ∞,
σλ → σ. Since Uϕ(cl − 2|σh; vl) − Uϕ(cl + 1|σh; vl) = cl + 1 − cl = 1, there is Λ ∈ N such
that for each λ ≥ Λ, Uϕ(cl − 2|σ
λ
h ; vl) > Uϕ(cl + 1|σ
λ
h; vl). By payoff monotonicity, for each
λ ≥ Λ, σλl (cl − 2) ≥ σ
λ
l (cl + 1). By convergence, σl(cl − 2) ≥ σl(cl + 1). Notice also that
for each λ ∈ N, Uϕ(cl − 1|σ
λ
h ; vl) ≥ Uϕ(cl|σ
λ
h; vl). Thus, for each λ ∈ N, σ
λ
l (cl − 1) ≥ σ
λ
l (cl).
Now,
Uϕ(cl + 1|σ
λ
l ; vh)− Uϕ(cl|σ
λ
l ; vh) =
∑
b<cl
σλl (b)(−1) + σ
λ
l (cl)(ch + 1− ch)
+σλl (cl + 1)(ch − (ch − 1))
≤ −σλl (cl − 2)− σ
λ
l (cl − 1)
+σλl (cl) + σ
λ
l (cl + 1) ≤ 0.
By payoff monotonicity, for each λ ≥ Λ, σλh(cl) ≥ σ
λ
h(cl + 1). By convergence, σh(cl) ≥
σh(cl + 1). Since p = cl + 1 is in the support of σh, cl is also in the support of σh. This
contradicts that the support of σh is contained in {p, ..., p}.
Finally, suppose that ES(v) > 2 and vl ≥ 7vh/12 − 7/6. We claim that p < vl/2 +
(vh/2− vl/2)/5 +4/5. Suppose by contradiction that p = vl/2+ (vh/2− vl/2)/5 +4/5. Let
y := cl − 3(p − 1 − cl). Direct calculation yields that since vl ≥ 7vh/12 − 7/6, y ≥ ch − p.
By (7), there is Λ ∈ N such that for each λ ≥ Λ, and each max{0, cl − 3t} ≤ b < p − 1,
σλl (b) ≥ σ
λ
l (p− 1). Since p is in the support of σh and p > cl, for each b < p, Uϕ(b|σh; vl) >
Uϕ(p|σh; vl). Thus, we can suppose without loss of generality that for each λ ≥ Λ and each
b < p, Uϕ(b|σ
λ
h ; vl) > Uϕ(p|σ
λ
h ; vl) and consequently σ
λ
l (b) ≥ σ
λ
l (p). Thus,
Uϕ(p|σ
λ
l ; vh)− Uϕ(p− 1|σ
λ
l ; vh) =
∑
b<p−1 σ
λ
l (b)(−1)
+σλl (p− 1)(2ch − p− ch) + σ
λ
l (p)(ch − p)
= σλl (p)(ch − p)−
∑
0≤b<y σ
λ
l (b)
−
∑
y≤b≤p−1 σ
λ
l (b) + (ch − p+ 1)σ
λ
l (p− 1)
≤ σλl (p)(ch − p)−
∑
0≤b<y σ
λ
l (p)
−
∑
y≤b≤p−1 σ
λ
l (p − 1) + (ch − p+ 1)σ
λ
l (p − 1)
≤ σλl (p − 1)(ch − p+ 1− 4(p − 1− cl)− 1) = 0,
where the last equality follows from direct calculation given that p = cl+(ch−cl)/5+4/5. By
payoff monotonicity, for each λ ≥ Λ, σλh(p−1) ≥ σ
λ
h(p). By convergence, σh(p−1) ≥ σh(p).
Thus, p − 1 is in the support of σh. This contradicts that the support of σh belongs to
{p, ..., p}.
Proof of sufficiency in Theorem 1. Let (M,ϕ) be an EPA and v ∈ Θ such that vl <
26
Statement Case 1 Case 2 Case 3 Case 4 Case 5
1 +
2 + + +
3 + + +
4 + + + +
Table 1: Cases in the proof of sufficiency in Theorem 1: If the case in the proof (column) applies to the
statement of the theorem (row), there is a + in the corresponding cell.
vh. We prove that each Nash equilibrium of (M,ϕ, v) satisfying the bounds in the statement
of the theorem is an empirical equilibrium. We prove it for WB. The proof for LB is
symmetric. We consider five cases for which the same type of construction applies. These
cases are exhaustive. Table 1 summarizes how they apply to the different statements in the
theorem.
Case 1: Let (πl, πh) be such that πl = cl and πh = ch+ES(v). We construct a sequence
of weakly payoff monotone distributions that converges to σ where σl = δcl−1 and σh = δcl .
One can easily see that σ is a Nash equilibrium of (M,ϕ, v). Note that for each b ≤ cl − 1,
Uϕ(cl|σl; vh)−Uϕ(b|σl; vh) ≥ Uϕ(cl|σl; vh)−Uϕ(cl−1|σl; vh) = vh−cl−ch = ch−cl > 0, (10)
and for each b > cl,
Uϕ(cl|σl; vh)− Uϕ(b|σl; vh) = b− cl > 0. (11)
Thus, cl is h’s unique best response to σl. Note also that for each b > cl,
Uϕ(cl − 1|σh; vl)− Uϕ(b|σh; vl) = cl − (vl − b) = b− cl > 0. (12)
Let ε > 0 and t ∈ N. For each xh ∈ R
Mh and xl ∈ R
Mh , let
f
ε,t
h (xh) = (1− ε)δcl + εl
t(x),
f
ε,t
l (xh) = (1− ε)δcl−1 + εl
t(x).
Let {γt}t∈N be such that for each t ∈ N, γ
t is a fixed point of the composition of (f
ε,t
h , f
ε,t
l )
and the expected payoff operator, i.e., the continuous mapping that assigns to each profile
of mixed strategies γ,
F ε,t(γ) := (f
ε,t
l
(Uϕ(b|γh; vl)b∈{0,...,p}), f
ε,t
h
(Uϕ(b|γl; vh)b∈{0,...,p})).
Existence of γt is guaranteed by Brower’s fixed point theorem.
We will now describe the intuition of the reminder of the proof of this case. We will spell
27
the details after. The construction of empirical equilibria in other cases will rest partially
on arguments that are related to this first case. Observe that as ε vanishes and t diverges to
infinity, the distributions γε,t converge to σ. Thus, we need to show that for appropriately
chosen parameters ε and t, γε,t is weakly payoff monotone for v. We will show that it is
indeed payoff monotone for v. Note that by continuity, we will be able to select ε small
enough that (10-12) hold. Since γε,t is interior, cl − 1 actually has higher expected utility
than cl for l (l gains one dollar when h bids below cl − 1; note that by our assumption
vl ≥ 4, cl − 1 > 0). To to complete the construction, it is enough to show that one can
select ε and t such that cl − 1 has higher utility than any lower bid for l given that h plays
γ
ε,t
h
. Note that the difference in utility between cl−1 and any bid below it for h is bounded
below by a positive number. This is so because when h bids cl − 1 the auction ends in a
tie with probability close to one. Bidding below cl − 1 guarantees h loses almost for sure.
Since the probabilities that γ
ε,t
h
places in cl − 1 and any bid below it are determined by a
logistic QRF, the ratio of this probabilities also vanishes, i.e., h plays any bid below cl − 1
with much less probability than cl − 1. Now, conditional on h bidding cl − 1, the difference
in utility for l between bidding cl − 1 or below this bid is at least one unit. If what h bids
below cl− 1 is small compared to what she bids cl− 1, l prefers to bid cl − 1 than any lower
bid.
