Why Unbounded Stopping Requires a Limit Argument
For a bounded stopping time, optional sampling relates the expected value of a martingale at the stopping time to its initial value. If the stopping time can be arbitrarily large, the usual route is to truncate it: for a stopping time \(\tau\), consider \(\tau\wedge n\). Each truncated time is bounded, so bounded optional sampling applies. The remaining issue is whether the expectations at \(\tau\wedge n\) converge to the expectation at \(\tau\). Almost-sure convergence alone does not justify that passage. Uniform integrability supplies the needed control.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), when Brownian motion is used below. A martingale \(M_t\) is assumed adapted, integrable at each deterministic time, and to satisfy \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). We take martingale paths to be right-continuous. A stopping time \(\tau\) is finite almost surely if \(\mathbb{P}(\tau<\infty)=1\).
Optional Stopping Under Uniform Integrability
Proof. We first recall the bounded optional sampling result: for a right-continuous martingale and bounded stopping times \(\rho\le\eta\), \(\mathbb{E}[M_\eta\mid\mathcal{F}_\rho]=M_\rho\). Apply it to \(\rho=s\wedge\tau\) and \(\eta=t\wedge\tau\), for deterministic \(s\le t\). Both are bounded, so \(N_t=M_{t\wedge\tau}\) is a martingale. By assumption its values form a uniformly integrable family. Since \(\tau\) is finite almost surely, \(N_t\to M_\tau\) almost surely as \(t\to\infty\).
We use the fact that a uniformly integrable sequence that converges almost surely also converges in \(L^1\). To see the relevant argument, truncate each variable at a fixed level \(K\). The truncated variables are bounded and converge almost surely, so their differences converge to zero in \(L^1\) by dominated convergence. The expected tails can be made uniformly small by uniform integrability; Fatou's lemma gives the same tail control for the limit. Taking \(K\) large and then the sequence index large proves \(L^1\) convergence. Applying this to \(N_n\) gives \(N_n\to M_\tau\) in \(L^1\). Since \(\mathbb{E}[N_n]=\mathbb{E}[M_0]\) for every integer \(n\), \(L^1\) convergence gives \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\).
For the conditional assertion, write \(X=M_\tau\). For each deterministic \(t\), the martingale property of \(N\), followed by its \(L^1\) convergence, gives \(N_t=\mathbb{E}[X\mid\mathcal{F}_t]\). The bounded optional sampling result then gives \(N_{\sigma\wedge n}=\mathbb{E}[X\mid\mathcal{F}_{\sigma\wedge n}]\). These conditional expectations are uniformly integrable: for any integrable \(X\), the family \(\{\mathbb{E}[X\mid\mathcal{G}]\}\), over sub-\(\sigma\)-fields \(\mathcal{G}\), is uniformly integrable. Indeed, if \(Y=\mathbb{E}[X\mid\mathcal{G}]\), then \(|Y|\le\mathbb{E}[|X|\mid\mathcal{G}]\); splitting \(|X|\) at a fixed cutoff controls the tail of \(Y\), uniformly in \(\mathcal{G}\). Since \(\sigma\) is finite, \(N_{\sigma\wedge n}\to N_\sigma=M_\sigma\) almost surely and therefore in \(L^1\). For every \(A\in\mathcal{F}_\sigma\), \(A\cap\{\sigma\le n\}\in\mathcal{F}_{\sigma\wedge n}\). Thus \(\mathbb{E}[\mathbf{1}_{A\cap\{\sigma\le n\}}X]=\mathbb{E}[\mathbf{1}_{A\cap\{\sigma\le n\}}N_{\sigma\wedge n}]\). Let \(n\to\infty\); integrability of \(X\), \(L^1\) convergence, and \(\mathbb{P}(\sigma<\infty)=1\) yield \(\mathbb{E}[\mathbf{1}_A X]=\mathbb{E}[\mathbf{1}_A M_\sigma]\). This is the defining conditional-expectation identity, proving the claim. \(\square\)
Worked Example: Exit from a Symmetric Interval
Let \(B_t\) be Brownian motion and set \(\tau=\inf\{t\ge0:|B_t|=2\}\). Find the probability that the exit occurs at \(2\).
First check that \(\tau\) is finite almost surely. Put \(p=\mathbb{P}(Z>4)>0\), where \(Z\) has distribution \(N(0,1)\). If the path has not exited by integer time \(k\), then \(B_k\in(-2,2)\). An increment \(B_{k+1}-B_k>4\) forces \(B_{k+1}>2\), and continuity forces an exit by time \(k+1\). Independence of Brownian increments therefore gives \(\mathbb{P}(\tau>n)\le(1-p)^n\). This tends to zero, so \(\tau<\infty\) almost surely.
By continuity, \(B_\tau\in\{-2,2\}\). The stopped family \(B_{t\wedge\tau}\) is bounded in absolute value by \(2\), hence uniformly integrable. Apply optional stopping to the martingale \(M_t=B_t\), whose initial value is zero. If \(q=\mathbb{P}(B_\tau=2)\), then
Check: substituting \(q=\tfrac12\) gives \(\mathbb{E}[B_\tau]=2(\tfrac12)-2(\tfrac12)=0\), as required by the theorem.
