Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 3

The tail sigma-algebra isIts events depend only on arbitrarily late coordinates. The Kolmogorov zero-one law states that for a sequence of independent random variables, every tail event has probability zero or one:Identical probability distributions and integrability are not required for the Kolmogorov zero-one law.

Let ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$. Since $X_{n+j}=S_{n+j}-S_{n+j-1}$,Permuting $X_1,\dots ,X_n$ preserves their joint probability distribution and leaves every generator of ${\mathcal{G}}_n$ unchanged. Therefore, for each bounded ${\mathcal{G}}_n$-measurable $H$,This is the symmetry of exchangeable random variables. By the defining identity of conditional expectation, all $\mathbb{E}[X_j\mid {\mathcal{G}}_n]$ are equal almost surely. Summing these conditional expectations and using that $S_n$ is ${\mathcal{G}}_n$-measurable givesHence the conditional expectation of a summand given future partial sums isThe integrability of each $X_j$ justifies every conditional expectation above.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

Split the adjacent products into two subsequences:Within each subsequence the pairs use disjoint coordinates, so each subsequence consists of independent and identically distributed random variables. The two subsequences need not be independent of each other. By independence within a pair,Apply the strong law of large numbers from (c) separately to obtain $k^{-1}{\sum }_{j=1}^k{Y}_j\to m^2$ and $k^{-1}{\sum }_{j=1}^k{Z}_j\to m^2$ simultaneously almost surely. Sinceand both prefactors tend to $1/2$, the strong law for adjacent products gives