Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 1

For a discrete-time martingale $(M_n)$ and real $a<b$, let $U_N[a,b]$ be the upcrossing count: buy on the first observation at or below $a$, sell on the next observation at or above $b$, and repeat up to time $N$, counting only completed pairs. Writing $z^{-}=max(-z,0)$, the Doob upcrossing inequality isThis convention permits the first purchase at time zero. Equivalently, the right side is $\mathbb{E}(M_N-a{)}^{+}-\mathbb{E}(M_0-a)$, because the martingale has constant expected value.

For completeness, let $H_k\in \{0,1\}$ indicate whether the trading rule holds one unit during $(k-1,k]$. This is a predictable process, so the martingale transform $G_N={\sum }_{k=1}^N{H}_k(M_k-M_{k-1})$ has zero expected value. Every completed trade gains at least $b-a$, and an unfinished trade loses at most $(M_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(M_N-a{)}^{-}$, giving the inequality.

The martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

Set $M_n={\sum }_{j=0}^n{X}_j$ and ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. The independence and zero expected values give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$, so $(M_n)$ is a martingale. The variance additivity for independent random variables givesThe Cauchy-Schwarz inequality implies ${\sup }_n\mathbb{E}|M_n|\le \sqrt{V}$. Applying the martingale convergence theorem from (b),In fact the L2 martingale convergence theorem also gives convergence in L2. Independently, for $n>m$ the same variance additivity for independent random variables yields $\mathbb{E}|M_n-M_m{|}^2={\sum }_{j=m+1}^n\mathbb{E}[X_j^2]\to 0$, confirming that the partial sums form a Cauchy sequence in $L^2$.

Let ${\xi }_1,{\xi }_2,\dots$ be independent and identically distributed random variables with the fair Bernoulli distribution, and putRelative to ${\mathcal{F}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$, this fair-coin doubling martingale satisfies $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$: while alive it doubles with probability $1/2$ and becomes zero otherwise. Its expected value is $\mathbb{E}{M}_n=2^n{2}^{-n}=1$.

The probability of an infinite run of ones is ${\lim }_n{2}^{-n}=0$, so $M_n$ is eventually zero almost surely. However,Thus the martingale lacks convergence in L1 to its limit from almost sure convergence. Nor can it have convergence in L1 to any other limit: convergence in L1 implies convergence in probability, whose limit is unique up to almost sure equality.