If $\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N X_i = X$ a.s. with $X$ constant, $\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N \mathbb{E} X_i = X$?

Question

Let $(X_i)$ be a sequence of (possibly dependent) random variables in $\mathbf{R}_+$ such that $\mathbb{E}[X_i]<\infty$ and $\mathbb{E}[X_i^2]<\infty$. Assume that almost surely $$ \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N X_i = X $$ where $X$ is a (finite) deterministic number. I'm trying to understand if I can conclude that $$ \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N \mathbb{E} X_i = X. $$ And in case this is not true, which other conditions am I missing.

Answer 1

This is in general incorrect.

Let the $X_i$ be independent with $P(X_i = i^2) = i^{-2}$ and $P(X_i = 0) = 1 - i^{-2}$.

By the Borel-Cantelli lemma, we have $P(X_i = i^2 \text{ i.o.}) = 0$, from which we can conclude that $\frac{1}{n} \sum \limits_{i = 1}^n X_i$ converges to $0$ almost surely. But $E[X_i] = 1$, which means $\frac{1}{n} \sum \limits_{i = 1}^n E[X_i] = 1 \not\to 0$.