Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 26J (Probability and Measure)

The Cauchy-Schwarz inequality givesThus $Z{X}_n$ is integrable for every $n$.

The hypothesis says exactly that $X_n\rightharpoonup 0$ in the Hilbert space $L^2$. We give the constructive Weak Banach–Saks theorem in a Hilbert space argument. Choose $n_1<n_2<\cdots$ inductively so that, for $Y_k=X_{n_k}$,This is possible because each earlier $Y_j$ is an admissible test variable in the weak-convergence assumption. Since $\Vert Y_k{\Vert }_2\le 1$,Therefore

From convergence in probability, choose a subsequence $X_{n_k}\to X$ almost surely. The Fatou lemma then givesso $X\in L^2$.

The bounded second moments make $(X_n)$ uniformly integrable, and $X$ is integrable. Convergence in probability together with uniform integrability therefore givesConvergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, letThen $X_n\to 0$ in probability and $\mathbb{E}|X_n{|}^2=1$, but $\Vert X_n{\Vert }_2=1$ for every $n$. This is the bounded second moments do not upgrade convergence in probability to convergence in L2 example.

SetBy variance additivity for independent random variables,Thus $Y_N$ differs in $L^2$ by $o(1)$ from the deterministic scalar $m_N$. The L2 convergence of averages of independent variables with bounded second moments gives the necessary and sufficient condition