1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thank you but we haven't talked about uniform integrability yet, is there a more elementary proof? $\endgroup$
    – JohnK
    Feb 4 '16 at 14:01
  • $\begingroup$ I tried. See my edit. $\endgroup$
    – Jimmy R.
    Feb 4 '16 at 14:27
  • $\begingroup$ Thanks, that is better. The fact that $E|X_n - c | \to 0$, implies $E(X_n) \to c$, right? $\endgroup$
    – JohnK
    Feb 4 '16 at 14:29
  • $\begingroup$ Yes, right. Actually $E|X_n-c|\to 0$ implies $E|X_n|\to c$ also. $\endgroup$
    – Jimmy R.
    Feb 4 '16 at 14:36
  • 1
    $\begingroup$ Beautiful answer, thanks again. $\endgroup$
    – JohnK
    Feb 4 '16 at 14:37
2
$\begingroup$

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 $\epsilon>0$.

$\endgroup$
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.