To simplify notation, let u
ε,t
l
(b) := Uϕ(b|γ
ε,t
h
; vl) and u
ε,t
h
(b) := Uϕ(b|γ
ε,t
l
; vh). For each
t ∈ N,
||σ − γε,t||∞ := max
i∈{l,h},b∈{0,...,p}
|σi(b)− γ
ε,t
i (b)| < ε.
Thus, there is η > 0 such that for each 0 < ε < η, each t ∈ N, each b ∈ {0, ..., p} \ {cl},
u
ε,t
h (cl)− u
ε,t
h (b) > 0, (13)
and for each b > cl,
u
ε,t
l (cl − 1)− u
ε,t
l (cl − 1) > 0. (14)
For each t and ε, γε,t is interior. Thus,
u
ε,t
l
(cl − 1)− u
ε,t
l
(cl) =
∑
b<cl−1
γ
ε,t
h
(b) > 0. (15)
By (13), for each 0 < ε < η and each t, γ
ε,t
h is ordinally equivalent to (u
ε,t
h (b))b∈{0,...,p}
(note that γ
ε,t
h is determined outside the support of σh by a logistic QRF). Let 0 < ε <
min{η, 2/(p−2)}. We claim that we can fin T ∈ N such that for each t ≥ T , γ
ε,t
l is ordinally
equivalent to (u
ε,t
l (b))b∈{0,...,p}. By (14) and (15) it is enough to show that there is T such
28
that for each t ≥ T , for each b < cl−1, u
ε,t
l
(cl−1)−u
ε,t
l
(b) > 0 (note that γ
ε,t
l
is determined
outside the support of σl by a logistic QRF).
For each b < cl − 1,
u
ε,t
h (cl − 1)− u
ε,t
h (b) =
∑
d<b γ
t
l (d)(−(cl − 1− b)) + γ
t
l (b)(vh − cl + 1− ch)
+
∑
b<d<cl−1
γtl (d)(vh − cl + 1− d) + γ
t
l (cl − 1)(ch − cl + 1)
> −εp+ 2(1− ε) > 0.
where the last inequality holds because ε < 2/(p − 2).
Thus, for each b < cl − 1.
u
ε,t
l (cl)− u
ε,t
l (b) =
∑
d<b γ
t
h(d)(−(cl − b)) + γ
t
h(b)(vl − cl − cl)
+
∑
b<d<cl
γth(d)(vl − cl − d) + γ
t
h(cl)(cl − cl)
≥ −
∑
d<b γ
t
h(d)cl + γ
t
h(cl − 1)
= γth(cl − 1)
(
1−
∑
d<b
γt
h
(d)
γt
h
(cl−1)
cl
)
= γth(cl − 1)
(
1−
∑
d<cl−1
γt
h
(d)
γt
h
(cl−1)
cl
)
= γth(cl − 1)
(
1− cl
∑
d<cl−1
etuh(d)
etuh(cl−1)
cl
)
= γth(cl − 1)
(
1− cl
∑
d<cl−1
e−t(uh(cl−1)−uh(d))
)
≥ γth(cl − 1)
(
1− cl
∑
d<cl−1
e−t(−εp+2(1−ε))
)
.
Thus, there is T ∈ N such that for each t ≥ T , u
ε,t
l (cl)− u
ε,t
l (b) > 0.
Case 2: πl = cl +1 and πh = ch +ES(v)− 1 and ES(v) > 2. Let ε > 0 and t ∈ N. For
each xh ∈ R
Mh and xl ∈ R
Mh , let
f
ε,t
h (xh) = (1− ε)δcl+1 + εl
t(x).
f
ε,t
l
(xh) = (1/2 − ε)δcl−1 + (1/2 − ε)δcl + 2εl
t(x).
Let {γt}t∈N be such that for each t ∈ N, γ
t is a fixed point of the composition of (f
ε,t
h , f
ε,t
l )
and the expected payoff operator. As t → ∞, γth → δcl and γ
t
l → (1/2)δcl−1 + (1/2)δcl . Let
σ be this limit distribution profile. Then, πl(σ) = cl + 1 and πh(σ) = ch + ES(v) − 1. In
order to complete the construction as in Case 1, we need to show that ε can be selected
small enough such that for large t, the distribution is strictly monotone with respect to
expected payoffs. The expected payoff of cl + 1 given σ for h, is greater than the payoff of
any other bid. The expected payoff of cl − 1 and cl given σ for l is greater than the payoff
of any bid higher than cl. Thus, it is only necessary to show that the expected payoff of
cl + 1, for the distribution profile γ
t for h, is eventually greater than the payoff of cl. Note
29
that the expected payoff of cl − 1 is never less than that of cl for l. For each b < cl − 1,
u
ε,t
h (cl − 1)− u
ε,t
h (b) =
∑
d<b γ
t
l (d)(−(cl − 1− b)) + γ
t
l (b)(vh − cl + 1− ch)
+
∑
b<d<cl−1
γtl (d)(vh − cl + 1− d) + γ
t
l (cl − 1)(ch − cl + 1)
> −2εp+ 4(1/2 − ε).
Thus, the argument in Case 1 can be easily reproduced for the sequence of fixed points.
Case 3: cl ≤ 3ch/8, cl + 1 < p ≤ vh/4 + 1/2, πl = p and πh = vh − p and ES(v) > 2.
Let ε > 0 and t ∈ N. For each xh ∈ R
Mh and xl ∈ R
Mh , let
f
ε,t
h (xh) = (1− ε)δp + εl
t(x).
f
ε,t
l (xh) = (1/p − ε)
∑
b<p
δb
+ (p − 1)εδp + εl
t(x).
Let {γt}t∈N be such that for each t ∈ N, γ
t is a fixed point of the composition of (f
ε,t
h , f
ε,t
l )
and the expected payoff operator. As t → ∞, γth → δp and γ
t
l → (1/p)
∑
b<p δb. Let σ be
this limit distribution profile. Then, πl(σ) = p and πh(σ) = vh − p. In order to complete
the construction as in Case 1, we need to show that ε can be selected small enough such
that for large t, the distribution is strictly monotone with respect to expected payoffs. The
expected payoff of p given σ for h, is greater than the payoff of any bid greater than p. The
expected payoff of each bid less than or equal to p−1 given σ for l is greater than the payoff
of any bid greater than or equal to p. The expected payoff of p given σ for l is greater than
the payoff of any bid greater than p. Thus, it is only necessary to show that the expected
payoff of p, for the distribution profile γt for h, is eventually greater than the payoff of each
b < p. Since 2 ≤ cl + 1 < p ≤ vh/4 + 1/2 = ch + 1/2, (ch − p+ 1)/p ≥ 1,
u
ε,t
h
(p)− u
ε,t
h
(p− 1) = −
∑
d<p−1 γ
t
l (d) + γ
t
l (p − 1)(vh − p− ch)
+γtl (p)(ch − p)
> −(1/p − ε)(p − 1)− ε+ (1/p − ε)(ch − p) + (p− 1)ε(ch − p)
= −1 + 1/p + ε(p− 1)− ε+ (ch − p+ 1)/p − 1/p
−ε(ch − p) + (p− 1)ε(ch − p) ≥ 0.
