How do you calculate the expected value of $e^{-X}$? If $X$ is a random variable, I would like to be able to calculate something like $$E(e^{-X})$$
How can I do this? Thank you so much.
 A: As pointed out in the comments, your specific question can be solved by evaluating the moment generating function and $t=-1$, but it appears you may be asking the more general question of how to calculate the expected value of a function of a random variable.
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. In your 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*}
Discrete example: $X \sim {\rm Bernoulli}(p)$. Then
\begin{align*}
E \left( \exp(-X) \right) &= \sum_{i=0}^{1} e^{-i} P(X=i) \\ 
&= (1-p)e^0 + pe^{-1} \\ 
&= (1-p) + p/e 
\end{align*}
