Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / A / 2 / b

Conditional on ${\mathcal{F}}_n$, put $x=X_n$. The two possible values have conditional meanThe process is adapted and bounded, hence integrable. Consequently it is a martingale. Since ${\sup }_n\mathbb{E}|X_n|\le 1$, the martingale convergence theorem gives an almost surely finite limit $X_{\infty }$, necessarily in $[0,1]$.

Moreover $|X_n-X_{\infty }{|}^2\le 1$ and these squared differences tend to zero almost surely. The dominated convergence theorem gives $\mathbb{E}|X_n-X_{\infty }{|}^2\to 0$. Thus the Bernoulli polarization martingale converges almost surely and in $L^2$.

Conditional on ${\mathcal{F}}_n$, the increment is $-X_n/2$ with probability $1-X_n$, and $(1-X_n)/2$ with probability $X_n$. ThereforeTaking expectations using the tower property yieldsThis is the conditional variance that drives the Bernoulli polarization martingale toward its endpoints.

The martingale is bounded in $L^2$, so the increment-square series from part (a)(ii) is finite. The identity just obtained implies ${\sum }_n\mathbb{E}[X_n(1-X_n)]<\infty$, hence $\mathbb{E}[X_n(1-X_n)]\to 0$. By almost sure convergence and the dominated convergence theorem,The integrand is nonnegative because $X_{\infty }\in [0,1]$. It must vanish almost surely, giving $X_{\infty }\in \{0,1\}$.

The martingale has constant expectation $a$. Its $L^2$ convergence implies $L^1$ convergence by Cauchy-Schwarz inequality, and hence $\mathbb{E}{X}_{\infty }=a$. ThereforeThe limiting law is the Bernoulli distribution with parameter $a$.