1
$\begingroup$

Suppose $X_n \rightarrow X$ a.s. as $n \rightarrow \infty$ and that $X_n \geq0$ for all $n, \omega$. Furthermore, $\{X_1,X_2,...\}$ is uniformly integrable. Prove that $\mathbb{E}X_n \rightarrow \mathbb{E}X$ as $n \rightarrow \infty$.

I know that a sequence of random variables converges almost surely to $X$ if for every $\epsilon > 0$ we have \begin{equation} P(\lim_{n \rightarrow \infty}|X_n - X|< \epsilon) = 1 \end{equation}

But I'm not sure how to proceed. Some hints would be greatly appreciated.

$\endgroup$
2
  • $\begingroup$ Both answers below are repeating the proof of Vitali's Theorem, which is worth looking into/knowing (and applies directly to your problem). $\endgroup$ Mar 2 '21 at 23:56
  • $\begingroup$ True but, to be fair, this question is more-or-less asking to prove Vitali's theorem, so just stating this holds by Vitali's theorem wouldn't be very helpful. Although I agree, OP should read a proof of Vitali's theorem, as it is more general than what OP asked. $\endgroup$
    – nullUser
    Mar 3 '21 at 17:15
1
$\begingroup$

You can always write

$$ \mathbb{E}\big[ \vert X_n-X \vert \big]= \int_F\vert X_n-X\vert dP+ \int_{F^c}\vert X_n-X\vert dP\leq \int_{F} \vert X_n\vert dP+ \int_F \vert X\vert dP+ \int_{F^c}\vert X_n-X\vert dP$$

Since $X$ is integrable and $\{X_n\}$ is unifromly integrable, for every $\epsilon>0$ there exists $\delta_\epsilon>0$ such that

$$ \int_F\vert X_n\vert dP<\epsilon \quad \text{and} \quad \int_F\vert X\vert dP<\epsilon $$ if $P(F)<\delta_\epsilon$. Using almost sure convergence, we know that

$$ P( \vert X_n-X\vert>\epsilon )\to0. $$

Therefore, for $n$ large enough we have $ P( \vert X_n-X\vert>\epsilon )<\delta_\epsilon$ and therefore

$$ \mathbb{E}\big[ \vert X_n-X \vert \big] \leq \int_{ \{ \vert X_n-X\vert>\epsilon \}} \vert X_n\vert dP+ \int_{ \{ \vert X_n-X\vert>\epsilon \} } \vert X\vert dP+ \int_{\{ \vert X_n-X\vert\leq\epsilon \}}\vert X_n-X\vert dP < $$

$$ < 2\epsilon+ \int_{\{ \vert X_n-X\vert\leq\epsilon \}} \epsilon\cdot dP\leq 3\epsilon. $$

This is true for all $\epsilon>0$, so $\mathbb{E}[X_n]\to \mathbb{E}[X]$.

$\endgroup$
2
  • $\begingroup$ What is the definition of the set $F$? $\endgroup$
    – Math_Day
    Mar 7 '21 at 0:09
  • $\begingroup$ @Math_Day $F$ is a general measurable set. In the last computation I take $F$ as $\{ \vert X_n-X\vert>\epsilon \}$. $\endgroup$ Mar 7 '21 at 6:46
1
$\begingroup$

Another (similar) approach is as follows.

For every $\epsilon > 0$, choose $K(\epsilon)$ so that $E[|X_n|1_{X_n\geq K(\epsilon)}]<\epsilon$ and $E[|X|1_{X\geq K(\epsilon)}] < \epsilon$ (this is simply applying the definition of UI to $\{ X_n \}_n \cup \{X\}$. Compute

$$ \begin{align} E|X_n-X| \leq & E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\ + &E[|X_n|1_{X_n > K(\epsilon), X \leq K(\epsilon)}] \\ +& E[|X_n|1_{X_n \leq K(\epsilon), X > K(\epsilon)}] \\ +& E[|X|1_{X_n > K(\epsilon), X \leq K(\epsilon)}]\\ +& E[|X|1_{X_n \leq K(\epsilon), X > K(\epsilon)}]\\ \leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] \\ + & E[|X_n|1_{X_n > K(\epsilon)}] \\ +&E[|X|1_{X > K(\epsilon)}] \\ +& E[|X_n|1_{X_n > K(\epsilon)}] \\ +&E[|X|1_{X > K(\epsilon)}] \\ \leq &E[|X_n-X|1_{X_n, X \leq K(\epsilon)}] + 4\epsilon. \end{align} $$

Sending $n \to \infty$ we find $$ \limsup_n E|X_n-X| \leq 4\epsilon. $$ Now send $\epsilon \to 0$.

$\endgroup$

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.