How to prove expectation exists (or improper integral converges)

Question

How can I prove this improper integral converges, or give a counterexample?

$$\int_{-\infty}^{\infty}x^n p(x)dx$$ where the only thing we know about $p(x)$ is $$\int_{-\infty}^{\infty}p(x)dx = 1 $$

In other words, does $\mathbb{E}X^n$ always exists?

Answer 1

Let $p$ be the function $$p(x)=\begin{cases} 1,&\exists n\in\mathbb N: x\in[2^n,2^n+2^{-n}]\\ 0,&\text{else.} \end{cases}$$

Then we have $\int_{\mathbb R}p(x)dx=1$ but $$\int_{\mathbb R}xp(x)dx = \sum_{n=1}^\infty \int_{2^n}^{2^n+2^{-n}}xp(x)dx\ge \sum_{n=1}^\infty \int_{2^n}^{2^n+2^{-n}} 2^n dx = \sum_{n=1}^\infty 1=\infty.$$

Answer 2

The answer is negative. For any $n\in\mathbb{N}_{>0}$, you may consider: $$ p(x) = \frac{1}{C_n}\cdot \frac{1}{1+|x|^{n+2}} $$ where: $$ C_n = \int_{-\infty}^{+\infty}\frac{dx}{1+|x|^{n+2}}. $$ Then $p(x)$ is a positive function with unit integral, but $$ \int_{-\infty}^{+\infty} x^m\,p(x)\,dx $$ is finite iff $m\leq n$.