Why Stopping-Time Finiteness Is Not Enough
For a martingale, the bounded-time optional stopping theorem gives an exact expectation identity. It is tempting to apply that identity to an unbounded stopping time by truncating the time, then taking a limit. The delicate step is the limit: almost-sure convergence of the stopped values does not in general imply convergence of their expectations. Even finite expected stopping time does not repair the problem without further conditions.
Throughout, processes are defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\). A stopping time \(\tau\) takes values in \(\{0,1,\ldots\}\cup\{\infty\}\), with \(\{\tau\le n\}\in\mathcal{F}_n\). For a martingale \((M_n)\), the truncated time \(\tau\wedge n=\min(\tau,n)\) is bounded. When \(M_0\) is integrable, bounded optional stopping gives \(\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]\). This identity alone says nothing about whether the expectations converge to \(\mathbb{E}[M_\tau]\).
A Limit Criterion
A direct sufficient condition for passing expectations through the truncation limit is convergence in \(L^1\), also called convergence in mean: \(\mathbb{E}|M_{\tau\wedge n}-M_\tau|\to0\). This is a useful diagnostic because it identifies exactly the missing step in the naive truncation argument.
Proof. By the assumed \(L^1\) convergence, \(\mathbb{E}|M_{\tau\wedge n}-M_\tau|\to0\). The inequality \(|\mathbb{E}[M_{\tau\wedge n}]-\mathbb{E}[M_\tau]|\le \mathbb{E}|M_{\tau\wedge n}-M_\tau|\) therefore implies \(\mathbb{E}[M_{\tau\wedge n}]\to\mathbb{E}[M_\tau]\). Each expectation on the left equals \(\mathbb{E}[M_0]\), so \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\). \(\square\)
The proposition is a sufficient condition, not a claim that it is necessary for every individual stopping problem. A common stronger way to guarantee this kind of passage is uniform integrability of the stopped values; that condition is treated in detail later. The examples below show concretely what can go wrong when no convergence-in-mean control is available.
Worked Example: A Bounded Stopping Time as a Contrast
Let \((X_n)_{n\ge1}\) be independent fair signs, so \(\mathbb{P}(X_n=1)=\mathbb{P}(X_n=-1)=\tfrac12\). On their probability space, let \(\mathcal{F}_n=\sigma(X_1,\ldots,X_n)\), \(S_0=0\), and \(S_n=\sum_{k=1}^nX_k\). Independence and \(\mathbb{E}X_{n+1}=0\) make \((S_n)\) a martingale. Define \(\tau=1\) if \(X_1=1\), and \(\tau=2\) if \(X_1=-1\). This is a stopping time bounded by \(2\).
If \(X_1=1\), which has probability \(\tfrac12\), then \(S_\tau=1\). If \(X_1=-1\), then \(\tau=2\) and \(S_\tau=-1+X_2\); the two possible values are \(-2\) and \(0\), each with probability \(\tfrac14\). Thus
Also, \(\mathbb{E}\tau=1\cdot\tfrac12+2\cdot\tfrac12=\tfrac32\). Here the equality of expectations is guaranteed by bounded optional stopping, not merely by the fact that the expected stopping time is finite.
Almost-Sure Finiteness: The One-Sided Symmetric Walk
Return to the fair random walk \(S_n\) from the example. Start it at \(S_0=0\), and let \(\tau=\inf\{n\ge0:S_n=1\}\), the first time it reaches \(1\). This is a stopping time for the natural filtration. We will show that \(\tau\) is finite almost surely, but \(\mathbb{E}\tau=\infty\). Since \(S_\tau=1\) almost surely while \(S_0=0\), this gives a counterexample to unrestricted optional stopping.
Proof. Fix an integer \(m\ge1\), and let \(h_i\) be the probability, starting from \(i\in\{-m,-m+1,\ldots,1\}\), of hitting \(1\) before \(-m\). The walk exits this finite interval almost surely: from any interior state, \(m+1\) consecutive \(+1\) steps force it to hit \(1\), and this event has probability \(2^{-(m+1)}>0\). In successive disjoint blocks of \(m+1\) steps, the probability of avoiding such a block is at most \(1-2^{-(m+1)}\), so the probability of avoiding all such blocks is zero.
