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#1
05-09-2008, 01:52 AM
 Doc Holiday Member Join Date: Aug 2006 Location: Tombstone Favorite beer: Anything that defiles myself Posts: 1,004
SOA 165

I dont get how E[X ^ 3] is calculated. Some help please?
Is this easy and I'm just burnt out, or am I screwed?
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#2
05-09-2008, 02:12 AM
 ElDucky Free Mason Join Date: Jul 2004 Location: In a van, down by the river Studying for Let me worry about blank Favorite beer: Trappistes Rochefort 8 Posts: 40,809

There's another one later on that is similar. I don't get it either. I didn't see much about in the manual but I am having another look.
#3
05-09-2008, 03:21 AM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,304

$X \wedge 3 = 3 - (3 - X)_+$ (just draw graph of y = 3 and y = x ^ 3)

But $E[3] = 3$, and since $(3 - X)_+ = 0$ for $X \geq 3$ we can compute the expectation as
$E[(3-X)_+] = (3-0) f_S(0) + (3-1) f_S(1) + (3-2) f_S(2)$

which is one way of seeing that
$E[X \wedge 3] = 3 - [3 f_S(0) + 2 f_S(1) + 1 f_S(2)]$

[Conceptually, the only difference between X ^ 3 and 3 occurs when x = 0, 1 or 2 and so you just need to subtract off the right stuff to account for that.]
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#4
05-09-2008, 10:14 AM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,607

It's easy to compute E[S ^ 3] directly. The possible values of S are 0, 1, 2, 3, 4, 5, ... and so the corresponding possible values of S ^ 3 are 0, 1, 2, 3, 3, 3, ... . So

$E[S\wedge 3]=f_S(0)0+f_S(1)1+f_S(2)2+[1-f_S(0)-f_S(1)-f_S(2)]3$

Jim Daniel
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#5
05-09-2008, 12:33 PM
 Doc Holiday Member Join Date: Aug 2006 Location: Tombstone Favorite beer: Anything that defiles myself Posts: 1,004

Quote:
 Originally Posted by Jim Daniel It's easy to compute E[S ^ 3] directly. The possible values of S are 0, 1, 2, 3, 4, 5, ... and so the corresponding possible values of S ^ 3 are 0, 1, 2, 3, 3, 3, ... . So $E[S\wedge 3]=f_S(0)0+f_S(1)1+f_S(2)2+[1-f_S(0)-f_S(1)-f_S(2)]3$ Jim Daniel
This makes sense. Thanks!
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#6
10-01-2008, 08:10 AM
 bmathew22 Member Join Date: Jan 2006 Posts: 126

Ok, so now I understand how to do this calculation once I have computed the fs', but how were the fs' computed?
#7
10-01-2008, 08:48 AM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,607

Just examine how S can eual ech of those values, and then compute the probability for each case.

Jim Daniel
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#8
10-01-2008, 05:31 PM
 bmathew22 Member Join Date: Jan 2006 Posts: 126

Ok, so fs(2) would be one loss of 2 or 2 losses of 1. Makes sense, thanks!
#9
09-08-2016, 11:11 PM
 umich Member SOA AAA Join Date: Mar 2016 Location: Detroit, MI College: University of Michigan Alum Posts: 142

Quote:
 Originally Posted by Jim Daniel It's easy to compute E[S ^ 3] directly. The possible values of S are 0, 1, 2, 3, 4, 5, ... and so the corresponding possible values of S ^ 3 are 0, 1, 2, 3, 3, 3, ... . So $E[S\wedge 3]=f_S(0)0+f_S(1)1+f_S(2)2+[1-f_S(0)-f_S(1)-f_S(2)]3$ Jim Daniel
Jim, I'm having trouble deciding whether to put condition to the expectation or not. I was able to calculate E(S-3)+ but I then divided it by Pr (S>3). Sometimes I can get it right but other times I use the wrong one. So why is it wrong to use the conditional expectation? Thank you!
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#10
09-09-2016, 01:07 AM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,607

Quote:
 Originally Posted by umich Jim, I'm having trouble deciding whether to put condition to the expectation or not. I was able to calculate E(S-3)+ but I then divided it by Pr (S>3). Sometimes I can get it right but other times I use the wrong one. So why is it wrong to use the conditional expectation? Thank you!
I don't understand why you would even THINK about conditioning on something. Does the expression E[X ^ 3] or E[(X - 3)+] have a "conditioned on" symbol | ? I certainly don't see one. You ask why is it wrong to use conditional expectation. I instead ask where you see anything that indicates that you SHOULD use conditional expectation. E[X ^ 3] simply asks you to compute the average value of the smaller of X and 3. No conditioning, none, nada, niet, nyet, nee.
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