# Expected value of a function of a random variable: help!

I am trying to show the following:

\begin{equation*} E[e^{-\gamma W}]=e^{-\gamma(E[W]-\frac{\gamma}{2}Var [W])} \end{equation*}

but I really can't remember what I am supposed to do to get from the LHS to the RHS. I have tried using integration this way

\begin{equation*} \int We^{-\gamma W}dW \end{equation*}

and then use integration by parts, but even though what I get resembles it, it can't be correct (because $e^{-\gamma W}$ is not the distribution of W).

I have also tried using Taylor series expansion, but I think I am way off, and I don't think an approximation here is what I need, because the equality above is exact.

FYI, this is not homework, I am working through a paper (page 10) and I would really like to know how every step was derived.

Can anyone at least point me to the right direction?

EDIT: This expectation on the RHS is very similar to the moment generating function formula (with a negative exponent). If you check here, you will see that the moment generating function for the normal distribution is like the LHS (but with a positive sign). So in a way I have my answer, but I still would like to know how to derive it, if there is a way. I know little if anything at all about moment generating functions, so maybe I shouldn't try and derive it but rather just use the result? Does it even make sense to try and derive it?

• @Isaac, have you read this: meta.math.stackexchange.com/questions/256/what-are-the-aa-tags
– Vivi
Commented Jul 27, 2010 at 9:17
• Yes and I'm inclined to agree with it (especially since I know nothing about the arXiv codes), but "pr.probability-theory" is on 12 other questions and "probability-theory" is not on any question. Since it was said there that a moderator would be easily able to globally change a tag, I figured it was better for now to use the form of the tag that would more easily bring up related questions. Commented Jul 27, 2010 at 9:30
• @Isaac: fair enough!
– Vivi
Commented Jul 27, 2010 at 9:31
• I looked at the paper, and I can't tell what the pdf associated with the expected value is. As KennyTM pointed out that is key to being able to rewrite things. You might want contact the authors. Commented Jul 27, 2010 at 18:12
• @Jonathan Fischoff: Oh, that I figured out. W3 = w2 + (p3 - p2)(y +z), but p3=v3, and v3 = v2 + sigma*e, where e is standard normal (see page 2208 for relevant equations and statement of distribution). So I know that in the end W3 will have a normal distribution because it is a linear function of a normally distributed variable.
– Vivi
Commented Jul 27, 2010 at 23:58

If W is randomly chosen with the PDF P(x), then the expectation value should be

$E[e^{-\gamma W}]=\int_{-\infty}^\infty P(x) e^{-\gamma x} dx$ http://mathcache.appspot.com/?tex=%5cpng%5c%5bE%5Be%5E%7B-%5Cgamma%20W%7D%5D%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20P%28x%29%20e%5E%7B-%5Cgamma%20x%7D%20dx%5c%5d

And I think that equation (E[e-γW] = e-γ(E[W] - ½γVar[W])) is correct only when W is a normal distribution.

• I don't see how you would still get X in the end if you are integrating wrt x, but I will try and solve and get back to you. I can tell you that W is not itself the RV, but it is a function of an RV which is normally distributed, so this may be right. I am going home now, when I get there I will try, if I can't I will keep trying tomorrow (I am in Australia, and it is night here). Thanks for you help, hey?
– Vivi
Commented Jul 27, 2010 at 10:19
• @Vivi: You shouldn't be getting W in the end, you should be getting E[W] and Var[W], both of which are expressions where W (or x) was integrated out. Commented Jul 27, 2010 at 10:44
• @Kaestur Hakari: yeah, but when I take the limit as x goes to plus or minus infinity, everything goes. I was unsuccessful. It is possible I am making a mistake (more than possible), so I will give it another go tomorrow, but for now, nothing.
– Vivi
Commented Jul 27, 2010 at 11:51