Past exam of the martingales and financial mathematics course of the University of Oxford / 2010 / A (Martingales Through Measure Theory)

A measurable space is a set $E$ together with a sigma-algebra $\mathcal{E}$ of its subsets: it contains $E$ and is closed under complements relative to $E$ and countable unions. A measure is a map $\mu :\mathcal{E}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity on disjoint measurable sets:A finite measure additionally satisfies $\mu (E)<\infty$.

For $\epsilon >0$, define the measurable decreasing tail eventsIf $f_n\to f$ almost everywhere, every point outside its null exceptional set eventually leaves each such tail event, so $\mu ({\bigcap }_n{A}_n(\epsilon ))=0$. Continuity from above of a measure applies because $\mu (E)<\infty$, giving $\mu (A_n(\epsilon ))\to 0$.

Conversely, assume these measures tend to zero. For each positive integer $k$, the intersection ${\bigcap }_n{A}_n(1/k)$ has measure zero. Outside the union of these countably many null sets, every $k$ has an index after which $|f_m-f|\le 1/k$ for all later $m$. Hence $f_m\to f$ there. This proves the desired equivalence. The argument uses all positive thresholds; replacing the intersection of tail events by a single strict limsup inequality at the same threshold would mishandle equality cases.

The triangle inequality gives the event inclusionCountable subadditivity of a measure bounds its measure by the sum of the two terms on the right. Both tend to zero as $m,n\to \infty$ by convergence in measure. Thus convergence in measure implies the Cauchy property in measure.

The Cauchy assumption lets us choose $n_k$ recursively, increasing strictly, so that $\mu (|f_n-f_m|>2^{-k})<2^{-k}$ for all $m,n\ge n_k$. In particular the measurable eventshave summable measures. The measure of ${\bigcup }_{k\ge K}{E}_k$ is at most ${\sum }_{k\ge K}{2}^{-k}\to 0$, so $\mathrm{limsup}{E}_k$ is a null set. Outside it, the successive differences are eventually bounded by a summable geometric series. Hence $f_{n_k}$ is pointwise Cauchy and has a finite real limit.

Define $f$ to be that limit wherever the sequence converges, and zero elsewhere. The convergence set is measurable by the countable Cauchy criterion, and the limit is measurable as a limit of measurable functions on that set. Thus there is no need to assume the measure is complete. Part (a)(ii), or the corresponding finite-measure implication, gives $f_{n_k}\to f$ in measure.

For the full sequence useChoose a sufficiently late subsequence index so the second term is small; the Cauchy property then bounds the first uniformly for all large $n$. Therefore $f_n\to f$ in measure. This proves completeness of convergence in measure on a finite measure space.

Kolmogorov zero-one law states that every event in the tail sigma-algebra of an independent sequence has probability zero or one. Removing any fixed number of summands changes $S_n/n$ by a finite random constant divided by $n$, so does not change its extended-real limsup. Therefore $\overline{X}$ is measurable with respect to the tail sigma-algebra. Each event $\{\overline{X}\le q\}$ for rational $q$ has probability zero or one. Monotonicity of these events forces the distribution to be concentrated at one extended-real value, allowing either infinity.

The strong law of large numbers says that for independent identically distributed variables with $\mathbb{E}|X_1|<\infty$,

Pointwise, the positive integers $n$ with $Z\ge n$ are exactly $1,\dots ,\lfloor Z\rfloor$. Thus $Y=\lfloor Z\rfloor ={\sum }_{n\ge 1}{1}_{\{Z\ge n\}}$. The monotone convergence theorem gives $\mathbb{E}Y={\sum }_{n\ge 1}\mathbb{P}(Z\ge n)$. Since $Y\le Z<Y+1$,These inequalities also hold for infinite expectations. This is the tail-sum formula comparison for a nonnegative real variable.

For any $\lambda >0$, the variable $Z=|X_1|/\lambda$ has infinite expectation. Part (a)(ii) therefore forces ${\sum }_n\mathbb{P}(|X_1|\ge \lambda n)=\infty$. Identical distributions give

For each positive integer $m$, the events $\{|X_n|\ge mn\}$ are independent and have divergent total probability by part (i). The Borel-Cantelli second lemma makes them occur infinitely often. Taking the countable intersection over $m$ givesIf ${\mathrm{limsup}}_n|S_n|/n$ were finite along any such sample path, then for some finite path-dependent $C$, $|S_n|\le Cn$ eventually. But $X_n=S_n-S_{n-1}$ would give $|X_n|/n\le C+C(n-1)/n\le 2C$ eventually, a contradiction. ThereforeThis is the divergence of normalized iid partial sums without a first absolute moment.

The Borel-Cantelli second lemma says that independent events whose probabilities have divergent sum occur infinitely often almost surely. The events here are independent and have probabilities $\min \{1,\lambda /n\}$. Their sum diverges by the harmonic series, so

The printed update needs a one-step index correction to match its filtration: the draw for the transition from $n$ to $n+1$ must be the fresh variable $U_{n+1}$. Also the trivial initial sigma-algebra is $\{\varnothing ,\Omega \}$, rather than the printed $\{\varnothing ,\mathcal{F}\}$. With literal $U_n$, the next increment is already known at time $n$, so $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_{n+1}$; the nontrivial update differs from $X_n$. Thus the stated martingale claim is false with that indexing. The following uses the standard Pólya urn described in words.

Put $m=B_0+R_0$, so the total count is $m+n$. Conditional on ${\mathcal{F}}_n$, the next blue indicator $I_{n+1}$ has expectation $X_n$. HenceAdaptedness and integrability follow from the recursive construction and $0\le X_n\le 1$. Thus $X_n$ is a bounded martingale. The martingale convergence theorem gives an almost-sure limit $X_{\infty }\in [0,1]$. Boundedness and the dominated convergence theorem then give $\mathbb{E}|X_n-X_{\infty }{|}^2\to 0$, establishing convergence in $L^2$.

The blue count never decreases, so $X_n\ge B_0/(m+n)$. For all sufficiently large $n$, this is at least $B_0/[2(n+1)]$. Whenever the fresh draw satisfies $U_{n+1}\le B_0/[2(n+1)]$, a blue ball is added. Part (a), with $\lambda =B_0/2$, says that such events occur infinitely often almost surely. Therefore $B_n\to \infty$. Consequently

Since $X_n>0$, the pointwise inequalityholds: on that event both sides equal one, and elsewhere the left side is zero. Take conditional expectation and pull out the positive ${\mathcal{F}}_n$-measurable denominator to obtain the requested bound. The supplied variance estimate improves it toalmost surely, by part (ii). These conditional probabilities are bounded by one. The dominated convergence theorem and tower property of conditional expectation giveThis proves Positivity of the limiting Pólya urn proportion without needing its explicit beta distribution.