Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 1

Put $M_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$. Each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, with $|M_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]\le 1$. Since a filtration is increasing, the tower property of conditional expectation givesThus $(M_n)$ is the conditional-expectation martingale associated with $Z$.

The Martingale convergence theorem gives an almost-sure limit $M_{\infty }$, because $|M_n|\le 1$. The dominated convergence theorem also gives $M_n\to M_{\infty }$ in $L^1$.

To identify the limit, take $A\in {\bigcup }_n{\mathcal{F}}_n$. For some $N$, $A\in {\mathcal{F}}_N$, and for every $n\ge N$,Passing to the $L^1$ limit preserves this equality. The sets for which $\mathbb{E}[M_{\infty }{1}_A]=\mathbb{E}[Z{1}_A]$ form a monotone class containing the algebra ${\bigcup }_n{\mathcal{F}}_n$, so the equality holds throughout ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $M_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, it is $\mathbb{E}[Z\mid {\mathcal{F}}_{\infty }]$. This proves the conditional-expectation convergence along a filtration both almost surely and in $L^1$.

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.