Convergence of Expectations

Question

Suppose $\{X_n\}$ is a sequence of non-negative random variables such that

$$EX_n<\infty, \text{ }\lim_{n\rightarrow \infty}EX_n =\infty$$

and $\lim_{n\rightarrow \infty}X_n$ exists a.s. May I argue that $E[\lim_{n\rightarrow \infty}X_n]=\infty$?

Thank's!

Answer 1

Nope, take $U_n$ a i.i.d. sequence of uniform random variables on $[0,1]$ and

$$X_n = n^3 1_{U_n \leq \frac{1}{n^2}}$$

It's easy to see $EX_n = n \to +\infty$

Since $\sum_n P(U_n \leq \frac{1}{n^2}) = \sum_n \frac{1}{n^2} < +\infty$, by Borel-Cantelli lemma, we know that almost surely, there are only finite many $U_n$ such that $U_n \leq \frac{1}{n^2}$, which means $$\lim X_n = 0\text{ almost surely}$$

so $E(\lim X_n) = E(0) = 0$

Answer 2

Let $P$ be a probability on $\left(\mathbb{N},\wp\left(\mathbb{N}\right)\right)$ and let $f:\mathbb{N}\rightarrow\mathbb{R}$ be a nonnegative function such that $f\left(n\right)p_{n}\rightarrow\infty$. Here $p_{n}$ denotes $P\left(\left\{ n\right\} \right)$.

You could choose for $p_{n}>0$ and $f\left(n\right)=np_{n}^{-1}$.

Prescribe $X_{n}:\mathbb{N}\rightarrow\mathbb{R}$ by $k\mapsto f(n)$ if $k=n$ and $k\mapsto0$ otherwise.

Then $\mathbb{E}X_{n}=f\left(n\right)p_{n}\rightarrow\infty$ and $X_{n}\rightarrow0$ a.s.