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 $\mathbb{E}(\lim X_n)=0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.