30
Moreover, for b < p,
u
ε,t
h (b)− u
ε,t
h (b− 1) = −
∑
d<b−1 γ
t
l (d) + γ
t
l (b− 1)(vh − b− ch)
+γtl (b)(ch − b)
> −(1/p − ε)(b − 1)− ε+ 2(1/p − ε)(ch − b)
≥ −b/p + 1/p + ε(b− 1)− ε+ 2(ch − b+ 1)/p − 2/p − 2ε(ch − b)
≥ 1− 2ε(ch − b)− ε− 1/p.
Thus, the argument in Case 1 can be easily reproduced for the sequence of fixed points.
Case 4: vl > 3vh/8, cl + 1 < p < vl/2 + (vh/2 − vl/2)/5 + 4/5, πl = p and πh = vh − p
and ES(v) > 2. Let y := cl − 3(p− 1− cl). Since p− 1 ≤ cl +(vh/2− vl/2)/5 +4/5− 1 and
cl > 3ch/8, y > 0. Let n := p− y = 4(p−1− cl)+1. Since p < vl/2+(vh/2− vl/2)/5+4/5,
1/(4(p − 1 − cl) + 1) > 1/(ch − p + 1). Thus, 1/n > 1/(ch − p + 1). Let σ be defined by:
σh = δp and σl := (1/n)
∑
y≤b<p δb. Then, Uϕ(p|σl; vh) − Uϕ(p − 1|σl; vh) = −(n − 1)/n +
(1/n)(ch−p) > 0. Thus, p is a unique best response to σl for h. Clearly, σl is a best response
to σh for l. Thus, σ is a Nash equilibrium with payoffs πl(σ) = p and πh(σ) = vh − p. Now,
for y ≤ b− 1 < b ≤ p− 1,
Uϕ(b|σl; vh)− Uϕ(b− 1|σl; vh) ≥ −(n− 2)/n + (1/n)(ch − b) + (1/n)(ch − b) > 1.
Thus, Uϕ(b|σl; vh) strictly increases in the set b ∈ {y, ..., p}. Let r ∈ {p, ..., p} be the
maximum for which Uϕ(r|σl; vh) ≥ Uϕ(y|σl; vh). Thus, Uϕ(r + 1|σl; vh) < Uϕ(y|σl; vh).
Now, for each b < y,
Uϕ(y|σl; vh)− Uϕ(b|σl; vh) ≥ σl(y)(ch − y) ≥ ES(v)/n > 0.
Thus, there is ε > 0 for which for each distribution profile γ such that ||σ − γ||∞ < ε,
(i) there is a constant c > 0 such that for each b 6= p, Uϕ(p|γl; vh) − Uϕ(b|γl; vh) ≥ c;
(ii) for each b ∈ {y, ..., r} and each d < y or d > r, Uϕ(b|γl; vh) − Uϕ(d|γl; vh) > 0; (iii)
Uϕ(b|γl; vh) strictly increases in the set b ∈ {y, ..., p}; and (iv) for each b ≤ p − 1 < d,
Uϕ(b|γh; vl)− Uϕ(d|γh; vl) > 0.
Let t ∈ N and η := ε/2(r − y + 2). For each xh ∈ R
Mh and xl ∈ R
Mh , let
f
ε,t
h (xh) = (1− ε/2)δp + η
∑
y≤b≤r
δb + ηl
t(x).
f
ε,t
l (xh) = (1/n − ε)
∑
y≤b<p
δb
+ nεlt(x).
31
Let {γt}t∈N be such that for each t ∈ N, γ
t is a fixed point of the composition of (f
ε,t
h
, f
ε,t
l
)
and the expected payoff operator. Then, ||γt − σ||∞ < ε. Thus, for each t ∈ N, γ
t satisfies
conditions (i)-(iv) above. By (i) and (ii), for each t ∈ N and each pair {b, d} ⊆ {0, ..., p},
γth(b) ≥ γ
t
h(d) if and only if Uϕ(b|γ
t
l ; vh) − Uϕ(d|γ
tl; vh). By (iv), for each t ∈ N and each
b ≤ p − 1 < d, Uϕ(b|γ
t
h; vl) − Uϕ(d|γ
t
h; vl) > 0. By (iii) we can reproduce the argument in
our proof of necessity in Theorem 1 and show that for each t ∈ N and each y ≤ b < p − 1,
Uϕ(b|γ
t
h; vl) − Uϕ(p − 1|γ
t
h; vl) ≥ 0. Now, by (i), as t → ∞, γ
t
h → γh := (1 − ε/2 + η)δp +
η
∑
y≤b≤r δb. Direct calculation shows that for each b < y, Uϕ(p − 1|γh; vl)− Uϕ(b|γh; vl) =
(p − 1 − cl)η. Thus, there is T ∈ N such that for each t ≥ T , if d < y ≤ b ≤ p − 1,
Uϕ(b|γ
t
h; vl) − Uϕ(d|γ
t
h; vl) > 0. This implies that for each i ∈ N and {b, d} ⊆ {0, ..., p},
γtl (b) ≥ γ
t
l (d) if and only if Uϕ(b|γ
t
h; vl) ≥ Uϕ(d|γ
th; vl). Thus, one can complete the proof
as in Case 1.
Case 5: vl > 3vh/8, vl < 7vh/12−7/6, cl+1 < p = vl/2+(vh/2−vl/2)/5+4/5, πl = p,
πh = vh−p and ES(v) > 2. Let y := cl−3(p−1−cl). Let n = p−1−y+1 = 4(p−1−cl)+1.
Since p− 1− cl ≥ 1, n ≥ 5. Direct calculation yields,
n = ch − p+ 1. (16)
Now, (ch − p)− y = 6((7vh/12− 7/6) − vl)/5 > 0. Since ch, p, and y are integers,
(ch − p)− y ≥ 1. (17)
Let σh := δp and σl := (1/n)
∑
y≤b≤p−1 δb. Clearly, for each p ≤ b < d, Uϕ(b|σl; vh) ≥
Uϕ(d|σl; vh) + 1. Clearly, for each d < y, Uϕ(y|σl; vh) > Uϕ(b|σl; vh). Let y ≤ d < p − 1.
Then,
Uϕ(d+ 1|σl; vh)− Uϕ(d|σl; vh) = −(b− y)(1/n) + (1/n)(vh − (d+ 1)− ch)
+(1/n)(ch − (d+ 1))
= −(b− y + 2)/n + 2(ch − p+ 1)/n − 2/n
≥ 1− 2/n > 0.
Let r ∈ {p, ..., p} be the maximum for which Uϕ(r|σl; vh) ≥ Uϕ(y|σl; vh). Thus, for each
y ≤ b ≤ r and d > r, Uϕ(b|σl; vh) > Uϕ(d|σl; vh). Clearly, for each b < p < d, Uϕ(b|σh; vl) >
Uϕ(p|σh; vl) > Uϕ(d|σh; vl). Thus, there is ζ > 0 such that if ||γ − σ||∞ < ζ, then
(a) Uϕ(b|γl; vh) > Uϕ(d|γl; vh) whenever one of the following four conditions is satisfied
(i) p ≤ b < d; (ii) d < b = y; (iii) y ≤ d < p − 1 and b = d + 1; or (iv) y ≤ b ≤ r and
d > r.
32
(b) For each p < d, Uϕ(p|γl; vh) ≥ Uϕ(d|γl; vh) + 1/2.
(c) For each b < p < d, Uϕ(b|γh; vl) > Uϕ(p|γh; vl) > Uϕ(d|γh; vl).
