Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 1 / b

The Doob maximal inequality for a nonnegative submartingale states that, for a nonnegative submartingale $(M_j{)}_{0\le j\le N}$ and $u>0$,For completeness, partition the crossing event according to its first crossing time $j$. On that event $M_j\ge u$, while $\mathbb{E}[M_N\mid {\mathcal{F}}_j]\ge M_j$. Summing the corresponding expectations over $j$ proves $u\mathbb{P}(\max M_j\ge u)\le \mathbb{E}[M_N{1}_{\{\max M_j\ge u\}}]\le \mathbb{E}{M}_N$.

If $q>p$, the exponential $r^{S_j}$, with $r=q/p>1$, is increasing as a function of $S_j$. Apply the inequality to the martingale from part (a), with $u=r^k$:Letting $N\to \infty$ and using continuity of probability measures on increasing events givesThe same event is obtained if the supremum excludes time zero, since $k\ge 1$. If $p\ge q$, the right side is at least one and the bound is automatic; the nontrivial Doob maximal inequality argument is for $q>p$.

For $q>p$, let $M={\sup }_{n\ge 0}{S}_n$, including $S_0=0$. It is a nonnegative integer-valued random variable, possibly infinite before the following estimate. The tail-sum formula for expectation and part (i) giveThus $\boxed{\mathbb{E}\underset{n\ge 0}{\sup }{S}_n\le p/(q-p)}$, and the supremum is finite almost surely. The sum identity follows by writing $M={\sum }_{k\ge 1}{1}_{\{M\ge k\}}$ and applying monotone convergence theorem. If the printed supremum is read literally over $n\ge 1$, its positive part has exactly the same positive-level tails and its negative part is bounded by one. The same upper bound on its expected value remains valid.