Convergence in distribution to derive the expectation convergence

Question

If $X_n\longrightarrow X$ in distribution, $\mathbb{E}(X)\lt\infty$, Do we have the following conclusion: $\mathbb{E}(X_n)\longrightarrow\mathbb{E}(X)$?

Answer 1

We choose $X_n=2^{2n}\mathbb 1_{I_{n,1}}$, where $I_{n,k}$, $1 \leq k \leq 2^n$ denote the k-th half open dyadic subinterval of $[0,1)$ of order $n$.

Then for every $\omega\in (0,1)$ there is $m\in \mathbb N$ such that $\omega \notin \mathbb I_{n,1}$ if $n\geq m$. Thus

$$\lambda(\{\omega:\lim_{n\rightarrow \infty}X_n(\omega)=0\})=1$$

i.e. $X_n\rightarrow 0$ a.s. but $\mathbb E(X_n)=2^{2n}2^{-n}=2^n$

So we found a sequence of random variables with converges almost surely (hence also in distribution) but does not converge in $L^1$

Answer 2

No: the problem is that weak convergence does not prevent $X_n$ to take large values on small sets. For example, if we choose $X_n$ such that $\mathbb P(X_n=0)=1-1/n$ and $\mathbb P(X_n=n)=1/n$, then $X_n\to 0$ in distribution but $\mathbb E[X_n]=1$ for each $n$.

However, if the sequence $(X_n)_{n\geqslant 1}$ is uniformly integrable, then the convergence $\mathbb E[X_n]\to \mathbb E[X]$ takes place.