Let t ∈ N, ε > 0, η := ε/2(r − y + 2), τ := (y + 3/2)ε/n such that 2max{ε, τ} < ζ. For
each xh ∈ R
Mh and xl ∈ R
Mh , let
f
ε,t
h (xh) = (1− ε/2)δp + η
∑
y≤b≤r
δb + ηl
t(x).
f
ε,t
l
(xl) = (1/n − τ)
∑
y≤b≤p−1
δb
+ ε
∑
0≤b≤y−1
δb
+ εδp + (ε/2)l
t(x).
Let {γt}t∈N be such that for each t ∈ N, γ
t is a fixed point of the composition of (f
ε,t
h
, f
ε,t
l
)
and the expected payoff operator. Then, ||γt − σ||∞ < ζ. Thus, for each t ∈ N, γ
t satisfies
(a)-(c) above. Now,
Uϕ(p|γ
t
l ; vh)− Uϕ(p− 1|γ
t
l ; vh) =
∑
0≤b≤p−2 γ
t
l (b)(−1)
+γtl (p− 1)(vh − p− ch) + γ
t
l (p)(ch − p)
≥ −yε− (n− 1)(1/n − τ)
−ε/2 + (1/n − τ)(ch − p) + ε(ch − p)
= ε(ch − p− y − 1/2)
+(1/n − τ)(ch − p+ 1− n).
By (16) and (17), Uϕ(p|γ
t
l ; vh) − Uϕ(p− 1|γ
t
l ; vh) ≥ ε/2 > 0. Thus, for each t ∈ N and
each {b, d} ⊆ {0, ..., p}, γth(b) ≥ γ
t
h(d) if and only if Uϕ(b|γ
t
l ; vh) − Uϕ(d|γ
t
l ; vh). Moreover,
since for each t ∈ N and each b 6= p, Uϕ(p|γ
t
l ; vh) ≥ Uϕ(p− 1|γ
t
l ; vh) + ε/2, we have that as
t → ∞,
γth → γh := (1− ε/2 + η)δp + η
∑
y≤b≤r
δb.
The construction can be completed as in Case 6.
Proof of Lemma 2. (M,ϕ) be a mechanism satisfying the property in the statement of
the lemma. Let v ∈ Θ, such that vl < vh, 0 ≤ t ≤ t + 1 ≤ ES(v). Let v
∗
l := 2(vl/2 + t)
and v∗h := v
∗
l +2. Thus, ES(v
∗) = 1 and the set of efficient and equitable allocations for v∗
is that in which agent h receives the object and the payoff of the lower valuation agent is
between vl/2+ t and vl/2+ t+1. By the hypothesis of the lemma, there is σ ∈ N(M,ϕ, v
∗)
that obtains an efficient allocation for v∗, i.e., agent h receives the object for sure, and
vl/2 + t ≤ πl(σ) ≤ vl/2 + t+ 1. We claim that σ ∈ N(M,ϕ, v). We prove first that σl is a
33
best response to σh for l with type vl. Let ml ∈ Ml be in the support of σl and m
′
l ∈ Ml.
Then
Uϕ(ml|σh; vl) = Uϕ(ml|σh; v
∗
l ) ≥ Uϕ(m
′
l|σh; v
∗
l ) ≥ Uϕ(m
′
l|σh; vl).
Where the first equality holds because in equilibrium σ, l never receives the object, so
both expressions are the same integrals of expected compensation; the first inequality is
the equilibrium condition for l with type v∗l ; and the third inequality holds because the
expected utility index of an allotment for l with value vl is less than or equal than the
expected utility index of the allotment for l with value v∗l , because v
∗
l ≥ vl.
Letmh ∈ Mh be in the support of σh andm
′
h ∈ Mh. Since σ assigns the object to agent h
with certainty, Uϕ(mh|σl; vh) − Uϕ(mh|σl; v
∗
h) = vh − v
∗
h. Since the utility index of h is in-
variant when receiving an amount of money and no object, Uϕ(m
′
h|σl; vh)−Uϕ(m
′
h|σl; v
∗
h) ≤
vh − v
∗
h. Thus,
Uϕ(m
′
h|σl; vh)− Uϕ(mh|σl; vh) ≤ Uϕ(m
′
h|σl; v
∗
h)− Uϕ(mh|σl; v
∗
h) ≤ 0,
where the last inequality is the equilibrium condition for σ with type v∗h. Thus, σh is a best
response to σl for h with type vh.
Thus, σ ∈ N(M,ϕ, v) is efficient and such that the lower valuation agent’s payoff is
between vl/2 + t and vl/2 + t+ 1.
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https://arxiv.org/abs/1907.12408
1 Introduction
2 Related literature
3 Model
4 Results
4.1 Extreme-price auctions and Nash equilibrium
4.2 Empirical equilibrium
4.2.1 Empirical equilibrium payoffs
4.2.2 The intuition
4.3 Empirical equilibrium analysis and full implementation theory
5 Discussion
6 Conclussion
7 Appendix
| 0non-cybersec
| arXiv |
I tuck my knees to my chest and lean forward until gravity takes me.. That's just how I roll. | 0non-cybersec
| Reddit |
redirecting output of /dev/fb0 to /dev/fb2. <p>I am using IMX6 Saberauto board on Linux OS. I have 2 displays primary is <code>HDMI</code> and secondary is <code>LVDS</code>. I want to run 2 applications.</p>
<p>One QT on primary display, which I am able to do. Second application is airplay application I want this application to run on seondary display ie <code>/dev/fb2</code>. But I am not able to do that.</p>
<p>It is coming on primary display only. Please let me know the command to direct airplay output to <code>/dev/fb2</code>.</p>
| 0non-cybersec
| Stackexchange |
Java: Float Formatting depends on Locale. <p>I live in Belgium. And generally, in mathematics, we write our decimals with a comma like this: 3,141592<br>
And that is also the result when I <strong>format</strong> the float.</p>
<pre><code>System.out.println(String.format("%f", 3.141592));
</code></pre>
<p>So, the <code>.</code> is replaced by a <code>,</code> like so: <code>3,141592</code>. So always when I need a point instead I have to add something like this: <code>String.format("%f", 3.14).replace(',','.');</code></p>
<p>So, the question is: is there a way to change the Locale which makes every formatter in Java use a point, instead of comma?</p>
<p>Thanks</p>
<hr>
<pre><code>System.out.println(Locale.getDefault());
</code></pre>
<p>prints</p>
<pre><code>nl_BE
</code></pre>
| 0non-cybersec
| Stackexchange |
Pro-Democracy Websites in Hong Kong Targeted with and Serving Malware (Java Exploits & PlugX). | 1cybersec
| Reddit |
Outlook 2016 for mac - migrate calendar to new exchange account. <p>I'm migrating a user from an old exhange host to a new one.
What I did was export everything from his old exchange account (mail, contacts, calendar, etc) to a .olm-file.
Then I created a new profile, set up the new account, and imported everything back into Outlook.</p>
<p>Buut.. the import didn't put everything back into the new account.
Rather, the import made a new local folder with all the contents from the .olm-file.</p>
<p>For mail and contacs I could drag&drop or copy&paste everything into the new exchange account, but for the calendar I don't know what to do?
Has anyone encountered this problem before?</p>
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| Stackexchange |
Access field of outer base class. <p>In Java, inner classes can normally access outer classes' private members.
While writing an Android app, I had a static inner class which extended it's outer class.
