Martingales and Optional Stopping

Stationary distributions describe state probabilities. Here we use conditional expectation to forecast a process. Let F_n be the sigma-algebra of information through time n, with F_n contained in F_(n+1). A martingale M_n is adapted (M_n is F_n-measurable), has E[|M_n|] < infinity, and satisfies E[M_(n+1) | F_n] = M_n almost surely. Its conditional next-value mean is its current value. The tower property gives E[M_(n+1)] = E[M_n].

For the fair unit walk at s, ((s+1)+(s-1))/2=s, so its conditional drift is 0. The walk is adapted and integrable, hence a martingale. Its next squared-value mean is ((s+1)^2+(s-1)^2)/2=s^2+1. Thus S_n^2 is not a martingale, but S_n^2-n is: subtracting n cancels the drift of 1. Adding cn to a martingale creates conditional drift c. Absolute value is not a martingale for a walk starting at 0: its next mean is 1, not 0.

Symmetric increments are not required. Let B_n start at 0 with independent +1 increments of probability p and -1 increments of probability 1-p. Its conditional mean increment is mu = 2p-1. The adapted, integrable process B_n-n mu has conditional mean change 0, so it is a martingale. For p = 0.6, subtract 0.2n. The same test applies: calculate and cancel the conditional drift.

For a stopping time T, {T <= n} is determined by F_n for every n. The decision whether stopping has occurred uses only information available by then. For an integrable martingale, if T <= N for a fixed finite N, optional stopping gives E[M_T] = E[M_0]. Bounding E[T] alone does not meet this condition. One proof writes the stopped change as a finite sum of increments included only when play has not yet stopped.

A fixed time bound is one sufficient condition. Another allows unbounded waiting: T is a stopping time, finite almost surely, and |M_(T min n)| <= C for every n with one fixed finite C. Here T min n means the earlier time. Apply bounded-time stopping to T min n, then bounded convergence gives E[M_T] = E[M_0]. Separate integrability at each finite time does not justify this limit. These are sufficient conditions, not all versions of the theorem.

Stopping a fair unit walk at a fixed finite time preserves its starting mean k. At first exit from integer barriers L<U, stopped positions stay in [L,U]. Put d=U-L. From any interior state, d consecutive +1 steps force exit, with probability 2^(-d). Survival through r blocks is at most (1-2^(-d))^r, tending to 0. Exit is thus finite almost surely; bounded stopped values preserve mean k. Adjacent interior sites allow arbitrarily long alternating paths.

Non-anticipation alone also fails to preserve means. For a fair unit walk starting at 6, let T be the first hit of 8. This stopping time is finite almost surely but has no fixed time bound. S_T = 8, so E[S_T] = 8. Yet E[S_(T min n)] = 6 for every finite n. The capped stopped positions have no common bound because the walk can travel arbitrarily far below 6 before hitting 8.

In the doubling model, bet 1 on independent fair flips, double after each loss and stop at the first win. After r losses, cumulative profit is -(2^r-1), and the next stake is 2^r. A win recovers all losses and leaves profit 1. Capping play at r flips gives profit 1 with probability 1-2^(-r), or -(2^r-1) with probability 2^(-r). Multiplying and adding gives mean 0.

With unlimited funds, the first win occurs almost surely and terminal profit is 1. Every finite capped profit has mean 0, but the limit has mean 1. The rare large losses belong to the capped games; they are not terminal losses in the unlimited model. The stopped profits have no common absolute bound, so the bounded-convergence argument does not apply. A finite bankroll cannot cover every possible losing streak.

A bounded stopping result also applies to forecasts after an observation. If T is a stopping time with 1 <= T <= N for fixed finite N, then E[M_T | F_1] = M_1 for a martingale. For a fair walk starting at 0, observe a first step of +1. Its conditional expected position at any such stopping time is +1. The unconditional stopped mean remains 0; these are different averages.