Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 1

For a martingale and a stopping time $T\le N$, expand the stopped value using its martingale differences:The event $\{T\ge k\}=\{T>k-1\}$ belongs to the filtration ${\mathcal{F}}_{k-1}$. Each summand is integrable and its conditional expectation given ${\mathcal{F}}_{k-1}$ is zero. Taking expectations of this finite sum therefore proves $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ for every bounded stopping time. This proves the forward direction of the bounded-stopping-time characterization of a martingale without assuming an optional stopping theorem.

For $A\in {\mathcal{F}}_n$, define the bounded stopping time $T=n+1$ on $A$ and $T=n$ on $A^c$. Before time $n$ its stopping events are empty, at time $n$ the event $\{T\le n\}=A^c$ belongs to ${\mathcal{F}}_n$, and subsequently it has stopped everywhere. Subtract the assumed mean identity for this $T$ from that for the constant stopping time $n$ to obtainSince $M_n$ is integrable and ${\mathcal{F}}_n$-measurable, this is precisely the defining property of conditional expectation giving $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus the adapted process is a martingale, completing the bounded-stopping-time characterization of a martingale.

The required condition is uniform integrability, written entirely in terms of the given probability laws:Together with the given almost sure convergence, this implies L1 convergence to an integrable $M_{\infty }$. Passing to the limit in $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$, using the contraction property of conditional expectation in L1 space, gives $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For every bounded stopping time $S$, the same identity holds with ${\mathcal{F}}_S$: split an event of its stopping-time sigma-algebra according to $\{S=k\}$ and use the deterministic-time identity. In particular $M_{T\wedge n}=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_{T\wedge n}]$. These conditional expectations are uniformly integrable: for an integrable $Y$ and $Z=\mathbb{E}[Y\mid \mathcal{G}]$, the event $A=\{|Z|>K\}$ satisfies $\mathbb{P}(A)\le \mathbb{E}|Y|/K$, andFirst choose $R$, then $K$, uniformly in $\mathcal{G}$. Define $M_T=M_{\infty }$ on $\{T=\infty \}$. Now $M_{T\wedge n}\to M_T$ almost surely and in L1 space, so the bounded-time mean identities imply $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ even when $T$ can be infinite.

With the natural filtration ${\mathcal{F}}_n=\sigma (Z_1,\dots ,Z_n)$, the sums form a martingale. Put $v_n={\sum }_{k\le n}{a}_k^2$. Independence and the characteristic function of a standard normal random variable giveIf $M_n\to M_{\infty }$ almost surely with a finite limit, dominated convergence gives convergence of these characteristic functions to $\mathbb{E}{e}^{iu{M}_{\infty }}$. If $v_n\to \infty$, that limit would be $1$ at zero and $0$ at every nonzero $u$. A characteristic function is continuous at zero: apply dominated convergence to $e^{iu{M}_{\infty }}$ as $u\to 0$. This contradiction provesThis is also sufficient by the L2-bounded martingale convergence theorem, since $\mathbb{E}{M}_n^2=v_n$. Thus the almost sure convergence criterion for an independent Gaussian series is an equivalence.

The exact condition is again ${\sum }_k{a}_k^2<\infty$, with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case ${\sup }_n\mathbb{E}{M}_n^2<\infty$, andproves uniform integrability. The preceding unbounded optional stopping argument yields $\boxed{\mathbb{E}{M}_T=0}$.

For necessity, suppose $v_n\to \infty$. For every fixed $R>0$, the normal distribution of $M_n$ gives $\mathbb{P}(M_n\ge R)\to 1/2$. Hence $\mathbb{P}({\sup }_n{M}_n\ge R)\ge 1/2$, and intersecting over positive integer $R$ shows $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. Unboundedness above is unchanged by altering finitely many of the independent $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a tail event; the Kolmogorov zero-one law makes its probability one. Consequently $T<\infty$ almost surely and $M_T\ge 1$. Its expectation is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in threshold stopping of an independent Gaussian series.