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

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.