Past exam of the martingales and financial mathematics course of the University of Oxford / 2007 / A / 4

A family of sigma-algebras is independent if for every finite collection of distinct indices $n_1,\dots ,n_k$ and sets $A_j\in {\mathcal{G}}_{n_j}$,A family of random variables is independent if its generated sigma-algebras $\sigma (X_n)$ are independent. Equivalently, the displayed product identity holds for the events $\{X_{n_j}\in B_j\}$ with arbitrary Borel $B_j$. Pairwise independence alone is weaker and does not suffice for a family.

The pi-systems criterion says that if ${\mathcal{I}}_1,{\mathcal{I}}_2$ are pi-systems and $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ for all $A\in {\mathcal{I}}_1$, $B\in {\mathcal{I}}_2$, then their generated sigma-algebras are independent. Necessity is immediate. We may add $\Omega$ to either pi-system, since it does not alter the generated sigma-algebra or the product identities.

For sufficiency, fix $A\in {\mathcal{I}}_1$ and defineIt contains $\Omega$, is closed under complements by subtraction from $\mathbb{P}(A)$, and is closed under countable disjoint unions by countable additivity. Thus it is a lambda-system. It contains ${\mathcal{I}}_2$, so the pi-lambda theorem gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and apply the same lambda-system argument to the sets $A\in {\mathcal{G}}_1$ satisfying the identity. The first step shows that this lambda-system contains ${\mathcal{I}}_1$, so it contains all ${\mathcal{G}}_1$. Thus the product identity holds on both entire sigma-algebras, proving their independence.

The tail sigma-algebra isFix $S\in \mathcal{T}$. independence of the variables and the pi-systems criterion imply independence of $\sigma (X_1,\dots ,X_n)$ from $\sigma (X_{n+1},X_{n+2},\dots )$. Hence $S$ is independent of every finite-head sigma-algebra ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$.

The increasing union ${\bigcup }_n{\mathcal{H}}_n$ is an algebra, hence a pi-system, and generates ${\mathcal{H}}_{\infty }=\sigma (X_1,X_2,\dots )$. Another pi-lambda argument extends the equality $\mathbb{P}(S\cap A)=\mathbb{P}(S)\mathbb{P}(A)$ from that union to all of ${\mathcal{H}}_{\infty }$. Since $S$ itself belongs to ${\mathcal{H}}_{\infty }$, take $A=S$:ThereforeThis is Kolmogorov zero-one law; no moment assumptions on the variables are needed.

Let $C$ be the event that the sample averages have a limit in the extended real number line. It is measurable: the bounded sequence $\arctan S_n$ converges exactly when it is Cauchy, and the Cauchy condition is a countable combination of measurable inequalities. Limits at $\pm \pi /2$ correspond to extended limits $\pm \infty$.

For each fixed $m$, remove the first $m-1$ terms and setThe difference $S_n-S_n^{(m)}=n^{-1}{\sum }_{k<m}{X}_k$ tends to zero for every outcome, since the finitely many variables are real-valued. Thus these two sequences have an extended limit simultaneously. Equivalently their arctangents differ by a quantity tending to zero, because arctangent is Lipschitz. The event $C$ is therefore expressible using $X_m,X_{m+1},\dots$ alone. Since this holds for every $m$, $C\in \mathcal{T}$.

Kolmogorov's law gives $\mathbb{P}(C)=0$ or one. In the latter case the measurable extended variable $\mathrm{limsup}{S}_n$, redefined on a null set if needed, is the almost sure convergence. In the former there is almost surely no extended-real limit. Hence extended convergence occurs almost surely, or divergence occurs almost surely. The law need not make the value of a possible limit finite.