Exercise 1.27 (Scheffé's lemma)

Question

Let $X_1,X_2,\dots$ be a sequence of absolutely integrable scalar random variables that converge almost surely to another absolutely integrable scalar random variable $X$. Suppose also that ${\bf E} |X_n|$ converges to ${\bf E} |X|$ as $n \rightarrow \infty$. Show that ${\bf E} |X-X_n|$ converges to zero as $n \rightarrow \infty$.

Elu Thingol · Accepted Answer

The trick to proving this assertion is to recall a famous identity
$$|X_n - X| = X_n + X - 2\min\left\{X_n,Xight\}$$
which we will verify at the end of this proof. Put integrals at the both sides of the equation and take limits as $n	o\infty$. We obtain, by linearity of expectation and limits, that
\begin{align}
\lim_{n	o\infty}\mathop{\mathbb{E}}|X_n - X|
	&= \lim_{n	o\infty}\mathop{\mathbb{E}}X_n + \lim_{n	o\infty}\mathop{\mathbb{E}}X - 2 \lim_{n	o\infty}\mathop{\mathbb{E}}\min\{X_n,X\}\
	&= X + X - 2\mathop{\mathbb{E}}\lim_{n	o\infty}\min\{X_n,X\}\
	&= 0
\end{align}
as desired.\par

We now verify the identity in the beginning of the proof.
\begin{itemize}
\item $X_n(\omega) \leq X(\omega)$, in which case
	$$|X_n(\omega) - X(\omega)| = X(\omega)-X_n(\omega) = X_n(\omega) + X(\omega) - 2 X_n(\omega).$$
\item Similarly, if $X_n(\omega) > X(\omega)$, we have
	$$|X_n(\omega) - X(\omega)| = X_n(\omega)-X(\omega) = X_n(\omega) + X(\omega) - 2 X(\omega).$$
\end{itemize}

Exercise 1.27 (Scheffé's lemma)

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Exercise 1.27 (Scheffé's lemma)

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