Turns out, private fields of the outer class could not be accessed:</p>
<pre><code>class Outer {
private int m_field = 1;
static class Inner extends Outer {
Inner() {
m_field = 2;
}
}
}
</code></pre>
<p>It gives a confusing error message:</p>
<blockquote>
<p>error: non-static variable m_field cannot be referenced from a static context</p>
</blockquote>
<p>Even though nothing is static except the class itself.</p>
<p>When the field m_field is made protected, it compiles without a problem.
But also, when doing this:</p>
<pre><code>class Outer {
private int m_field = 1;
static class Inner extends Outer {
Inner() {
((Outer)this).m_field = 2;
}
}
}
</code></pre>
<p>It works without problems.
Is this a bug in the compiler? Why would you need to cast to the outer class, which you are already an instance of?</p>
<hr>
<p>EDIT:</p>
<p>For a real use-case of this, consider a class like this:</p>
<pre><code>public abstract class MyItem {
private int m_counter = 0;
public abstract int updateSomething();
public static class CountItem extends MyItem {
public int updateSomething() { m_counter++; }
}
public static class DoubleCountItem extends MyItem {
public int updateSomething() { m_counter += 2; }
}
}
</code></pre>
<p>Pretty abstract example, but it can be used to provide basic implementations for abstract classes which don't need a lot of code on their own.</p>
<p>EDIT2:</p>
<p>As @Nathan suggests, it seems like this problem can be recreated by 2 classes without nesting:</p>
<pre><code>class Base {
private int x = 0;
void a(Extended b) {
((Base)b).x = 1; //<-- with cast: compiles, without: error
}
}
class Extended extends Base {
}
</code></pre>
<p>Which gives better error message:</p>
<blockquote>
<p>error: x has private access in Base</p>
</blockquote>
| 0non-cybersec
| Stackexchange |
Compiler chooses erroneous overload instead of valid overload. <p>Take a look at this code:</p>
<pre><code>#include <vector>
#include <functional>
template<typename RandIt, typename T, typename Pred>
auto search_with(RandIt begin, RandIt end, const T& value, Pred&& pred) noexcept {
//...
return begin;
}
template<typename RandIt, typename T>
auto search_with(RandIt begin, RandIt end, const T& value) noexcept {
return search_with(begin, end, value, std::less<T>{});
}
template<typename Array, typename T, typename Pred>
auto search_with(const Array& array, const T& value, Pred&& pred) noexcept {
return search_with(std::begin(array), std::end(array), value, std::forward<Pred>(pred));
}
int main() {
std::vector<int> v = { 1, 2, 3 };
search_with(v, 10, std::less<int>{}); // ok
search_with(v.begin(), v.end(), 10); // fail!
}
</code></pre>
<p>I do not understand why in the second <code>search_with</code> call, the compiler selects the third overload. If I comment out the third overload, then the code compiles fine. This indicates that the second overload is not discarded as it does compile, and it should be a valid overload.</p>
<p>However, the third overload is chosen, which fails, as there is no specialization of <code>std::begin</code> (and <code>std::end</code>) for iterators:</p>
<pre><code>main.cpp: In instantiation of 'auto search_with(const Array&, const T&, Pred&&) [with Array = __gnu_cxx::__normal_iterator<int*, std::vector<int> >; T = __gnu_cxx::__normal_iterator<int*, std::vector<int> >; Pred = int]':
main.cpp:23:39: required from here
main.cpp:17:34: error: no matching function for call to 'begin(const __gnu_cxx::__normal_iterator<int*, std::vector<int> >&)'
return search_with(std::begin(array), std::end(array), value, std::forward<Pred>(pred));
~~~~~~~~~~^~~~~~~
</code></pre>
<p>I would have thought that the opposite happens: the third overload is discarded because it fails to compile, and the second one is chosen.</p>
<p>But that is obviously not the case, so what is happening here? Why is the wrong overload being chosen? Why is the third overload a better match then the second one?</p>
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Nothing gets me Randy like a bunch of fully promoted Apocalypse Tanks. (Yuris Revenge). | 0non-cybersec
| Reddit |
iphone not charging after 7.0.6. My friend just updated his iPhone to 7.0.6 and now his iphone isn't charging. He's used apples charging cables and other off brand chargers and neither are working. Any help would be greatly appreciated. Thanks. | 0non-cybersec
| Reddit |
Which encoding supports this symbol ƒ?. <p>I have this encoding in my file:</p>
<pre><code>\usepackage[utf8]{inputenc} %unicode support
</code></pre>
<p>but I still get the error </p>
<pre><code>Package inputenc Error: Unicode char \u8... not set up for use with ...
</code></pre>
<p><strong>How can you make LaTeX understand the symbol ƒ?</strong></p>
| 0non-cybersec
| Stackexchange |
Complete immutability and Object Oriented Programming. <p>In most OOP languages, objects are generally mutable with a limited set of exceptions (like e.g. tuples and strings in python). In most functional languages, data is immutable.</p>
<p>Both mutable and immutable objects bring a whole list of advantages and disadvantages of their own.</p>
<p>There are languages that try to marry both concepts like e.g. scala where you have (explicitly declared) mutable and immutable data (please correct me if I am wrong, my knowledge of scala is more than limited).</p>
<p>My question is: <strong>Does <em>complete</em> (sic!) immutability -i.e. no object can mutate once it has been created- make any sense in an OOP context?</strong></p>
<p><strong>Are there designs or implementations of such a model?</strong></p>
<p><strong>Basically, are (complete) immutability and OOP opposites or orthogonal?</strong></p>
<p>Motivation: In OOP you normally operate <strong>on</strong> data, changing (mutating) the underlying information, keeping references between those objects. E.g. an object of class <code>Person</code> with a member <code>father</code> referencing another <code>Person</code> object. If you change the name of the father, this is immediately visible to the child object with no need for update. Being immutable you would need to construct new objects for both father and child. But you would have a lot less kerfuffle with shared objects, multi-threading, GIL, etc.</p>
| 0non-cybersec
| Stackexchange |
Cannot boot, ending in (initramfs), and kernel panic. <p>After getting some errors upgrading the Linux kernel packages with <code>apt</code> (including insufficient disk space on the boot partition, where the images are stored), I couldn't boot any more.</p>
<p>First, on my setup: I have one hard disk, <code>/dev/sda</code> with one boot partition <code>/dev/sda1</code> (this is where the kernel images are stored, and was mounted to /boot). The "root" partition is <code>/dev/mapper/ubuntu--vg--usbkey-root</code>.</p>
<p>More precisely:</p>
<pre><code>$ sudo fdisk -l /dev/sda
Disk /dev/sda: 465.8 GiB, 500107862016 bytes, 976773168 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disklabel type: dos
Disk identifier: 0x00000000
Device Boot Start End Sectors Size Id Type
/dev/sda1 * 2048 499711 497664 243M 83 Linux
/dev/sda2 501758 976771071 976269314 465.5G 5 Extended
/dev/sda5 501760 976771071 976269312 465.5G 8e Linux LVM
</code></pre>
<hr>
<pre><code>$ ls /dev/mapper
control ubuntu--vg--usbkey-root ubuntu--vg--usbkey-swap_1
</code></pre>
<hr>
<pre><code>$ sudo lvs
LV VG Attr LSize Pool Origin Data% Meta% Move Log Cpy%Sync Convert
root ubuntu-vg-usbkey -wi-a----- 457.51g
swap_1 ubuntu-vg-usbkey -wi-a----- <7.96g
</code></pre>
<hr>
<pre><code>$ lsblk
NAME MAJ:MIN RM SIZE RO TYPE MOUNTPOINT
loop0 7:0 0 1.7G 1 loop /rofs
loop1 7:1 0 86.6M 1 loop /snap/core/4486
loop2 7:2 0 140M 1 loop /snap/gnome-3-26-1604/59
loop3 7:3 0 1.6M 1 loop /snap/gnome-calculator/154
loop4 7:4 0 12.2M 1 loop /snap/gnome-characters/69
loop5 7:5 0 21M 1 loop /snap/gnome-logs/25
loop6 7:6 0 3.3M 1 loop /snap/gnome-system-monitor/36
sda 8:0 0 465.8G 0 disk
├─sda1 8:1 0 243M 0 part
├─sda2 8:2 0 1K 0 part
└─sda5 8:5 0 465.5G 0 part
├─ubuntu--vg--usbkey-root
│ 253:0 0 457.5G 0 lvm /mnt
└─ubuntu--vg--usbkey-swap_1
253:1 0 8G 0 lvm
sdb 8:16 1 1.9G 0 disk /cdrom
├─sdb1 8:17 1 1.8G 0 part
└─sdb2 8:18 1 2.3M 0 part
sr0 11:0 1 1024M 0 rom
</code></pre>
<p>My last attempt was to follow the instructions from <a href="https://zeldor.biz/2010/12/install-grub-from-chroot/" rel="nofollow noreferrer">this article</a>.</p>
<p>So I did the following:</p>
<pre><code>$ sudo mount /dev/sda1 /mnt/boot/
$ sudo mount /dev/mapper/ubuntu--vg--usbkey-root /mnt/
$ sudo mount -t proc none /mnt/proc
$ sudo mount -o bind /dev /mnt/dev
$ sudo mount -t sysfs sys /mnt/sys
$ sudo chroot /mnt
# update-grub
Generating grub configuration file ...
