Past exam of the probability, measure and martingales course of the University of Oxford / 2023 / 1

Fix $A\in {\mathcal{A}}_1$ and defineThis is a Dynkin system: it contains $\Omega$; taking complements preserves the equality because $\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$; and countable additivity preserves it under disjoint unions. The hypothesis says that ${\mathcal{D}}_A$ contains the pi-system ${\mathcal{A}}_2$. Hence Dynkin lemma gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and putThe same complement and disjoint-union calculations make this a Dynkin system. The first step shows that it contains ${\mathcal{A}}_1$, so Dynkin lemma gives ${\mathcal{E}}_B={\mathcal{G}}_1$. ThusThis is precisely the definition of independent sigma-algebras, and proves independence extended from generating pi-systems.

For $m\ge 1$, fix signs $x_1,\dots ,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\dots ,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0{x}_1$ and $Y_j=x_{j-1}{x}_j$ for $2\le j\le m$. Each input tuple has probability $2^{-(m+1)}$ by independence. ConsequentlyEvery finite subfamily is a marginal distribution of one such initial segment, so the entire sequence is independent, with fair sign probability distributions.

Using the usual everywhere sign-valued representatives, for every $n\ge 1$ we haveThe right side is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. Therefore $Y_0$ is measurable with respect to their intersection $\mathcal{G}$.

Write $\mathcal{T}={\bigcap }_n{\mathcal{T}}_n$. The Kolmogorov zero-one law applies to the independent random variables $(X_n)$, so every event $C\in \mathcal{T}$ has probability zero or one. Also $Y_0$ is independent of $\mathcal{Y}$: this follows first for finite intersections of sign events from the given independence, then for their generated sigma-algebra by part (a).

The sets $D\cap C$ with $D\in \mathcal{Y}$ and $C\in \mathcal{T}$ form a pi-system generating $\mathcal{H}$. If $\mathbb{P}(C)=0$, the event $D\cap C$ has probability zero and is independent of every event in $\sigma (Y_0)$. If $\mathbb{P}(C)=1$, it differs from $D$ by a null set, so the same independence follows from that of $Y_0$ and $\mathcal{Y}$. Another application of part (a) provesIn particular $\mathcal{H}$ cannot contain the nontrivial event $\{Y_0=1\}$, whereas $\mathcal{G}$ does. Thus adjoining a sigma-algebra need not commute with a decreasing intersection.

There is a small version qualification if the assumptions specify only the probability distributions, allowing exceptional values other than signs. Put $q_n=X_n/(Y_1\cdots Y_n)$ when the denominator is nonzero, and $q_n=0$ otherwise. For $k\ge n$, $q_k$ is measurable with respect to $\sigma (\mathcal{Y},{\mathcal{T}}_n)$. The limit superior of $(q_k)$, after scalar clipping to $[-1,1]$, is therefore $\mathcal{G}$-measurable and equals $Y_0$ almost surely. Literal measurability of an arbitrary version of $Y_0$ need not follow without choosing sign-valued representatives or adjoining the ambient null sets. All the independence conclusions are unaffected.

The first entrance time ${\tau }_K$ is a stopping time, because $\{{\tau }_K\le n\}={\bigcup }_{j=0}^n\{M_j\le -K\}\in {\mathcal{F}}_n$. Before this time $M_n>-K$; at entrance, the bounded increments give $M_{{\tau }_K}>-K-L$. Thus the stopped martingaleis a nonnegative martingale. Here the stopped process is a martingale since its increment is $1_{\{{\tau }_K>n\}}(M_{n+1}-M_n)$, whose conditional expectation given ${\mathcal{F}}_n$ is zero.

The martingale convergence theorem states that a nonnegative martingale converges almost surely to a finite integrable random variable; indeed its expectations are constant, giving the required bound in the L1 norm. Apply this theorem to $(N_n)$. On $\{{\tau }_K=\infty \}$ the stopped process equals the original one at every time. HenceThe overshoot bound is why the assumption of bounded increments matters.

Apply part (i) for every positive integer $K$ and discard the countable union of its exceptional null sets. A sample path bounded below never reaches $-K$ for some such $K$, and must therefore converge to a finite limit of a sequence.

Apply the same argument to the martingale $(-M_n)$. Outside one more null set, every sample path bounded above also converges finitely. Thus a sample path which does not converge finitely must be unbounded in both directions. Each finite initial segment has finite values, so removing any such segment preserves unboundedness. It follows that the limit superior is $+\infty$ and the limit inferior is $-\infty$. ConsequentlyThis proves the bounded-increment martingale convergence-or-oscillation dichotomy; in particular one-sided divergence to infinity is impossible with positive probability.

Choose versionsThe stochastic process $(A_n)$ is increasing, and its $k$th increment is ${\mathcal{F}}_{k-1}$-measurable, so it is predictable. The Doob decomposition theorem therefore givesTo check the martingale property directly, each summand is integrable and has zero conditional expectation given ${\mathcal{F}}_{k-1}$. The sums are an adapted process for the given filtration. Thus $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$. Its bounded increments satisfy $|M_n-M_{n-1}|\le 1$.

Both $X_n$ and $A_n$ are increasing, so each has a limit of a sequence in $[0,\infty ]$. Apply part (c)(ii) to the martingale $M_n=X_n-A_n$, whose bounded increments were verified in part (i).

On $\{X_{\infty }<\infty ,A_{\infty }=\infty \}$ we would have $M_n\to -\infty$. On $\{X_{\infty }=\infty ,A_{\infty }<\infty \}$ we would have $M_n\to +\infty$. Both events have probability zero by the convergence-or-oscillation conclusion. Hence the two divergence events agree up to a null set. Since $X_{\infty }=\infty$ means precisely that infinitely many $B_n$ occur,Taking probabilities proves the displayed equality in the question. This stronger agreement of events is the Conditional Borel-Cantelli lemma.

Suppose the events $(B_n)$ are mutually independent events. Then $B_k$ is independent of ${\mathcal{F}}_{k-1}$, first by factorization against finite intersections of previous events and then by part (a). ThereforeIf ${\sum }_k\mathbb{P}(B_k)=\infty$, the sum in part (ii) is infinite almost surely, and consequentlyThis is the Second Borel-Cantelli lemma. In fact part (ii) also yields probability zero in the convergent case for these independent events.