Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 2

A sequence of integrable random variables $(X_n)$ has uniform integrability exactly whenThe uniformly integrable martingale convergence theorem states that a uniformly integrable discrete-time martingale $(M_n)$ has an integrable random variable $M_{\infty }$ withIndeed, uniform integrability implies ${\sup }_n\mathbb{E}|M_n|<\infty$, so the martingale convergence theorem gives almost sure convergence. The combination of uniform integrability and almost sure convergence gives convergence in L1. For $m\ge n$, the martingale identity $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ passes to the limit by the L1 contraction of conditional expectation.

Conversely, uniform integrability of conditional expectations shows that a martingale of the form $M_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for one integrable random variable $Y$ is uniformly integrable.

The original PDF has $\mathbb{E}|X_T|<\infty$ and $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$. The TeX transcription drops the expected value and absolute value in the first condition, and the absolute value in the second. The proof uses the PDF's conditions.

Because the stopping time $T$ is finite almost surely, $X_{n\wedge T}\to X_T$ almost surely. More strongly,The first term tends to zero by hypothesis; the second does so by the dominated convergence theorem. Thus there is convergence in L1.

Here is the needed L1 convergence implies uniform integrability argument. For any integrable random variables $Z,Y$ and $K>0$, splitting according to $|Y|>K/2$ givesTake $Z=X_{n\wedge T}$ and $Y=X_T$. For large $n$, the first term is uniformly small by convergence in L1, and the second is small for large $K$ by integrability. The finitely many remaining $n$ are handled individually by integrability. Hence the stopped process is uniformly integrable. This proves the stopped-martingale uniform integrability criterion; the martingale assumption is not needed for this particular implication.

Pointwise, an integer-valued nonnegative random variable has the representationwhen $T=k$, exactly the terms $n=0,\dots ,k-1$ equal one. Applying the Tonelli theorem, or the monotone convergence theorem to the finite sums of indicator random variables, gives the tail-sum formula for expectationBoth sides may be infinite; no integrability assumption is necessary.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.