Warning: Setting GRUB_TIMEOUT to a non-zero value when GRUB_HIDDEN_TIMEOUT is set is no longer supported.
Found linux image: /boot/vmlinuz-4.4.0-127-generic
Found initrd image: /boot/initrd.img-4.4.0-127-generic
Found linux image: /boot/vmlinuz-4.4.0-124-generic
Found initrd image: /boot/initrd.img-4.4.0-124-generic
Found linux image: /boot/vmlinuz-4.4.0-116-generic
Found initrd image: /boot/initrd.img-4.4.0-116-generic
Found memtest86+ image: /memtest86+.elf
Found memtest86+ image: /memtest86+.bin
/run/lvm/lvmetad.socket: connect failed: No such file or directory
WARNING: Failed to connect to lvmetad. Falling back to internal scanning.
done
</code></pre>
<p>Is this warning an issue? Anyway, I then did the following:</p>
<pre><code># /usr/sbin/grub-install --recheck --no-floppy /dev/sda
Installing for i386-pc platform.
Installation finished. No error reported.
</code></pre>
<p>Then I rebooted, and I was lead to a <code>(initramfs)</code> prompt. The screen contained the following error message:</p>
<pre><code>fsck: error 2 (No such file or directory) while executing fsck.ext2 for /dev/sda1
</code></pre>
<p>However, I checked <code>/dev/sda1</code> with <code>fsck</code> from the bootable USB key and it reports no errors... Same for <code>/dev/sda5/</code>.</p>
<p>Also, the command </p>
<p>(initramfs) ls /root</p>
<p>reports the contents of <code>/dev/sda1</code>.
However, besides the expected content, there is also a directory <code>/root/boot/grub</code>:</p>
<pre><code>(initramfs) ls /root/boot/grub
fonts locale grubenv i386-pc
</code></pre>
<p>Running </p>
<pre><code>(initramfs) exit
</code></pre>
<p>brings me to a screen which ends with</p>
<pre><code>end Kernel panic - not syncing: Attempted to kill init!
</code></pre>
<p>All this is quite confusing to me. Any suggestions are appreciated.</p>
| 0non-cybersec
| Stackexchange |
Playful girlfriend. | 0non-cybersec
| Reddit |
Why the braves do the tomahock chop!. | 0non-cybersec
| Reddit |
4 parties and a funeral: A day on the trail with Sanders and Biden. | 0non-cybersec
| Reddit |
Postgresql truncate table with foreign key constraint. <p>Currently I am trying to truncate tables which have foreign key constraint on Postgresql 11.3.</p>
<p>I tried doing this </p>
<pre><code>BEGIN;
SET CONSTRAINTS ALL DEFERRED;
TRUNCATE tableA;
COMMIT;
</code></pre>
<p>but received error</p>
<pre><code>ERROR: cannot truncate a table referenced in a foreign key constraint
DETAIL: Table "xxx" references "tableA".
HINT: Truncate table "xxx" at the same time, or use TRUNCATE ... CASCADE.
</code></pre>
<p>Doesn't <code>SET CONSTRAINTS ALL DEFERRED</code> would turn off the foreign key constraint check? Are there anyway to truncate a table without triggering foreign key constraint check and not involving CASCADE?</p>
| 0non-cybersec
| Stackexchange |
Creator of the Jolicraft texture pack completely redoes it as Jolicraft 2. | 0non-cybersec
| Reddit |
Can I use __qualname__ in a Python type hint with postponed annotation evaluation?. <p>I like using <code>__qualname__</code> for the return-type annotation of factory-style class methods, because it doesn't hardcode the classname and therefore keeps working subclasses (cf. <a href="https://stackoverflow.com/a/58789374/857390">this answer</a>).</p>
<pre class="lang-py prettyprint-override"><code>class Foo:
@classmethod
def make(cls) -> __qualname__:
return cls()
</code></pre>
<p>Currently this seems to work fine, but I am not sure whether this will still be possible with the postponed evaluation of annotations (<a href="https://www.python.org/dev/peps/pep-0563/" rel="nofollow noreferrer">PEP 563</a>): the PEP <a href="https://www.python.org/dev/peps/pep-0563/#id5" rel="nofollow noreferrer">says</a> that</p>
<blockquote>
<p>Annotations can only use names present in the module scope as postponed evaluation using local names is not reliable (with the sole exception of class-level names resolved by <code>typing.get_type_hints()</code>).</p>
</blockquote>
<p>The PEP also <a href="https://www.python.org/dev/peps/pep-0563/#id7" rel="nofollow noreferrer">says</a>:</p>
<blockquote>
<p>The <code>get_type_hints()</code> function automatically resolves the correct value of <code>globalns</code> for functions and classes. It also automatically provides the correct <code>localns</code> for classes.</p>
</blockquote>
<p>However, the following code</p>
<pre class="lang-py prettyprint-override"><code>from __future__ import annotations
import typing
class Foo():
@classmethod
def make(cls) -> __qualname__:
return cls()
print(typing.get_type_hints(Foo.make))
</code></pre>
<p>fails with</p>
<pre><code> File "qualname_test.py", line 11, in <module>
print(typing.get_type_hints(Foo.make))
File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 1004, in get_type_hints
value = _eval_type(value, globalns, localns)
File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 263, in _eval_type
return t._evaluate(globalns, localns)
File "/var/local/conda/envs/py37/lib/python3.7/typing.py", line 467, in _evaluate
eval(self.__forward_code__, globalns, localns),
File "<string>", line 1, in <module>
NameError: name '__qualname__' is not defined
</code></pre>
<p>Commenting the <code>__future__</code> import makes the code work again, in that case it outputs</p>
<pre><code>{'return': <class '__main__.Foo'>}
</code></pre>
<p>as expected.</p>
<p><strong>Is my use case <code>__qualname__</code> supported (which would mean that <code>get_type_hints</code> is buggy), or is this approach not possible with PEP 563?</strong></p>
| 0non-cybersec
| Stackexchange |
President Obama Endorses Gay Marriage on ABC Nightly News. | 0non-cybersec
| Reddit |
Elevator malfunction. | 0non-cybersec
| Reddit |
The measurements on this jug. | 0non-cybersec
| Reddit |
Any good mp3 volume normalizer?. <p>I tried to use <a href="https://superuser.com/questions/14635/windows-command-line-program-to-change-the-volume-of-an-mp3-file">MP3Gain</a> to keep mp3 files with the same volume. </p>
<p>The problem is: when I re synchronize all files into my IPod, some musics get too loud.</p>
<p>So, it seems MP3Gain is not fixing volume at all or there is a problem with synchronization software.</p>
<p>Perhaps I should "re add" files into ITunes 10.1 library again.</p>
<p>Meanwhile, is there a another good mp3 volume normalizer for testing playback in IPod?</p>
| 0non-cybersec
| Stackexchange |
Graph of the AFCS playoff chances throughout the season. | 0non-cybersec
| Reddit |
I am C. Robert Cargill, long time film critic turned co-writer of SINISTER. AMA.. I'll be back around 5pm Central (sooner if my schedule permits) to answer your questions. Let 'er rip, Dreadit!
