Past exam of the martingales and financial mathematics course of the University of Oxford / 2011 / A / 3

A filtration is a nondecreasing family of sub-sigma-algebras: ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}\subseteq \mathcal{F}$. It represents information available up to each time.

A martingale relative to the filtration is a sequence $M_n$ such that $M_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|M_n|<\infty$, andIterating conditional expectation gives the analogous identity between any two ordered times.

A stopping time is a random time $\tau$ taking values in the nonnegative integers together with infinity and satisfying $\{\tau \le n\}\in {\mathcal{F}}_n$ for each $n$. Whether stopping has occurred by time $n$ can therefore be decided from information available then.

The optional stopping theorem says that for a martingale and bounded stopping times $\sigma \le \tau$,For unbounded almost surely finite stopping times, additional hypotheses are essential. Sufficient versions include uniform integrability of the stopped martingale, or an integrable stopping time together with uniformly bounded increments. In the latter case, stop at $\tau \wedge n$ and use $|M_{\tau \wedge n}|\le |M_0|+C\tau$ and the dominated convergence theorem. This is bounded-increment optional stopping with integrable time.

Put $q=1-ϵ<1$. The hypothesis implies $\mathbb{P}(\tau >n+N\mid {\mathcal{F}}_n)\le q$. Since $\{\tau >n\}\in {\mathcal{F}}_n$ and survival to $n+N$ implies survival to $n$, the tower property of conditional expectation givesApply this at $n=0,N,2N,\dots$ and use $\mathbb{P}(\tau >0)\le 1$ to obtainThis is a geometric tail bound from a uniform escape probability. The tail-sum formula for a nonnegative integer-valued variable now gives

Let $T$ be the first completion time of the word. A fresh gambler arrives before every throw with capital one, betting it all on the successive symbols $1,2,1,2$. Each correct bet multiplies capital by six and each wrong bet loses it, so every bet is fair. A gambler who completes the word keeps the winnings.

At time $T$, the gambler who bet on all four final throws has capital $6^4$. The gambler who bet on the final two throws has correctly matched the suffix $12$ and has capital $6^2$. There are no other survivors: the only nonempty suffixes of $1212$ that are also prefixes are $12$ and $1212$. Thus the total surviving capital is $6^4+6^2$.

To justify the expectation identity, independent blocks of four throws each have probability $6^{-4}$ of equalling the pattern, so $\mathbb{P}(T>4k)\le (1-6^{-4}{)}^k$ and $\mathbb{E}T<\infty$. Total gambling capital minus the number of arriving unit stakes is a martingale with bounded increments before $T$. Optional stopping therefore equates expected terminal capital with expected total stakes:This is the waiting time for a word in independent uniform symbols. The extra $6^2$ reflects the overlapping occurrence of the prefix $12$ at the word's end.