At the boundaries, \(h_{-m}=0\) and \(h_1=1\). At each interior state, conditioning on the first step gives \(h_i=\tfrac12h_{i-1}+\tfrac12h_{i+1}\). Rearranging shows \(h_{i+1}-h_i=h_i-h_{i-1}\), so the successive differences are constant. There are \(m+1\) equal increments from \(-m\) to \(1\), and their total is \(h_1-h_{-m}=1\). Each increment is therefore \(\tfrac1{m+1}\). Starting from \(0\), which is \(m\) increments above \(-m\), gives \(h_0=\tfrac{m}{m+1}\).
The events “hit \(1\) before \(-m\)” increase as \(m\) increases. Their union is exactly \(\{\tau<\infty\}\): if the walk hits \(1\) at a finite time, its minimum before that time is finite, so for sufficiently large \(m\) it hits \(1\) before \(-m\). Hence \(\mathbb{P}(\tau<\infty)=\lim_{m\to\infty}\tfrac{m}{m+1}=1\).
To show that \(\mathbb{E}\tau\) is infinite, suppose instead that \(\mathbb{E}\tau<\infty\). The bounded-time identity gives \(\mathbb{E}[S_{\tau\wedge n}]=0\) for every \(n\). Each step has absolute size \(1\), so pathwise \(|S_\tau-S_{\tau\wedge n}|\le(\tau-n)^+\), where \(x^+=\max(x,0)\). Since \((\tau-n)^+\le\tau\mathbf{1}_{\{\tau>n\}}\), and \(\tau\) is integrable, dominated convergence gives \(\mathbb{E}[(\tau-n)^+]\to0\). It follows that \(\mathbb{E}|S_\tau-S_{\tau\wedge n}|\to0\), so the proposition on convergence in \(L^1\) would imply \(\mathbb{E}S_\tau=0\). But \(\tau<\infty\) almost surely and the walk moves in steps of size \(1\), so \(S_\tau=1\) almost surely and \(\mathbb{E}S_\tau=1\). This contradiction proves \(\mathbb{E}\tau=\infty\). Finally, \(\mathbb{E}[S_\tau]=1\ne0=\mathbb{E}[S_0]\). \(\square\)
The truncations explain the failure: \(S_{\tau\wedge n}\to1\) almost surely, but every truncated expectation is zero. Thus the truncations cannot converge in \(L^1\) to \(1\). Almost-sure convergence alone does not allow expectations to be passed through a limit.
Worked Example: Almost-Sure Limit, Different Limiting Expectation
For the same walk and stopping time, the truncated value has the exact form \(S_{\tau\wedge n}=1\) on \(\{\tau\le n\}\), and \(S_{\tau\wedge n}=S_n\) on \(\{\tau>n\}\). Bounded optional stopping gives \(\mathbb{E}[S_{\tau\wedge n}]=0\) for each \(n\). Yet \(\tau<\infty\) almost surely, so \(S_{\tau\wedge n}\to1\) almost surely.
If one incorrectly interchanges limit and expectation, one obtains the false calculation
The correction is that almost-sure convergence does not justify this interchange. In fact, the left side is \(\mathbb{E}[1]=1\), whereas the limit on the right is \(0\). This example also rules out \(L^1\) convergence: if the truncations converged in \(L^1\) to \(1\), their expectations would converge to \(1\), contrary to their exact expectations of \(0\).
Finite Expected Time Can Also Fail
The one-sided walk has infinite expected stopping time, but that is not the only obstruction. The next martingale has a stopping time with finite expectation and still violates the optional-stopping conclusion. Its stopped values become increasingly large on increasingly rare events; those rare values preserve the martingale expectation at each fixed time, but disappear from the almost-sure terminal value.
