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Hi I have been struggling to find the variance of the $\exp(-x)$ in terms of $\exp$.

For the function Y = exp (-x) where X is N (0,1) show that the variance of Y = $\exp(\exp-1)$

This is what I think need to be done: f(x) =$exp(-x)$ where X~N(0,1). work out the expectation (mean say).

specific question can be solved by evaluating the [moment generating function][1] and $t=-1$,

In general, if $X$ has density function $p$, then

$$ E \left( f(X) \right) = \int_{D} f(x) p(x) dx $$

where $D$ denotes the support of the random variable. For discrete random variables, the corresponding expectation is

$$ E \left( f(X) \right) = \sum_{x \in D} f(x) P(X=x) $$

These identities follow from the [definition of expected value][2]. In my example $f(X) = \exp(-X)$, so you would just plug that into the definition above.

Continuous example: Suppose $X \sim N(0,1)$, then

\begin{align*} E \left (\exp(-X) \right) &= \int_{-\infty}^{\infty} e^{-x} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-(x^2 + 2x)/2} dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-(x^2 + 2x + 1)/2} e^{1/2} dx \\ &= e^{1/2} \int_{-\infty}^{\infty} \underbrace{\frac{1}{\sqrt{2 \pi}} e^{-(x+1)^2/2}}_{{\rm density \ of \ a \ N(-1,1)}} dx \\ &= e^{1/2} \end{align*}

I am now struggling with the variance part? I think I need to Mult this with the pdf of the standard normal, and integrate over –inf and +inf. complete the square on the exponent.

Is this correct, how will I work out the variance?

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Your question is confusing. I would remove the first paragraph, and modify the last question to define $Y = e^{-X}$, and find the variance of $Y$. Is this what you mean? –  Stefan Smith Apr 13 '13 at 16:48
    
Yes, sorry for the confusion, your interpretation is absolutely correct –  sleepybob Apr 13 '13 at 17:03
    
Please, would you be able to help, any advice would be of immense help. –  sleepybob Apr 13 '13 at 17:56
    
I thought the variance of $e^{ax}$ was $\frac{1}{a^2}$. –  Alyosha Apr 13 '13 at 19:58
    
you are right, I have edited the question to clarify, any help would be amazing –  sleepybob Apr 13 '13 at 20:03
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1 Answer

up vote 1 down vote accepted

Hint: $$ \mathbf{Var}(Y)=\mathbf{E}(Y^2)-\mathbf{E}(Y)^2 $$

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Hurray! got it..thanks! –  sleepybob Apr 13 '13 at 23:33
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