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

The finite sigma-algebra ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$ is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is ${\mathcal{F}}_n$-measurable and integrable. Every $A\in {\mathcal{F}}_n$ is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.

Solved by gpt-5.6-sol high.

Uniqueness means almost sure equality. Suppose that two ${\mathcal{F}}_n$-measurable integrable random variables $Y$ and $Z$ satisfy the integral identity from part (a). The event $D=\{Y>Z\}$ belongs to ${\mathcal{F}}_n$, and henceThe integrand is nonnegative and is positive precisely on $D$, so $\mathbb{P}(D)=0$. Interchanging $Y$ and $Z$ gives $\mathbb{P}(Z>Y)=0$, and therefore $Y=Z$ almost surely.

Solved by gpt-5.6-sol high.

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (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 provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

Solved by gpt-5.6-sol high.

Let $\mathcal{G}=\sigma (A_1,A_2,\dots )$. The limit $Y$ from part (c) is $\mathcal{G}$-measurable because each $X_n$ is. If $A\in {\bigcup }_n{\mathcal{F}}_n$, then $A\in {\mathcal{F}}_N$ for some $N$, and for every $n\ge N$,The $L^1$ convergence lets us pass to the limit and obtain $\mathbb{E}[Y{1}_A]=\mathbb{E}[X{1}_A]$.

The sets on which this identity holds form a Dynkin system, and ${\bigcup }_n{\mathcal{F}}_n$ is a generating pi-system. The Dynkin lemma therefore extends the identity to every $A\in \mathcal{G}$. Consequently $Y$ has both defining properties of $\mathbb{E}[X\mid \mathcal{G}]$, established here without appealing to the general existence theorem.

Solved by gpt-5.6-sol high.