# Help with an exponential distribution problem

• Nov 7th 2011, 07:09 PM
tayman2037
Help with an exponential distribution problem
I've been trying at this problem for the last hour, and I'm not sure what I'm doing wrong. Here is the question:

Suppose a research paper states that the distribution of the daily sea-ice advance/retreat from each sensor is similar and is approximately double exponential. The proposed double exponential distribution has density function f(x) = 0.5λe^(−λ|x|)
for −http://www.webassign.net/images/infinity.gif < x < http://www.webassign.net/images/infinity.gif. The standard deviation is given as 42.1 km. (Round your answers to four decimal places.)

I need to find the value of the parameter (λ) . I tried doing 1/42.1, (since the standard deviation equals 1/λ, but that didn't get me the right answer. Now I'm trying to integrate xf(x) to find the expected value, but I get infinity in my answer. Any help would be greatly appreciated.

(EDIT) I'm not getting infinity anymore when I work out my integrals, but I still end up getting E(X)=standard deviation=1/
λ. And like I said, 1/42.1 does not give me the right answer. I can practice another version of the problem where the standard deviation is 38.8, and the answer is .0364. Hope this helps some people.

(EDIT 2) I'm beginning to think my online homework program is wrong. I mean, I've worked so hard on this problem, and tried finding the variance by the definition (E(x^2)-(E(X))^2) but that just brings me right back to 1/
λ^2 for the variance. I'm going to talk to my professor tomorrow, but if anyone else can spot out something, that would be great.
• Nov 9th 2011, 03:48 AM
SpringFan25
Re: Help with an exponential distribution problem
My notation:
Y: a "regular" exponential variable with the same parameter (pdf = $\lambda e^{-\lambda y}$)

the pdf of X is symmetric about 0 so E(X) = 0

because of symmetry E(X^2) will be the same as for a regular exponential variable (check you can prove / understand this, your teacher will prboably expect you to show the integral rather than jsut saying "symmetry").

so we have $E(X^2) = E(Y^2)$ (RESULT 1)

The mean and variance of Y are well known, so E(Y^2) can be computed using:
$Var(Y) = E(Y^2) - E^2(Y)$
substitute the known mean and variance of Y:
$\lambda^{-2} = E(Y^2) - \lambda^{-2}$
$2\lambda^{-2} = E(Y^2)$

using this with result 1, we now have:
$E(X^2) = E(Y^2) = \frac{2}{\lambda^2}$

Now compute the variance of X
As always: $Var(X) = E(X^2) - E^2(X)$

Substitute known values:
$Var(X) = \frac{2}{\lambda^2} - 0$

so the standard deviation (sd) satisfies:

$sd(x) = \sqrt{Var(x)} = \sqrt{\frac{2}{\lambda}}$

Re-arrange for $\lambda$:
$\lambda = \frac{\sqrt{2}}{sd(x)}$

substituting numbers in gives a parameter value of .364 if the standard deviation is 38.8, as per your example.

Post if you need a bit more explanation. If you are not comfortable using symmettry or referring to the properties of variable Y, it should be possible to do these integrals directly (but you will have to break them up, treating positive and negative values of X seperately).
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