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