Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 3

Here $F=1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-sigma-algebra is the orthogonal projection in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the probability measure.

The increasing coordinate sigma-algebras ${\mathcal{E}}_n$ generate the whole product sigma-algebra $\mathcal{F}$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal{F})$: their cylinder indicators generate $\mathcal{F}$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is ${\mathcal{E}}_N$-measurable, then for $n\ge N$ the $L^2$ projection contraction givesApproximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.

For the decreasing tail sigma-algebras ${\mathcal{G}}_n$, put $Y_n=\mathbb{E}[F\mid {\mathcal{G}}_n]$. If $m\ge n$, the tower property gives $\mathbb{E}[Y_n\mid {\mathcal{G}}_m]=Y_m$. Projection orthogonality therefore yieldsThe norms on the right decrease and are nonnegative, so $(Y_n)$ is Cauchy in $L^2$ and has a limit $Y$. Choose an almost-surely convergent subsequence and take its pointwise upper limit. For every fixed $n$ its tail consists of ${\mathcal{G}}_n$-measurable functions, so this supplies a version of $Y$ measurable with respect to every ${\mathcal{G}}_n$, hence to $\mathcal{G}={\bigcap }_n{\mathcal{G}}_n$. For any $C\in \mathcal{G}$, $\mathbb{E}[Y_n{1}_C]=\mathbb{E}[F{1}_C]$; passing to the limit identifies $Y=\mathbb{E}[F\mid \mathcal{G}]$.

Finally $\Vert W{\Vert }_1\le \Vert W{\Vert }_2$ on a probability space, by the Cauchy-Schwarz inequality. ThusThese are the increasing-filtration and reverse martingale convergence theorem limits; the argument above establishes the requested norm convergence directly.

Let $T$ swap coordinate blocks $0,\dots ,n$ and $n+1,\dots ,2n+1$, fixing all other coordinates. The symmetric law makes $T$ measure preserving, and invariance of the event $A$ gives $F\circ T=F$ pointwise. Put ${\mathcal{B}}_n=\sigma (X_{n+1},\dots ,X_{2n+1})$.

Conditional expectation transforms covariantly under this block swap:To check it, the left side is ${\mathcal{B}}_n$-measurable, and for $C\in {\mathcal{B}}_n$ the change of variables by $T$ sends its defining integral to the conditional-expectation identity on $T(C)\in {\mathcal{E}}_n$, with $F\circ T=F$. Thus the two versions agree. Measure preservation now givesThis is the first requested estimate.

The L1 contraction of conditional expectation follows from the pointwise bound $|\mathbb{E}[Z\mid \mathcal{H}]|\le \mathbb{E}[|Z|\mid \mathcal{H}]$: take expectations to get $\Vert \mathbb{E}[Z\mid \mathcal{H}]{\Vert }_1\le \Vert Z{\Vert }_1$. Applied to a difference, it makes conditional expectation a contraction between any two integrable inputs.

Let $U_n=\mathbb{E}[F\mid {\mathcal{B}}_n]$. Since ${\mathcal{B}}_n\subseteq {\mathcal{G}}_n$, $U_n$ is ${\mathcal{G}}_n$-measurable. HenceCombining this with the other given approximation bound yieldsThe left side does not depend on $n$, and $\epsilon >0$ was arbitrary. It is therefore zero. For a $\mathcal{G}$-measurable version $Y=\mathbb{E}[F\mid \mathcal{G}]$, the event $C=\{Y>1/2\}$ belongs to $\mathcal{G}$ and satisfiesThis proves that exchangeable invariant events are tail events modulo null sets. It does not assert literal pointwise equality of the two events or sigma-algebras before completing them with null sets.