Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 29 / 1

Write $z^{-}=\max (-z,0)$. For $a<b$, let $U_N[a,b]$ be the upcrossing count obtained by successively buying at an observation at or below $a$ and selling at an observation at or above $b$, with the first purchase allowed at time zero. The Doob upcrossings lemma for an integrable discrete-time martingale isThe weaker last bound is also the usual Doob upcrossing inequality.

To prove it, hold one unit between each purchase and sale, and zero units otherwise. If $H_k$ records the holding over $(k,k+1]$, the decision is ${\mathcal{F}}_k$-measurable, with $H_k\in \{0,1\}$. The resulting martingale transform has gainEach completed upcrossing earns at least $b-a$. An unfinished holding loses at most $(X_N-a{)}^{-}$. If a purchase occurs at zero, its price is $X_0\le a$, giving the additional discount $(X_0-a{)}^{-}$, whether that holding has been sold or remains open. If no purchase occurs at zero, this extra term is zero. Therefore, path by path,Taking expectations proves the bound.

The Martingale convergence theorem, stated without proof, says that a discrete-time martingale satisfying ${\sup }_n\mathbb{E}|X_n|<\infty$ converges almost surely to a finite integrable random variable $X_{\infty }$. In particular every nonnegative martingale converges almost surely to a finite limit. Under uniform integrability, the uniformly integrable martingale convergence theorem additionally gives convergence in L1 and $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$.

A martingale has uniform integrability when its entire family of time values has uniformly small integrable tails:Here every $X_n$ is an integrable random variable. This implies ${\sup }_n\mathbb{E}|X_n|<\infty$, because the contribution from $|X_n|\le K$ is at most $K$. Mere boundedness of these expectations does not imply uniform integrability.

Put $Z_n=X_{n\wedge T}$. The stopped martingale in discrete time identityshows that $Z$ is a martingale: the indicator function is ${\mathcal{F}}_n$-measurable and the increment has zero conditional expectation. Integrability follows because $Z_n$ is selected from the finitely many integrable values $X_0,\dots ,X_n$.

For the bounded stopping time $\tau =n\wedge T$, the optional stopping theorem in its conditional form givesHere ${\mathcal{F}}_{\tau }$ is the stopping-time sigma-algebra. Indeed, for $A\in {\mathcal{F}}_{\tau }$, partition $A$ into $A\cap \{\tau =k\}\in {\mathcal{F}}_k$ and apply the martingale identity $\mathbb{E}[X_n\mid {\mathcal{F}}_k]=X_k$ on each piece. This proves the displayed conditional expectation identity directly.

Let $C={\sup }_n\mathbb{E}|X_n|<\infty$ and $A_n=\{|Z_n|>K\}\in {\mathcal{F}}_{n\wedge T}$. Conditional absolute-value domination and the Markov inequality give $\mathbb{P}(A_n)\le C/K$ and, for any $R>0$,First choose $R$ using the uniform integrability of $X$, and then choose $K$. The estimate is uniform in $n$, proving uniform integrability of a stopped uniformly integrable martingale. No finiteness assumption on $T$ is needed.