Worked Example: Exit from an Asymmetric Interval
Let \(\tau=\inf\{t\ge0:B_t\notin(-1,2)\}\). Compute the probability of exiting at the upper endpoint.
As in the preceding argument, if \(B_k\in(-1,2)\), an increment \(B_{k+1}-B_k>3\) forces exit by time \(k+1\). The probability \(p=\mathbb{P}(Z>3)\) is strictly positive, so \(\mathbb{P}(\tau>n)\le(1-p)^n\). Thus \(\tau\) is finite almost surely. Continuity gives \(B_\tau\in\{-1,2\}\), and \(|B_{t\wedge\tau}|\le2\), so the stopped Brownian family is uniformly integrable.
Let \(q=\mathbb{P}(B_\tau=2)\). Optional stopping gives zero expected exit position. Writing out both possible values gives
Check: the lower-exit probability is \(1-\tfrac13=\tfrac23\), and the expected position is \(2(\tfrac13)-1(\tfrac23)=\tfrac23-\tfrac23=0\). The upper and lower probabilities are not equal because the endpoints are not equally distant from the starting point.
Using a Second Martingale to Find an Exit Time
Expectation calculations need not stop at hitting probabilities. Brownian motion has the martingale \(B_t^2-t\). Its martingale property follows from the independent increment \(B_t-B_s\), which has mean zero and variance \(t-s\): expanding \(B_t^2=(B_s+(B_t-B_s))^2\) and conditioning on \(\mathcal{F}_s\) gives \(\mathbb{E}[B_t^2-t\mid\mathcal{F}_s]=B_s^2-s\).
Worked Example: Expected Time to Exit a Bounded Interval
For \(\tau=\inf\{t\ge0:|B_t|=\tfrac32\}\), compute \(\mathbb{E}[\tau]\).
The same Gaussian-increment argument used above gives \(\mathbb{P}(\tau>n)\le(1-p)^n\), now with \(p=\mathbb{P}(Z>3)>0\): from any point in \((-\tfrac32,\tfrac32)\), an increment greater than \(3\) forces exit by the next integer time. Consequently \(\mathbb{E}[\tau]=\int_0^\infty\mathbb{P}(\tau>t)\,dt<\infty\); for example, on each interval \([n,n+1]\), the integrand is at most \(\mathbb{P}(\tau>n)\le(1-p)^n\), and the resulting geometric series is finite.
Now stop \(M_t=B_t^2-t\) at \(\tau\). For every \(t\), continuity and the definition of exit give \(|B_{t\wedge\tau}|\le\tfrac32\), while \(t\wedge\tau\le\tau\). Hence \(|M_{t\wedge\tau}|\le\tfrac94+\tau\). The right-hand side is integrable, so the stopped family is uniformly integrable. Optional stopping applies, and \(M_0=0\). Since \(B_\tau^2=\tfrac94\), it follows that
Check: the computed value gives \(\mathbb{E}[B_\tau^2-\tau]=\tfrac94-\tfrac94=0\). More generally, for exit from \((-a,a)\) the same argument gives \(\mathbb{E}[\tau]=a^2\), provided \(a>0\); the result for \(a=\tfrac32\) is \((\tfrac32)^2=\tfrac94\).
The Essential Pitfall: Passing Expectations Through a Limit
For each \(n\), bounded optional sampling gives \(\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]\). It is not valid to conclude \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\) from almost-sure convergence \(M_{\tau\wedge n}\to M_\tau\) alone. The theorem requires uniform integrability precisely to make this limiting step valid.
A concrete incorrect calculation illustrates the danger. Let \(\tau=\inf\{t\ge0:B_t=1\}\). If one applies bounded optional sampling at \(\tau\wedge n\), then \(\mathbb{E}[B_{\tau\wedge n}]=0\) for every \(n\). Since \(\tau\) is finite almost surely and \(B_{\tau\wedge n}\to B_\tau=1\), it is tempting—but wrong—to take expectations through the limit and infer \(0=\mathbb{E}[B_\tau]=1\). The error is the unjustified limit interchange: the stopped family in this argument is not uniformly integrable. Indeed, uniform integrability would imply convergence in \(L^1\), which would force the expectations to converge to \(1\), contradicting their constant value \(0\).
| What is available | What it gives |
|---|---|
| \(\tau\) bounded | Bounded optional sampling applies directly. |
| \(\tau\) finite almost surely and the stopped family uniformly integrable | The truncations converge in \(L^1\); optional stopping holds at \(\tau\). |
| Almost-sure convergence of stopped values only | Not enough to pass expectations to the limit. |
Check Your Understanding
For each question, identify which hypothesis permits the passage from bounded truncations to the unbounded stopping time.
- State the conclusion of optional stopping when \(\tau\) is finite almost surely and \(\{M_{t\wedge\tau}:t\ge0\}\) is uniformly integrable.
- Why does boundedness of \(B_{t\wedge\tau}\) imply uniform integrability in the symmetric-interval example?
- For exit from \((-a,a)\), with \(a>0\), use the martingales \(B_t\) and \(B_t^2-t\) to find the upper-exit probability and the expected exit time.
- In the asymmetric-interval example, write the equation for the expected exit position using the two possible endpoint values.
- Why does almost-sure convergence \(M_{\tau\wedge n}\to M_\tau\) not by itself imply convergence of their expectations?