Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 33 / 4

For a nonnegative submartingale $(M_k{)}_{0\le k\le n}$ and $M_n^{\ast }={\max }_{0\le k\le n}{M}_k$, the weak form of the Doob maximal inequality isTo prove it, let $A_k$ be the event that $k$ is the first index with $M_k\ge u$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $A=\{M_n^{\ast }\ge u\}$. The submartingale property gives $\mathbb{E}[M_n\mid {\mathcal{F}}_k]\ge M_k$. Multiplying by $1_{A_k}$, taking expectations and summing yields $\mathbb{E}[M_n{1}_A]\ge {\sum }_k\mathbb{E}[M_k{1}_{A_k}]\ge u\mathbb{P}(A)$. Nonnegativity gives the second inequality.

The associated Doob Lp maximal inequality, for $p>1$ and $M_n\in L^p$, isIndeed $M_k\le \mathbb{E}[M_n\mid {\mathcal{F}}_k]$, so conditional Jensen inequality first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $p{u}^{p-2}du$ and using the Tonelli theorem givesThe last step is Holder inequality. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary martingale, its absolute value is a nonnegative submartingale by conditional Jensen inequality, so both corresponding absolute-maximum bounds follow.

The moment-generating function of a normal distribution givesIndependence of $X_{n+1}$ from ${\mathcal{F}}_n$ impliesEvery exponential here is integrable. Thus the process is a martingale exactly when $-\lambda +{\lambda }^2/2=0$. Its two roots are $0$ and $2$, and the unique positive root isThe resulting exponential martingale of a random walk is nonnegative and has expectation $1$ at every finite time.

The strong law of large numbers gives $S_n/n\to -1$ almost surely, so $S_n\to -\infty$. Therefore the infinite-horizon maximum $Z$ is finite and attained almost surely; also $Z\ge 0$ because $S_0=0$.

Apply the Doob maximal inequality to $M_k=e^{2S_k}$ through time $n$. Since $\mathbb{E}{M}_n=1$,These finite-horizon events increase to $\{Z>t\}$. Continuity of probability under increasing unions givesFor $t<0$ the same bound is trivial, because its right side exceeds $1$. Only bounded-horizon maximal inequalities were used before passing to the limit; no unjustified stopping at an infinite-horizon crossing time is needed.

For $0<b<\lambda =2$, integrate the identity $e^{bz}-1={\int }_0^{z}b{e}^{bt}dt$ for $z\ge 0$. The Tonelli theorem and the tail bound in (c) giveFor $b=0$ the expectation is $1$, and for $b<0$ the fact $Z\ge 0$ gives $0<e^{bZ}\le 1$. ThereforeThis is an exponential moment bound for a Gaussian random-walk maximum. The upper-tail bound proves the requested sufficient range; it alone does not establish a converse at the endpoint.