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

A real-valued random variable is a measurable function $Y:(\Omega ,\mathcal{F})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$: equivalently, $\{Y\le t\}\in \mathcal{F}$ for every real $t$.

For $p=1$ the asserted inequality is equality. For $p>1$, let $q=p/(p-1)$. If $\mathbb{E}{Y}^p=\infty$, the inequality is automatic. Otherwise Hölder's inequality on the probability space givesThus $\boxed{\Vert Y{\Vert }_1\le \Vert Y{\Vert }_p}$, using that the total measure is one.

Write ${\Delta }_0=X_0$ and ${\Delta }_j=X_j-X_{j-1}$ for $j\ge 1$. If $i<j$, then ${\Delta }_i$ is ${\mathcal{F}}_{j-1}$-measurable. The martingale property and linearity of conditional expectation giveThe product ${\Delta }_i{\Delta }_j$ is integrable by Cauchy-Schwarz inequality. The pull-out property of conditional expectation, followed by its tower property, therefore givesFor unbounded ${\Delta }_i$, the pull-out identity follows by truncation and Cauchy-Schwarz inequality. This proves martingale-difference orthogonality, including orthogonality to the initial value.

Expand $X_n={\sum }_{j=0}^n{\Delta }_j$ in the Hilbert space $L^2$. Orthogonality gives the exact identityThe terms are nonnegative and $\mathbb{E}{X}_0^2<\infty$, so boundedness of the left side over $n$ is equivalent to convergence of the series on the right.

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 $a$.