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.

  • $\begingroup$ 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

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 $\mathbb{E}[X_n]\to \mathbb{E}[X]$.

  • $\begingroup$ 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

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 $\epsilon \to 0$.


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