Worked Example: A Geometric Stopping Time with a Biased Terminal Value
Let \(\tau\) be a positive-integer-valued random variable with \(\mathbb{P}(\tau=k)=2^{-k}\) for \(k\ge1\). These probabilities sum to \(1\), since \(\sum_{k=1}^{\infty}2^{-k}=1\). Let \(\mathcal{F}_n=\sigma(\mathbf{1}_{\{\tau\le k\}}:1\le k\le n)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\), and define
On \(\{\tau\le n-1\}\), the process stays at \(1\). On \(\{\tau>n-1\}\), the conditional probability that \(\tau=n\) is \(\tfrac12\), because \(\mathbb{P}(\tau\ge n)=2^{-(n-1)}\) and \(\mathbb{P}(\tau=n)=2^{-n}\). The conditional probability that \(\tau>n\) is also \(\tfrac12\). Therefore, on this survival event,
For \(n=1\), the final expression is \(0=M_0\); on the absorbed event at later times, both current and next values equal \(1\). Thus \((M_n)\) is an integrable martingale. Integrability at each fixed \(n\) also follows directly from its two finite possible values.
The stopping time has finite mean. Using the tail-sum formula for a positive-integer-valued random variable, \(\mathbb{E}\tau=\sum_{k\ge1}\mathbb{P}(\tau\ge k)=\sum_{k\ge1}2^{-(k-1)}=2\). But \(M_\tau=1\) on every outcome, so \(\mathbb{E}M_\tau=1\ne0=\mathbb{E}M_0\).
Check the truncations. The probability of absorption by time \(n\) is \(1-2^{-n}\), and the probability of survival past \(n\) is \(2^{-n}\). Hence
Nevertheless, \(M_n\to1=M_\tau\) almost surely. The discrepancy is visible in the \(L^1\) distance: on \(\{\tau\le n\}\), \(|M_n-1|=0\); on \(\{\tau>n\}\), \(|M_n-1|=2^n\). Thus \(\mathbb{E}|M_n-M_\tau|=2^{-n}2^n=1\) for every \(n\), so the convergence is not in \(L^1\). This calculation verifies exactly why the limit criterion does not apply.
How to Diagnose an Unbounded-Stopping Argument
When an argument applies optional stopping at an unbounded time, the key question is not only whether \(\tau\) is finite, or whether \(\mathbb{E}\tau\) is finite. Check what happens to the sequence of truncated values \(M_{\tau\wedge n}\). The bounded-time identities concern that sequence; the desired conclusion concerns \(M_\tau\). A justification must connect their expectations, for example by proving convergence in \(L^1\), rather than only pointwise convergence.
Common mistake. The incorrect calculation is \(\mathbb{E}[M_\tau]=\mathbb{E}[\lim_n M_{\tau\wedge n}]=\lim_n\mathbb{E}[M_{\tau\wedge n}]\), justified only by \(\tau<\infty\) almost surely. The symmetric-walk example gives \(1\) on the left and \(0\) on the right. The geometric example shows that even \(\mathbb{E}\tau<\infty\) does not validate the interchange. The correction is to establish a condition that controls the limiting expectations; without it, stop-time finiteness alone does not yield optional stopping.
Check Your Understanding
For each question, distinguish the bounded-time identity from any claim about the unbounded terminal value.
- For the one-sided symmetric walk, what is \(\mathbb{E}[S_{\tau\wedge n}]\) for each finite \(n\), and what is \(\mathbb{E}[S_\tau]\)?
- Why do the events “hit \(1\) before \(-m\)” increase as \(m\) increases, and what is their probability when the walk starts at \(0\)?
- In the geometric example, compute \(\mathbb{E}\tau\) from the tail probabilities \(\mathbb{P}(\tau\ge k)\).
- For that same example, calculate \(\mathbb{E}|M_n-M_\tau|\). What does the result say about convergence in \(L^1\)?
- State the additional convergence condition in the proposition that permits the bounded-time expectation identities to pass to \(M_\tau\).