Measure theory expectation of $X\sim\exp(1)$

Question

I am studying measure theory expectation and I encountered the following problem.

Let $X\sim\exp(1)$. Calculate $EX^n$ for every $n\in \mathbb{N}$.

We know that the density function is $\exp(-x)$ when $x\ge0$ and zero otherwise. I am having a hard time understanding measure theory expectation and how I am supposed to take it into account in this problem. Shouldn't the expectation be

$$\int_\Omega x^n\exp(-x)dx$$

but to be honest, I don't know how to go on from here either.

Answer 1

The probability density function is $$f(x)=\begin{cases} e^{-x} \space\space \space \space \space x\geq0, \\0\space\space\space\space\space\space\space\space\space x<0\end{cases}$$

as you correctly mentioned. Now to compute the required expectation we use the Gamma function:

$$\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}dx,\space\space\space \Re(z)>0.$$

Moreover for any positive integer $z$ we have $\Gamma(z)=(z-1)!.$

Thus the required expectation is given by $$\mathbb E(X^{n})=\int_{-\infty}^{\infty}x^nf(x)dx=\int_{0}^{\infty}x^ne^{-x}dx$$ $$=\int_{0}^{\infty}x^{(n+1)-1}e^{-x}dx=\Gamma(n+1)=n! $$

since $n\in\mathbb N$.

Answer 2

The actual integral is $$ \int_0^{\infty} x^n \exp(-x) \text{d} x $$ which can be evaluated using integration by parts.

Answer 3

Let's think about the case $n=1$.

By definition:

$$ EX=\int_\Omega X(\omega)\ dP $$ where $(\Omega,P)$ is the underlying probability space. This is what you call the "measure theory expectation".

On the other hand, since you know the density of $X$, you have $$ EX=\int_\Omega X(\omega)\ dP=\int_{\mathbb{R}}xf(x)\ dx\tag{1} $$ where $f$ is the density function $f$. The integral in (1) is nothing but an (improper) Riemann integral.

---

In general, we have the law of the unconscious statistician.

Answer 4

Generally if $X$ has a pdf $p$ then $$\mathbb E(f(X)) = \int_{\Omega} f(X)d\mathbb P = \int_\mathbb R f(x) p(x) dx. $$ For instance, $\mathbb P(X<t) = \mathbb E\mathbf 1_{\{X<t\}} = \int_{-\infty}^t p(x) dx $ which is a formula that should feel familliar.

As you mentioned, in your case, $p(x) = \mathbf 1_{[0,\infty)} \exp (-x)$; plug it in, mumble something about Lebesgue integral = improper Riemann integral for continuous functions with lots of decay, and do the integral. The final answer should be $n!$.