**Edit 6:10pm** Alright gang! I'm off to try and get another job! But I'll drop back in a few hours to do another round of answers if you still have questions! Thanks for all the great discussion so far!
**Edit 9:00pm** Thanks Dreadit! Been a great day with some amazing questions. A real pleasure on my end to hang with you. NIGHT! | 0non-cybersec
| Reddit |
Limit of indicator function. <p>I am considering the measure space <span class="math-container">$(\mathbb{R}, \mathcal{B}(\mathbb{R},\lambda)$</span>. For every integer <span class="math-container">$\geq 1$</span> the function <span class="math-container">$f_{n}$</span> is defined as:</p>
<p><span class="math-container">$f_{n}=n\cdot \mathbb{1}_{[n,n+n^{-2}]}$</span></p>
<p>I am asked to show that <span class="math-container">$f_{n} \to \infty $</span> for <span class="math-container">$n \to \infty$</span> pointwise.</p>
<p><strong>My attempt</strong></p>
<p>By the definition of pointwise convergence we would like to analyse the limit of <span class="math-container">$f_{n}$</span> when <span class="math-container">$x$</span> is fixed and <span class="math-container">$n \to \infty$</span>.</p>
<p>Now fix <span class="math-container">$x_{0} \in \mathbb{R}$</span>. If <span class="math-container">$x_{0} \in [n,n+n^{-2}]$</span> for an <span class="math-container">$n_{0}>2$</span> then <span class="math-container">$f_{n_{0}}=1$</span> but then <span class="math-container">$x_{0} \notin [n,n+n^{-2}]$</span> for all <span class="math-container">$N>n_{0}$</span> since all intervals <span class="math-container">$[n,n+n^{-2}]$</span> for <span class="math-container">$n>2$</span> are disjoint. This means that all <span class="math-container">$f_{N>n_{0}}=0$</span> which shows that <span class="math-container">$f_{n} \to 0$</span> when <span class="math-container">$n \to \infty$</span>.</p>
<p>I have tried to simplify the above by only considering <span class="math-container">$n>2$</span>. I know that <span class="math-container">$f_{1}=1$</span> and <span class="math-container">$f_{2}=2$</span> at the same time only when <span class="math-container">$n=2$</span>.</p>
<p>Feedback is much appreciated.</p>
| 0non-cybersec
| Stackexchange |
Creating a 'sequence' of subsets using the subset axiom, then using induction to prove they are all equal. <p>Setup:</p>
<p><span class="math-container">$(N,\sigma)$</span> satisfies Peano axioms, i.e. <span class="math-container">$N \equiv ω$</span>.</p>
<p>Using the subset axiom we define, for each <span class="math-container">$k \in N$</span>, <span class="math-container">$F_k = \{x \in N \; | \; \phi_k(x) \}$</span>. </p>
<p>Next we define <span class="math-container">$D = \{d \in N \; | \; F_d = N\}$</span> and show that <span class="math-container">$1 \in D$</span> and when <span class="math-container">$d \in D$</span>, so is <span class="math-container">$d + 1$</span> (using the fact that <span class="math-container">$F_d = N$</span>). So <span class="math-container">$D = N$</span> and for all <span class="math-container">$k \in N$</span>, <span class="math-container">$F_k = N$</span>.</p>
<p>What axioms of ZF set theory come into play here?</p>
<p>The answer can be found by reading the comments, so I will list them here:</p>
<p><a href="https://en.wikipedia.org/wiki/Axiom_of_infinity" rel="nofollow noreferrer">Axiom_of_infinity</a></p>
<p><a href="https://en.wikipedia.org/wiki/Axiom_of_extensionality" rel="nofollow noreferrer">Axiom of extensionality</a></p>
<p><a href="https://en.wikipedia.org/wiki/Axiom_schema_of_specification" rel="nofollow noreferrer">Axiom schema of specification</a></p>
<hr>
<p>I noticed that there is a vote to close this question - that it is unclear what I am asking. Well, without details, I guess it isn't the best question. But, the question did 'collect' three very important axioms from ZF set theory.</p>
<hr>
<p>I thought that since induction isn't an axiom, it is nothing to make a 'fuss' about. But, it seems that I need to take a refresher course in induction. For example, have you used <em>simultaneous</em> induction lately?</p>
<p>For a review with exercises, see</p>
<p><a href="https://www.maths.ox.ac.uk/system/files/attachments/induction_0.pdf" rel="nofollow noreferrer">Induction</a>
by Richard Earl
Mathematical Institute, Oxford
August 2003</p>
<p>I am adding the induction tag to this posting.</p>
| 0non-cybersec
| Stackexchange |
Office bunny. | 0non-cybersec
| Reddit |
My [22F] boyfriend [25M] of 1.5 years didn't put any effort into my birthday but still asked for a "thank you blowjob".. To start, my boyfriend and I have been together almost 1.5 years and have an apartment together.
Yesterday was my birthday and my boyfriend started the day by posting to my facebook that it was going to be a fun day full of presents and surprises. It was excited because in the past he's hasn't been very good with special occasions.
Valentine's day 2013 he bought me a rose from a gas station when he went there to buy cigarettes, and that was it for the day. It was a nice gesture, but it left me feeling like he was too lazy to put any thought into the day. When we talked about it, he admitted that laziness was the case. I felt hurt but got over it, and explained to him that I feel like it's not about what he gets for me or does for me, what matters is the time and thought he put into things.
Two weeks later it was my birthday. I was honest with him on Valentine’s Day and didn't play head games with him, so it was clear that it would be nice if he put some thought into doing something (anything) for me. A card made from printer paper and crayons would have meant something, but instead he just said happy birthday to me and went about his day. I was upset and felt hurt again, so I calmly expressed my disappointment, and told him that I'll try better to express what I'd like from him in the future.
So yesterday after he spent the morning building me up, I was feeling really optimistic about the day. When noon time came around though, I found myself still waiting for the surprises. I inquired about our plans for the day and he said he had nothing planned. We can do whatever I want though. Well, of course I can do whatever I want. I'm an adult with a job; I didn't need to wait for my birthday to know that. I didn't say that, but I thought it. I just suggested a restaurant instead.
