Optional Sampling

The stopped-process theorem gave $\E[X_{N \wedge n}] = \E[X_0]$ for every fixed time $n$. The optional sampling theorem asks when this survives the limit $n \to \infty$, so that the fair-game identity holds at the stopping time itself: $\E[X_N] = \E[X_0]$. The gap is an interchange of limit and expectation, and the sufficient condition is uniform integrability, or, more simply, a bounded stopping time.

Optional sampling theorem

Recall that a sequence $\{X_n\}$ is uniformly integrable (U.I.) if

$$\sup_n \E\big[ |X_n| \, \mathbb{1}_{\{|X_n| > M\}} \big] \to 0 \qquad \text{as } M \to \infty,$$

that is, no mass escapes to infinity along the sequence. The point of the condition is the Vitali convergence theorem: almost-sure convergence together with uniform integrability upgrades to $L^1$ convergence, which is exactly what lets us pass a limit through an expectation.

Theorem: Optional Sampling

Statement

Let $\{X_n\}$ be a uniformly integrable martingale and let $N$ be a stopping time that is finite almost surely. Then

$$\E[X_N] = \E[X_0].$$

If $\{X_n\}$ is a submartingale instead, the equality weakens to $\E[X_0] \le \E[X_N]$.

Proof

We give the submartingale case; the martingale case is identical with every $\le$ an equality.

1. The stopped process gives the identity at finite times. By the stopping theorem, $\{X_{N \wedge n}\}$ is a submartingale, so its mean is non-decreasing from its starting value:

   $$\E[X_0] \le \E[X_{N \wedge n}] \qquad \text{for every } n. \tag{1}$$

2. Pointwise limit. Since $N < \infty$ almost surely, for almost every $\omega$ we eventually have $n > N(\omega)$, at which point $X_{N \wedge n}(\omega) = X_N(\omega)$. Hence $X_{N \wedge n} \to X_N$ almost surely.

3. The stopped sequence is uniformly integrable. Split it at the stopping time,

   $$X_{N \wedge n} = X_n \, \mathbb{1}_{\{n \le N\}} + X_N \, \mathbb{1}_{\{n > N\}}.$$

   The first term is dominated by $|X_n|$ drawn from the U.I. family $\{X_n\}$, and the second by the single integrable variable $|X_N|$. A sum of uniformly integrable families is uniformly integrable, so $\{X_{N \wedge n}\}_n$ is U.I.

4. Upgrade to $L^1$. Almost-sure convergence (step 2) plus uniform integrability (step 3) gives $L^1$ convergence by the Vitali convergence theorem. In particular the means converge:

   $$\E[X_{N \wedge n}] \to \E[X_N].$$

5. Pass to the limit. Letting $n \to \infty$ in $(1)$ gives $\E[X_0] \le \E[X_N]$, with equality throughout in the martingale case.

Remarks

1. The same argument with the inequalities reversed gives the supermartingale version, $\E[X_N] \le \E[X_0]$.
2. The proof used uniform integrability only of the stopped sequence $\{X_{N \wedge n}\}$, never of $\{X_n\}$ itself. This weaker requirement is what the corollary below exploits.

When the stopping time is bounded, the stopped sequence is uniformly integrable for free (it is eventually constant), and the theorem applies with no integrability hypothesis at all.

Corollary: Bounded Stopping Times

Statement

Let $\{X_n\}$ be a martingale (resp. submartingale) and let $N$ be a bounded stopping time, meaning $N \le M$ almost surely for some constant $M$. Then

$$\E[X_N] = \E[X_0] \qquad (\text{resp. } \E[X_0] \le \E[X_N]),$$

with no uniform integrability needed.

Proof

Since $N \le M$ almost surely, $N \wedge M = N$. Applying the stopped-process identity at the single fixed time $n = M$ gives

$$\E[X_0] \le \E[X_{N \wedge M}] = \E[X_N],$$

with equality in the martingale case. No limit is taken, so uniform integrability never enters.