# Prove that if $X_n \xrightarrow{P} c$, then $E(X_n) \to c$ for $X_n$ uniformly bounded

I have been trying to prove that for a random variable that is uniformly bounded, i.e. $|X_n - c| <M$ for all $n$, convergence in probability to $c$ implies that

$$E\left(X_n \right) \to c$$

as well. This is quite intuitive as all probability mass is bounded away from infinity in this case, but I am having a hard time formalizing this notion. Could you please give me some hints? Please keep it simple, though. Thank you.

A necessary and sufficient condition for $L^1$ convergence is $X_n\xrightarrow{P} X$ and the sequence $(X_n)$ is uniformly integrable.

So, it remains to show that $X_n$ is uniformly integrable (or equivalently that $Y_n=X_n-c$ is uniformly integrable). But it is straightforward that $$E[|X_n-c|\mathbf 1_{|X_n-c|\ge M}]=0\le ε$$ for any $ε>0$. So $E[X_n-c]\to 0$.

Edit: Since $|X_n-c|$ is a non-negative random variable you have that $$E|X_n-c|=\int_{0}^{+\infty}P(|X_n-c|\ge x)dx=\int_{0}^{M}P(|X_n-c|\ge x)dx$$ where the second equality is implied by $|X_n-c|<M$. So you can bound the integration limits on the RHS at $M$ (away from infinity) and hence you can take limits and interchange the order of limit and integration (I will do it with Fatou and $\le$ but you can do it directly, since you have convergence and bounded region of integration. Fatou works also with unbounded limits.): \begin{align}\limsup_{n\to+\infty} E|X_n-c|&=\limsup_{n\to+\infty}\int_{0}^{M}P(|X_n-c|\ge x)dx\\& \le \int_{0}^{M}\limsup_{n\to+\infty}P(|X_n-c|\ge x)dx=0\end{align} because $\limsup P(|X_n-c|\ge x)=\lim P(|X_n-c|\ge x)=0$ due to convergence in probability.

• Thank you but we haven't talked about uniform integrability yet, is there a more elementary proof? Feb 4 '16 at 14:01
• I tried. See my edit. Feb 4 '16 at 14:27
• Thanks, that is better. The fact that $E|X_n - c | \to 0$, implies $E(X_n) \to c$, right? Feb 4 '16 at 14:29
• Yes, right. Actually $E|X_n-c|\to 0$ implies $E|X_n|\to c$ also. Feb 4 '16 at 14:36
• Beautiful answer, thanks again. Feb 4 '16 at 14:37

Start with $$E|X_n-c|\le M\cdot P[|X_n-c|>\epsilon]+\epsilon\cdot P[|X_n-c|\le\epsilon]\le M\cdot P[|X_n-c|>\epsilon]+\epsilon.$$ Because $X_n$ converges in probability to $c$, $\lim_n P[|X_n-c|>\epsilon]=0$ for each $\epsilon>0$. It follows from the inequality displayed above that $$\limsup_nE|X_n-c|\le\epsilon,$$ for each $\epsilon>0$.