Show $\lim_{N \to \infty} \mathbb{E} [ X_N \mathbb{I}\{ X_N > b\}] = 0$.

Question

Assume that a sequence of positive random variables $X_N$ converges almost surely to some constant $a$. Moreover, their expectation also converges to $a$.

$\lim_{N \to \infty} X_N = a \hspace{3mm} \text{a.s.}, \hspace{3mm} \lim_{N \to \infty} \mathbb{E}[X_N] = a.$

Is it true that $\lim_{N \to \infty} \mathbb{E} \big[ X_N \mathbb{I}\{ X_N > b\} \big] = 0$ for a constant $b > a$?

Can we replace the almost sure convergence with the convergence in probability?

Answer 1

No. Consider the variables $X_N$ that takes value $2^N$ with probability $2^{-N}$, $-2^N$ with probability $2^{-N}$, and $0$ otherwise. Then

1. $\sum_N P(X_N \neq 0) < \infty$, so $X_N \to 0$ almost surely.

2. For any $b > 0$, we have $E(X_N1_{X_N > b}) = 1$ for sufficiently large $N$.

But if $X_N$ is non-negative then the proposition is true. For if we let $Y_N = X_N 1_{X_N \leq b}$, then $Y_N \to a$ almost surely, thus by bounded convergence we have $$\lim_{N \to \infty} E(Y_N) = a.$$ So $$\lim_{N \to \infty} E(X_N - Y_N) = 0$$ as desired.