Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 26G (Probability and Measure)

Uniform integrability meansIf ${\sup }_n\mathbb{E}|X_n{|}^p=C<\infty$ for $p>1$, thenproving uniform integrability. For a counterexample, let $X_n=n$ with probability $1/n$ and zero otherwise. Then $\mathbb{E}|X_n|=1$, but for every $K$ and $n>K$ the tail expectation is one.

Write $Y_n=n^{-1}{\sum }_{i=1}^n|X_i|$. Truncate each $|X_i|$ at a level $K$. The average of the truncated parts is at most $K$, while the expected average of the tails isindependently of $n$. Markov's inequality then shows that $(Y_n)$ is uniformly integrable as $K\to \infty$. Since $|S_n|/n\le Y_n$, the sequence $(S_n/n)$ is uniformly integrable.

Convergence in probability to the constant $\mathbb{E}{X}_1$, together with uniform integrability, implies convergence of first absolute moments by the Vitali convergence theorem. Hence

By the union bound,For an $L^2$ random variable, $t^2\mathbb{P}(|X_1|>t)\to 0$. Taking $t=\epsilon \sqrt{n}$ makes the right side tend to zero, proving $M_n/\sqrt{n}\to 0$ in probability.

For any $\epsilon >0$,ThereforeThe last term tends to zero. Taking the limsup and then $\epsilon \downarrow 0$ proves the claim.