How to find $E[e^{\lambda X}]$ for a random variable $X$?

Question

Let $X$ be a random variable, how can we find the expectation $E[e^{\lambda X}]$?

$e^{\lambda X}$ looks like a pdf of exponential distribution, but I have never seen how to find the expectation of a pdf? All I know is the expectation of a random variable.

If $e^{\lambda X}$ is not a pdf but a random variable instead, how can we find the expectation of an "exponential random variable"?

Thanks in advance for the help.

Answer 1

If is $f$ is the pdf of $X$ then $\mathbb{E}[e^{\lambda X}] = \int e^{\lambda x}f(x) dx$.

Answer 2

Don't be confused by the exponentiel, this is not the pdf of an exponential random variable. This is indeed the expectation of the random variable $Z=e^{\lambda X}$.

So if $f_X$ is the pdf of $X$, then $$E(Z)=E(e^{\lambda X})=\int_\mathbb{R}e^{\lambda x}f_X(x)dx$$

More generally, for a (measurable) function $h$, $$E(h(X))=\int_\mathbb{R}h(x)f_X(x)dx$$ (if this expectation exists)

Answer 3

Assuming you have a 'nice' pdf, you can take the Taylor expansion of the random variable, i.e.

$$E[e^{\lambda X}]= E\left[\sum_{n=0}^{\infty}\frac{\left(\lambda X\right)^n}{n!}\right] = \sum_{n=0}^{\infty}\frac{\lambda^n}{n!}E\left[X^n\right].$$

The first step is just the series, while the second step follows from linearity of the expectation value. If you have a closed form of $E\left[X^n\right]$, then you are set. Note that this method can be generalized to other functions as well!

Answer 4

The expectation of a function $g$ of a continuous random variable $X$ is $$E[g(X)] = \int g(x) f_X(x)\, \text dx$$ where $f_X(x)$ is the pdf of $X$. In the simplest case you have $g(x) = x$ and $$E[X] = \int x f_X(x)\, \text dx.$$ In your case $g(x) = e^{\lambda x}$ and hence $$E[e^{\lambda X}] = \int e^{\lambda x} f_X(x)\, \text dx.$$ What you are calculating here is the moment generating function (MGF) of the random variable $X$.