Suppose $X$ to be a real random variable with $E(X)=0$ and $E(X^2)=1$.
How can I prove:
$\lim_{n\to\infty}E(X^2\mathbb{I}_{\{X^2\ge n\}})=0$
I am thinking, with no success, about these results:
- monotone convergence for decreasing functions
- integral over a set of measure zero is zero for a non negative function
Thanks for the help.
As discussed in the comments, we know that since $E[X^2] < \infty$ we must have $X^2 < \infty$ a.s. This implies $\lim_{n \rightarrow \infty} X^2(\omega) 1_{X^2(\omega) \ge n} = 0$ for a.e. $\omega$. Since $X^2 1_{X^2 \ge n}$ is a decreasing sequence of random variables bounded from below by $0$, by the monotone convergence theorem we have $$\lim_{n \rightarrow \infty} E[X^2 1_{X^2 \ge n}] = E[\lim_{n \rightarrow \infty}X^2 1_{X^2 \ge n}] = 0. $$