Past exam of the martingales and financial mathematics course of the University of Oxford / 2011 / A (Martingales Through Measure Theory)

A pi-system is a collection of subsets closed under binary intersections: $A,B\in \mathcal{I}$ implies $A\cap B\in \mathcal{I}$. Induction then gives closure under every nonempty finite intersection. A pi-system need not contain complements or unions.

The sigma-algebra generated by a collection is the smallest sigma-algebra containing that collection:The family being intersected is nonempty, since it contains the power set of $\Omega$. Intersections of sigma-algebras are sigma-algebras, so this definition has the asserted minimality.

The set function is countably additive if, whenever $A_1,A_2,\dots \in \mathcal{I}$ are pairwise disjoint and their union $A$ also belongs to $\mathcal{I}$,The equality is in the extended nonnegative reals. To be a measure one also imposes $\mu (\varnothing )=0$ and an appropriate domain such as a sigma-algebra.

The intersection of two nonempty members iswhich belongs to the collection. Any intersection involving the empty set is empty and is included too. Hence $\pi (\mathbb{R})$ is a pi-system.

Each half-line $(-\infty ,x]$ is a closed Borel set, and the empty set is Borel. Thus $\sigma (\pi (\mathbb{R}))\subseteq \mathcal{B}(\mathbb{R})$.

Conversely,The generated sigma-algebra contains both sets on the right and is closed under their difference, so it contains every open interval. By the given definition of the Borel sigma-algebra, the reverse inclusion follows. Therefore

Two sigma-algebras are independent if every event in one is independent of every event in the other:The same formula defines independence of the generating collections when restricted to their members.

Independence of the full sigma-algebras immediately gives independence of their generating pi-systems. For the converse, fix $H\in {\mathcal{H}}_0$. On $\mathcal{G}$, define finite measuresThey have the same total mass $\mathbb{P}(H)$ and agree on ${\mathcal{G}}_0$ by assumption. The supplied measure-uniqueness theorem makes them equal on $\mathcal{G}$. Thus the independence formula holds for every $G\in \mathcal{G}$ and every $H\in {\mathcal{H}}_0$.

Now fix such an arbitrary $G$. The measures $H\mapsto \mathbb{P}(G\cap H)$ and $H\mapsto \mathbb{P}(G)\mathbb{P}(H)$ on $\mathcal{H}$ have equal total mass and agree on ${\mathcal{H}}_0$. A second application gives equality on all of $\mathcal{H}$. This proves independence extended from generating pi-systems.

For each $i$, the collection of events $\{X_i\le x\}$, together with the empty event, is a pi-system. Its generated sigma-algebra is $\sigma (X_i)$, because the half-lines generate the Borel sigma-algebra and inverse images preserve intersections and the sigma-algebra operations.

The displayed probability identity says precisely that these two pi-systems are independent. Part (ii) makes $\sigma (X_1)$ and $\sigma (X_2)$ independent, which is the definition of independent random variables. Conversely independence of the random variables applies to the half-line events and gives the identity. Thus the joint cumulative distribution function factors exactly when the variables are independent.

Convergence in probability meansThe probability concerns each index separately; it does not require eventual closeness along almost every sample path.

For $1\le p<\infty$, convergence in Lp means the variables belong to $L^p$ andEquivalently the $p$th absolute moment of the difference tends to zero. The analogous definition for $p=\infty$ uses the essential-supremum norm.

Almost sure convergence meansOutside a single null set, the numerical sequence converges pointwise.

For events $E_n$, the Borel-Cantelli first lemma states that ${\sum }_n\mathbb{P}(E_n)<\infty$ implies $\mathbb{P}(E_n\text{ infinitely often})=0$. Indeed,These unions decrease to $\mathrm{limsup}{E}_n$, so continuity from above proves the assertion.

The Borel-Cantelli second lemma states that independent events with ${\sum }_n\mathbb{P}(E_n)=\infty$ satisfy $\mathbb{P}(E_n\text{ infinitely often})=1$. For every fixed $m$, independence and $1-x\le e^{-x}$ giveContinuity from above shows there is probability zero of no event after $m$. A countable union over $m$ gives probability zero of only finitely many occurrences. Independence is needed for this second conclusion, but not for the first.

For fixed $\epsilon >0$ and $n>\epsilon$,so there is convergence in probability to one. The absolute difference is exactly $n$ on the exceptional event, henceFor every $p>1$ this tends to infinity, so convergence in Lp fails. In contrast its first moment is $1/\log n\to 0$, so convergence actually holds in $L^1$.

There are exactly $n$ summands in $S_n$, and the triangle inequality givesThe last step is convergence of arithmetic means of a sequence tending to zero. Thus $S_n/n\to 1$ in $L^1$, and Markov's inequality gives

Let $E_n=\{|X_n-1|=n\}$. These events are independent and ${\sum }_{n\ge 2}\mathbb{P}(E_n)={\sum }_{n\ge 2}1/(n\log n)=\infty$. By the Borel-Cantelli second lemma, they occur infinitely often almost surely.

If $S_n/n$ had any finite sample-path limit, thenBut on $E_{n+1}$, $|X_{n+1}|/n$ is either $1$ or $(n+2)/n$, contradicting this convergence infinitely often. Therefore the averages have no finite almost sure limit. This is a rare-jump counterexample to a strong law with convergent first-moment averages.

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.