Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 2 / b

For each $N\ge 1$, put $C_N={\bigcup }_{n\ge N}{A}_n$. Closure of a sigma-algebra under countable unions and countable intersections givesThe events $C_N$ decrease to $A$. Continuity of probability from above applies because $\mathbb{P}(C_1)\le 1<\infty$. Also, $A_n\subseteq C_N$ whenever $n\ge N$, soLetting $N\to \infty$ provesThis reverse Fatou inequality for events requires no independence.

Write $B_n=\{X_n>a_n\}$. If $B_n$ occurs, the running maximum at time $n$ exceeds $a_n$, so $B_n\subseteq A_n$ and ${\mathrm{limsup}}_n{B}_n\subseteq A$.

For the reverse inclusion, fix an outcome for which only finitely many $B_n$ occur. Choose $N$ such that $X_m\le a_m$ for every $m\ge N$. The finite prefix has a finite maximum $K={\max }_{1\le m<N}{X}_m$ when $N>1$; there is no prefix to check when $N=1$. Since $a_n\to \infty$, for every sufficiently large $n$ that prefix satisfies $K\le a_n$. For the remaining indices $N\le m\le n$, the nondecreasing thresholds giveThus ${\max }_{m\le n}{X}_m\le a_n$ for every sufficiently large $n$, so only finitely many $A_n$ occur. Taking the contrapositive proves the reverse inclusion, and hence the pointwise event equalityThis exceedance equivalence for increasing thresholds is deterministic. Its proof uses both monotonicity of $a_n$ and divergence to $+\infty$; it does not require independence or identical probability distributions.

The tail sigma-algebra of this sequence isFor each fixed $r$, the preceding event equality givesEvery event on the right is determined by coordinates with index at least $r$. Therefore $A\in \sigma (X_r,X_{r+1},\dots )$ for every $r$, and $A\in \mathcal{T}$. The random variables $X_n$ are independent, so Kolmogorov's zero–one law applies: every tail event has probability zero or one. Hence $\mathbb{P}(A)\in \{0,1\}$.

The role of independence can also be seen directly. A tail event is independent of each finite initial sigma-algebra $\sigma (X_1,\dots ,X_m)$. The Monotone class theorem extends this independence to $\sigma (X_1,X_2,\dots )$; since $A$ itself lies there, it is independent of itself. Thus $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.^[ref]

For $B_n=\{X_n>a_n\}$, the definition of the common cumulative distribution function givesThe strict exceedance matters here: an atom of a measure at $a_n$ belongs to $\{X_n\le a_n\}$, so it is already included in $F(a_n)$. The events $B_n$ are mutually independent events, since each depends on a separate member of the independent random variables $X_n$.

If ${\sum }_n(1-F(a_n))<\infty$, countable subadditivity gives, for every $N$,Continuity of probability then gives $\mathbb{P}({\mathrm{limsup}}_n{B}_n)=0$. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and $1-u\le e^{-u}$ give, for every fixed $N$,Taking $M\to \infty$ shows that the probability of no exceedance after $N$ is zero. A countable union over $N$ shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves