Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 25K (Probability and Measure)

For independent random variables $X_1,X_2,\dots$, let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, $\mathcal{F}=\sigma (X_1,X_2,\dots )$, and let the tail sigma-algebra be $\mathcal{T}={\bigcap }_n\sigma (X_{n+1},X_{n+2},\dots )$. Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Such an event is independent of each ${\mathcal{F}}_n$ by independence of the variables. The collection of events $B\in \mathcal{F}$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ is a Dynkin system: it contains the whole space and is closed under complements and disjoint countable unions. It contains the pi-system ${\bigcup }_n{\mathcal{F}}_n$, which generates $\mathcal{F}$. The pi-lambda theorem therefore makes $A$ independent of all of $\mathcal{F}$. Since $A$ itself belongs to $\mathcal{F}$,

Let $D_n=B_n\setminus B_{n+1}$. These sets are pairwise disjoint andBy countable additivity, ${\sum }_n\mu (D_n)=\mu (B_1)-\mu (B)<\infty$. The tails of this convergent nonnegative series tend to zero, soThis proves continuity from above of a measure and explains the finite-measure assumption.

A necessary qualification is ${\sigma }^2>0$. As printed, the assumptions also permit $X_i=0$ almost surely; then $\mathbb{P}(A_n)=0$ for every $K>0$, so the asserted positive lower bound is false. Assume the intended nondegenerate case.

The central limit theorem states that independent identically distributed variables with mean zero and finite positive variance satisfy $S_n/(\sigma \sqrt{n})\Rightarrow Z$, where $Z\sim N(0,1)$. Since the normal distribution has no atom at $K/\sigma$,For example,

Continue with ${\sigma }^2>0$. For each integer $K\ge 1$, let $B_m(K)={\bigcup }_{n\ge m}\{S_n/\sqrt{n}\ge K\}$. These events decrease with $m$, and their probabilities are at least $c_K>0$ by (iii). Continuity from above of a measure givesIt is important not to assume that this exact threshold event is automatically a tail event: a vanishing change of partial sums can affect infinitely many exact comparisons at the threshold. Instead useFor any fixed $m$, subtracting $S_m/\sqrt{n}\to 0$ shows that $L={\mathrm{limsup}}_{n\to \infty }(X_{m+1}+\cdots +X_n)/\sqrt{n}$. Thus $L$ is measurable with respect to every tail sigma-algebra. The event $\{L\ge K\}$ has positive probability by the preceding bound, so Kolmogorov zero-one law makes its probability one. Intersecting over all positive integers givesFor any specified real $K>0$, this conclusion implies infinitely many $n$ with $S_n/\sqrt{n}\ge K$, completing both requests. In the excluded zero-variance case the limsup is zero.