{c}
{wiki=Doob's_martingale_convergence_theorems}

A discrete-time martingale with $\sup_n\mathbb E|M_n|<\infty$ converges almost surely to an integrable random variable. If the martingale is <uniform integrability>[uniformly integrable], convergence also holds in $L^1$.

Martingale convergence theorem

For $A\in\mathcal F_n$, the defining identity for $X_{n+1}$ gives
$$
\mathbb E[X_{n+1}\mathbf1_A]=\mathbb E[X\mathbf1_A]
=\mathbb E[X_n\mathbf1_A].
$$
Part (b) therefore identifies $X_n$ with $\mathbb E[X_{n+1}\mid\mathcal F_n]$, so $(X_n,\mathcal F_n)$ is a <martingale>.

The atom formula also proves
$$
|X_n|\leq\mathbb E[|X|\mid\mathcal F_n].
$$
Let $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb P(B_{n,K})\leq\mathbb E|X|/K$ by <Markov inequality>, while
$$
\mathbb E[|X_n|\mathbf1_{B_{n,K}}]
\leq\mathbb E[|X|\mathbf1_{B_{n,K}}].
$$
The <uniform absolute continuity for a finite measure> makes the right-hand side uniformly small as $K\to\infty$. Thus $(X_n)$ is <uniform integrability>[uniformly integrable]. The <Martingale convergence theorem> now supplies an integrable random variable $Y$ such that $X_n\to Y$ both <almost sure convergence>[almost surely] and in $L^1$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 1 / c / Solution

{wiki}

A family of integrable random variables is uniformly integrable when
$$
\lim_{K\to\infty}\sup_{X}\mathbb E\bigl[|X|\mathbf1_{\{|X|>K\}}\bigr]=0.
$$
Almost-sure or probabilistic convergence together with uniform integrability upgrades to $L^1$ convergence.


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 Incoming links: Uniform integrability