Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 28 / 1

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Use ${\mathcal{F}}_m=\sigma ({\xi }_1,\dots ,{\xi }_m)$ and $S_0=0$. Independence and zero means imply $\mathbb{E}[S_{m+1}\mid {\mathcal{F}}_m]=S_m$, so the partial sums form a square-integrable martingale. More explicitly,Thus $S_m^2$ is a nonnegative submartingale. The Doob maximal inequality at threshold $x^2$ now proves the Kolmogorov maximal inequality:Finally, expanding the terminal square and using independence and the zero means gives $\mathbb{E}{S}_n^2={\sum }_{j=1}^n\mathbb{E}{\xi }_j^2$. Identical distributions are not required.

Put $V=\mathbb{E}{X}_n^2$. For $c\ge 0$, the process $(X_m+c{)}^2$ is a nonnegative submartingale, by the conditional Jensen inequality. A crossing of $x$ by $X$ implies a crossing of $(x+c{)}^2$ by this squared process. Since $\mathbb{E}{X}_n=\mathbb{E}{X}_0=0$, the Doob maximal inequality givesFor $V>0$, differentiation of the last expression gives $2(cx-V)/(x+c{)}^3$, so its minimum over $c\ge 0$ occurs at $c=V/x$. Substitution proves the one-sided martingale maximal inequalityIf $V=0$, then $X_n=0$ almost surely and $X_m=\mathbb{E}[X_n\mid {\mathcal{F}}_m]=0$ for every $m\le n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has mean zero, variance $V$, and achieves equality.