The place we went to had a menu mainly of things to share. The boyfriend is a picky eater and anything I liked didn't sound good to him. I got over it and let him order what he wanted for us. After we ate I asked again what the plan was, and there was none, so I asked him if he wanted to come to a liquidation store with me.
I need new shoes so I went was the shoe section to find a pair. I'll buy myself a damn birthday gift. We also need curtains for our house though, and he was learning towards getting the curtains he noticed. I put the shoes back because I agreed with him that we need curtains first. Sure it's my birthday but I don't want to be selfish so I'll get something that benefits us both. We went home after that.
A while later I inquire about birthday cake. He was supposed to get one but it wasn't on his list of priorities so he agreed to come to the grocery store with me to get a cake. I didn't get a magical birthday full of surprises like I thought, but picking out a cake could cheer me up. The boyfriend ended up picking it out though. He didn't like what I chose and wouldn't eat it if I got it. So he picked out the cake he wanted...and made sure he got the discount one. I'm still fine though. It's good. It's just a day. It's just a cake.
We go home and eat some cake and get ready for bed. I have a headache and don't feel so well anymore. The boyfriend asks me if my day was good besides the headache. It could have been worse. I'm not ungrateful for what I have, so I tell him I had a good day. He responds with:
**“Good. I was going to ask you for a 'Thank you blowjob' for everything, but since you have a headache I won't.”**
I responded with “You should not have said that. You don't have to tell me everything you think" and left it at that and rolled over to go to sleep.
I feel really insulted and let down. I'm getting really tired of explaining the idea of making the other person feel special. I don't expect to be made to feel special, but I hope for it. His comment last night was so insensitive and selfish, and here I am constantly reminding myself that I shouldn't expect to be treated like a princess just because my friends are and I certainly have no right to demand to be treated more special.
I don't know how to frame the topic anymore. I'm not going to end the relationship over it, but I'm starting to feel like maybe I should stop doing special things for him because he seems to expect and feel like he's earned those things without reciprocating. It's hurting my feelings. I could explain my feelings until I'm blue in the face and tell him exactly what I want from him and he still wouldn't do it. I need an outside perspective and some advice.
---
**tl;dr**: Boyfriend doesn't put effort into special occasions. I've explained my feelings multiple times but he's too lazy to change anything. Expects an elaborate thank you for his sub par efforts.
Edit: Oops. I'm actually 23. Like I said, yesterday was my birthday.
---
Update: I had no idea my post would reach so many people, but you guys have offered me some great advice and a few different perspective which was what I needed. Thank you. I talked to the boyfriend a little last night, but I want to finish getting through this thread and reply as much as I can before I finalize what I want to say to him.
---
Update #2: This is just an update to let you guys know that I haven't abandoned the thread...my laptop charger decided to stop working so computer access is currently limited. I don't have time to do a full update now, but I will get back to you guys within the week. | 0non-cybersec
| Reddit |
How do I subtract times?. <p>I have an embarrassingly basic modular arithmetic question. I understand that I can subtract, for example, 1 hour from 10 o'clock to get 9 o'clock, or even 2 hours from 1 o'clock to get 11 o'clock; but how do I subtract 2 <em>o'clock</em> from 11 o'clock to get 3 <em>hours</em>. </p>
<p>That is, I want a function that takes two times on the face of a clock, and gives me the interval between them.</p>
<hr>
<p>For example: I have two angular values measured in "hours". One value is the <a href="http://www.physics.csbsju.edu/astro/CS/CS.10.html" rel="nofollow">right ascension</a> of the apparent sun, $\alpha(t)$, while the other is the right ascension of the "<a href="http://www.thefreedictionary.com/mean+sun" rel="nofollow">mean sun</a>", $\langle\alpha\rangle(t)$; both are mod 24. I'm interested in <a href="http://en.wikipedia.org/wiki/Equation_of_time" rel="nofollow">the angular difference between these two values</a>
$$E(t) = \langle\alpha\rangle(t) - \alpha(t)$$
Is this just inherently ambiguous, so that I need to impose some additional constraint based on information about the system (e.g., here that the values can be positive or negative, and are always small, so that, say $\pm 22$ should be interpreted as $\mp 2$), or is there some simple systematic way to ensure that I get the correct values?</p>
| 0non-cybersec
| Stackexchange |
A bit of an embarrassing question. I'm a Korean student currently living in America for college so I didn't really know who to ask since my mom would think I was literally the devil if I brought this up. I've recently tried masturbating out of curiosity mainly, and every time I get close to what I think might be an orgasm I end up peeing a little. It's extremely embarrassing and completely kills the moment, so I've sorta just given up afterwards each time. Is this normal? If not, is there anything I can do about it? Thanks.
Sorry if this isn't the right place to post this, but I didn't really know where to ask. If this isn't the right place can someone please tell me a better subreddit to ask this question? Thanks. | 0non-cybersec
| Reddit |
[Image] Just do your best. | 0non-cybersec
| Reddit |
Audio to multiple HDMI sinks. <p>I have 2 HDMI monitors connected to my laptop, and both of them display just fine. However, I want to play audio on both of them at the same time as well. I saw this post: <a href="https://unix.stackexchange.com/questions/401492/how-to-output-unique-audio-to-multiple-ports-profiles-of-a-pulseaudio-card/401562#401562?newreg=a2c8d4c42a4944e1935687f0f0251994">How to output unique audio to multiple ports/profiles of a PulseAudio card?</a></p>
<p>I have attempted the solution described in the answer, of editing the pulseaudio confs to create a new profile. However, when I tried setting it, it said no such profile:</p>
<pre><code>No such profile: output:analog-stereo+output:hdmi-stereo+output:hdmi-stereo+output:hdmi-stereo
</code></pre>
<p>I was wondering if anyone else had a similar experience.</p>
| 0non-cybersec
| Stackexchange |
Sometimes Android allows for a bit too much customization.... | 0non-cybersec
| Reddit |
Relative performance of 2006 Mac Pro and 2007 iMac. <p>I'm trying to evaluate a purchase but am having trouble finding apples-to-apples performance comparisons for these two machines. A year and and many details separate them. Are there any common benchmarks that consider disk and graphics performance? Which is preferable for software development (dual-head, but otherwise light on the GPU)?</p>
<p><strong>Mac Pro (2006)</strong> <a href="http://www.everymac.com/systems/apple/mac_pro/stats/mac-pro-quad-2.66-specs.html" rel="nofollow">[Detailed specs]</a></p>
<ul>
<li>Two dual-core Xeon processors at 2.66GHz (1.33GHz bus)</li>
<li>3GB RAM</li>
<li>NVIDIA GeForce 7300 GT with 256 MB RAM</li>
</ul>
<p><strong>iMac (2007)</strong> [<a href="http://www.everymac.com/systems/apple/imac/stats/imac-core-2-duo-2.4-24-inch-aluminum-specs.html" rel="nofollow">Detailed specs</a>]</p>
<ul>
<li>Core 2 Duo (obviously dual-core) at 2.4GHz (800MHz bus)</li>
<li>3GB RAM</li>
<li>ATI Radeon 2600 with 256 MB RAM</li>
</ul>
<p>Both machines have their original hard drives (SATA).</p>
<p>Thanks in advance for your input.</p>
| 0non-cybersec
| Stackexchange |
Moon Moon doesn't care much for water.. | 0non-cybersec
| Reddit |
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