Does anyone know how to prove $\lim E[X(n)]=E[\lim X(n)]$???
Here I need to prove $\lim E[X(n)]\le E[\lim X(n)]$ and $\lim E[X(n)]\ge E[\lim X(n)]$.
Based on "Fatou Lemma", I can get that $E[\liminf X(n)]\le \liminf E[X(n)]$ and $\limsup E[X(n)]\ge E[\limsup X(n)]$.
Since we have $\liminf E[X(n)] \le \lim E[X(n)]$ and $\lim E[X(n)] \le \limsup E[X(n)]$, then we get $$ E[\liminf X(n)]\le\lim E[Xn] \le E[\limsup X(n)]. $$ Here I am stuck: if I can prove that $E[\liminf X(n)]= E[\limsup X(n)]$, it's done!! But I don't know how to prove it, does anyone know that??
$\lim \mathbb{E}(X_n)$ is not equal to $\mathbb{E}(\lim X_n)$ in general. Take
$$X_n=\begin{cases}n\hspace{1cm}\text{with probability }\dfrac{1}{n}\\0\hspace{1cm}\text{with probability }1-\dfrac{1}{n}.\end{cases}$$
Then $X_n\stackrel{p}{\to}0,$ but $\mathbb{E}(X_n)=1$ for all $n.$ That is, $\lim \mathbb{E}(X_n)=1$ but $\mathbb{E}(\lim X_n)=0.$