Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 2

For the forward implication, let $(M_n)$ be a martingale and let the stopping time $T$ satisfy $0\le T\le N$ for a deterministic integer $N$. The stopped variable is integrable because it uses only finitely many integrable values. The pathwise identityexpresses it through martingale increments. Since $\{T\ge j\}=\{T>j-1\}\in {\mathcal{F}}_{j-1}$, each summand has expectation zero by conditional expectation. ConsequentlyThis proves the bounded-time case of the optional stopping theorem directly; no assumptions about an unbounded stopping time are being used.

For the converse in the characterization of a martingale by stopped expectations, fix $n\ge 0$ and $A\in {\mathcal{F}}_n$. Define the bounded stopping timeIndeed, its only nontrivial sublevel event is $\{T_A\le n\}=A^c\in {\mathcal{F}}_n$. Applying the assumed stopped-expectation equality to $T_A$ and to the deterministic stopping time $n$ givesThis holds for every $A\in {\mathcal{F}}_n$. Since $M_n$ is adapted and both variables are integrable, the defining test-event property of conditional expectation yieldsTogether with the given adaptation and integrability, this is exactly the martingale property.

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.

The mean increment is $\mathbb{E}{X}_1=-6/7+2/7=-4/7$, so $M_n=S_n+(4/7)n$ is a martingale. Applying the bounded optional stopping theorem at $n\wedge T$ givesThe terminal sums are bounded by the exit-state bounds from (b), while $n\wedge T$ increases to the integrable stopping time $T$. The dominated convergence theorem on the left and the monotone convergence theorem on the right therefore proveThis is also Wald's equation here. In fact, the stopped identity already implies $\mathbb{E}(n\wedge T)\le 7m/4$ from $S_{n\wedge T}\ge -m$, giving another direct finite-expectation justification.

Write $p_m=\mathbb{P}(S_T=m)$, $p_{m+1}=\mathbb{P}(S_T=m+1)$, and $q_m=p_m+p_{m+1}$. Part (b) givesso $q_m\le 2^{-m}$. The exit-state decomposition also givesCombining this with part (c), including the upper overshoot, yieldsFor example, the same calculation supplies the quantitative boundThe asymptotic linear growth is determined by the negative mean increment; the exponentially unlikely upper exit gives a vanishing correction.