Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 1

The Martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.