Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 2 / c

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.

Use the nonnegative martingale $(Y_n)$ and filtration $({\mathcal{H}}_n)$ established above. Fix a finite $n$ and letFor $j<n$, $\{T\le j\}$ is the union of the threshold-crossing events up to $j$, so $T$ is a bounded stopping time. The optional stopping theorem for a martingale at a bounded stopping time gives $\mathbb{E}{Y}_T=\mathbb{E}{Y}_0=1$. On $H$ one has $Y_T\ge \lambda$; on its complement one still has $Y_T\ge 0$. ThereforeOnly a bounded stopping time has been used, so no extra uniform integrability assumption at an unbounded hitting time is implicit. This is also the Doob maximal inequality for a nonnegative submartingale specialized to a mean-one martingale.

For $p>1$, the bound $X_k\le 2$ ensures the finite moment $m_p=\mathbb{E}{X}_1^p\le 2^p$. By Jensen's inequality, $m_p\ge (\mathbb{E}{X}_1{)}^p=1$. The process $Z_k=Y_k^p$ is an adapted, nonnegative integrable random variable at each time, and independence givesThus $(Z_k)$ is a submartingale, and $\mathbb{E}{Z}_n=m_p^n$. The Doob maximal inequality for a nonnegative submartingale states that $c\mathbb{P}({\max }_{k\le n}{Z}_k\ge c)\le \mathbb{E}{Z}_n$ for $c>0$. Apply it with $c={\lambda }^p$. Since the power function is increasing on $[0,\infty )$,Combining this with the preceding bound givesThe result is a bound on the running maximum, which is why a maximal inequality for a submartingale, rather than an estimate of the single terminal random variable, is needed.^[ref]