Now assume $X_n\to X$ <almost surely> and $\mathbb E[X_n^2]\to\mathbb E[X^2]$. In particular, $(X_n)$ is bounded in $L^2$, so it is <uniform integrability from bounded second moments>[uniformly integrable] in $L^1$. Almost-sure convergence and <uniform integrability> therefore give
$$
\mathbb E|X_n-X|\longrightarrow0.
$$

We next show that $X_n$ converges weakly to $X$ in the <L2 space>. For $Y\in L^2$, let $Y_M=Y\mathbf1_{\{|Y|\leq M\}}$. Then
$$
|\mathbb E[(X_n-X)Y_M]|\leq M\mathbb E|X_n-X|\longrightarrow0,
$$
while the <Cauchy-Schwarz inequality> gives, uniformly in $n$,
$$
|\mathbb E[(X_n-X)(Y-Y_M)]|
\leq(\lVert X_n\rVert_2+\lVert X\rVert_2)\lVert Y-Y_M\rVert_2.
$$
Letting first $n\to\infty$ and then $M\to\infty$ proves $\mathbb E[X_nY]\to\mathbb E[XY]$. Taking $Y=X$ and expanding the square now gives
$$
\mathbb E[(X_n-X)^2]
=\mathbb E[X_n^2]+\mathbb E[X^2]-2\mathbb E[X_nX]
\longrightarrow0.
$$
Consequently
$$
\boxed{X_n\longrightarrow X\text{ in }L^2.}
$$
This proves the <almost-sure convergence and convergence of second moments> criterion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 26K / a / ii / Solution

If square-integrable random variables satisfy $X_n\to X$ <almost surely>, then $X_n\to X$ in $L^2$ exactly when $\mathbb E[X_n^2]\to\mathbb E[X^2]$. Boundedness in $L^2$ and almost-sure convergence first give weak convergence in the Hilbert space $L^2$; weak convergence together with convergence of norms gives strong convergence.

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