Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 25J (Probability and Measure)

The first Borel-Cantelli lemma says that ${\sum }_n\mathrm{Pr}(A_n)<\infty$ implies $\mathrm{Pr}(A_n\text{ infinitely often})=0$, without an independence assumption. Indeed, for every $N$,The decreasing intersection of these tail unions is the infinitely-often event; continuity of probability from above proves the assertion. The second lemma says that for independent $A_n$, divergence of the probability sum implies $\mathrm{Pr}(A_n\text{ infinitely often})=1$.

Convergence in probability supplies a subsequence $X_{n_k}$ converging almost surely to $X$: choose it so $\mathrm{Pr}(|X_{n_k}-X|>2^{-k})<2^{-k}$, then apply the first Borel-Cantelli lemma. For each fixed $N$, the tail of this subsequence lies in $\sigma (X_n:n\ge N)$, and its almost sure limit is $X$. Thus $X$ has a version measurable with respect to the completed tail sigma-field of the independent sequence.

By Kolmogorov's zero–one law, every event in that tail sigma-field has probability zero or one. Consequently $\mathrm{Pr}(X\le r)\in \{0,1\}$ for every rational $r$. Since $X$ is a finite real random variable, its distribution function has limits zero and one at the two infinities; monotonicity then identifies a single threshold $c$ with $\mathrm{Pr}(X=c)=1$. Hence $X$ is almost surely constant. Completion only adds null events and does not alter the zero-one conclusion.

Under complete convergence of random variables, for each integer $k\ge 1$ the first Borel-Cantelli lemma gives $\mathrm{Pr}(|X_n-X|>1/k\text{ infinitely often})=0$. Intersecting these probability-one conclusions over the countably many $k$ shows that every tolerance is eventually satisfied. Therefore complete convergence implies almost sure convergence, without independence.

Almost sure convergence implies convergence in probability, so part (b) identifies the limit as a constant $c$ almost surely. For each $ϵ>0$, the events $A_n=\{|X_n-c|>ϵ\}$ are independent. Almost sure convergence gives $\mathrm{Pr}(A_n\text{ infinitely often})=0$. If their probability sum diverged, the second Borel-Cantelli lemma would instead give probability one. Hence the sum is finite for every $ϵ$, proving almost sure convergence implies complete convergence for independent sequences.

Let $U$ be one uniform random variable on $(0,1)$, and defineFor every $U>0$ the sequence is eventually zero, so $X_n\to 0$ almost surely. But $\mathrm{Pr}(|X_n|>1/2)=1/n$, whose sum diverges. This sequence is not completely convergent; the exceedance events are nested and dependent.