Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 1

Put $A=\{|X|\ge Z\}$. Both $A$ and its indicator function are measurable, and $0\le Z{1}_A\le |X|$. Thus $Z{1}_A$ is an integrable random variable. Monotonicity of conditional expectation givesThe measurable-factor property of conditional expectation says that a bounded $\mathcal{G}$-measurable function $H$ satisfies $\mathbb{E}[HW\mid \mathcal{G}]=H\mathbb{E}[W\mid \mathcal{G}]$ for every integrable random variable $W$. Here the factor $Z$ need not be bounded. Apply that property first to $Z\wedge m$ and $W=1_A$, then use monotone convergence on the nonnegative products, including the conditional version, to obtainAll the left-hand conditional expectations are well defined; the final one is finite almost surely because $Z{1}_A$ is an integrable random variable. Since $Z>0$ almost surely, division now yieldsThis proves the conditional Markov inequality with a measurable threshold without assuming that $Z^{-1}|X|$ is an integrable random variable. The measurable-factor rule, with its product-integrability condition, is also stated in the standard conditional expectation theory.^[ref]

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

For every $n$, ${\mathcal{F}}_n$ is a sigma-algebra contained in $\mathcal{F}$, and including one more random variable among its generators can only enlarge it. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$; the trivial ${\mathcal{F}}_0$ is contained in every sigma-algebra. These are exactly the requirements for a filtration.

The constant $X_0=x$ is ${\mathcal{F}}_0$-measurable. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable. The pair $(X_n,U_{n+1})$ is then ${\mathcal{F}}_{n+1}$-measurable, and the event $\{U_{n+1}\le X_n\}$ is ${\mathcal{F}}_{n+1}$-measurable because comparison of two real coordinates defines a Borel set. Define its indicator function by $B_{n+1}$. The recursion becomesIt is a measurable function of that pair. Mathematical induction proves $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$, so $(X_n)$ is an adapted process. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later integrability arguments.

Retain the binary random variables $B_j=1_{\{U_j\le X_{j-1}\}}$. Iterating the recursion givesEach binary history determines the full vector $(X_0,\dots ,X_n)$, so its sigma-algebra is finite by the sigma-algebra of a finite-valued random vector. Conversely, because $0<x<1$, the integer $2^n{X}_n-x$ uniquely determines the digits $B_1,\dots ,B_n$; also $B_j=2X_j-X_{j-1}$. ThusAll $2^n$ binary histories have positive probability. By independence of the driving uniform random variables, along a fixed finite history every intermediate state is in $(0,1)$, so each subsequent conditional branch probability, either $X_{j-1}$ or $1-X_{j-1}$, is strictly positive. It follows that ${\mathcal{F}}_n^X$ has $2^n$ nonempty fibres, each an atom of a sigma-algebra, and exactly $2^{2^n}$ elements.

Adaptedness from the preceding part gives ${\mathcal{F}}_n^X\subseteq {\mathcal{F}}_n$. For $n\ge 1$, however, ${\mathcal{F}}_n$ contains the events $\{U_1\le t\}$ for every $t\in [0,1]$. They have distinct probabilities $t$, so there are infinitely many distinct such events. Consequently the finite sigma-algebra ${\mathcal{F}}_n^X$ cannot equal ${\mathcal{F}}_n$. The inclusion is strict for every $n\ge 1$.

The adapted process is bounded by $1$, so every $X_n$ is an integrable random variable, indeed a square-integrable random variable. Independence of $U_{n+1}$ from ${\mathcal{F}}_n$ and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the ${\mathcal{F}}_n$-measurable value of $X_n$: for a fixed $t\in [0,1]$, the Lebesgue integral ${\int }_0^1{1}_{\{u\le t\}}du$ equals $t$. More formally, independence integrates the bounded Borel measurable function $(t,u)\mapsto 1_{\{u\le t\}}$ over its second coordinate. Thereforewhich verifies the martingale property for this filtration.

Since $B_{n+1}^2=B_{n+1}$, the conditional expectation of the squared martingale difference isThe law of total expectation yields $\mathbb{E}[(X_{n+1}-X_n{)}^2]=\frac{1}{4}\mathbb{E}[X_n(1-X_n)]$. In particular, $\mathbb{E}{X}_n=x$ for every $n$.

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give