Why Approximate a Continuous-Time Stopping Rule?
In discrete time, a bounded stopping time can take only finitely many values, so optional stopping reduces to a finite sum of martingale increments. In continuous time, a bounded stopping time may take infinitely many values. A useful route is to round it up to a finite grid, apply the discrete-time result, and then pass to the limit. Right-continuity makes the stopped values converge; a separate integrability argument is needed to ensure their expectations converge too.
Work on a probability space \((\Omega,\mathcal F,\mathbb P)\) with filtration \((\mathcal F_t)_{t\ge0}\). A process \((M_t)_{t\ge0}\) is a martingale on \([0,T]\) if it is adapted, each \(M_t\) is integrable, and \(\mathbb E[M_t\mid\mathcal F_s]=M_s\) almost surely whenever \(0\le s\le t\le T\). A random time \(\tau\) is a stopping time if \(\{\tau\le t\}\in\mathcal F_t\) for every \(t\ge0\). Right-continuity of \(M\) means that, outside one null set, \(M_{t_n}\to M_t\) whenever \(t_n\downarrow t\).
The boundedness assumption is on the stopping time, not on the values of the martingale. The martingale itself may be unbounded. The proof below explains why right-continuity alone is not the whole convergence argument: we also need a uniform integrability bound.
Finite Grids and the Convergence Argument
A family \(\mathcal X\) of integrable random variables is uniformly integrable if \(\sup_{X\in\mathcal X}\mathbb E[|X|\mathbf 1_{\{|X|>K\}}]\to0\) as \(K\to\infty\). We first record the particular convergence fact needed here, including its justification.
Proof. For the first assertion, put \(Z=\mathbb E[X\mid\mathcal G]\) and \(A=\{|Z|>K\}\), where \(K>0\). Since \(A\in\mathcal G\), \(\mathbb E[|Z|\mathbf 1_A]\le\mathbb E[|X|\mathbf 1_A]\). Splitting according to whether \(|X|\le K/2\) gives \(\mathbb E[|Z|\mathbf 1_A]\le (K/2)\mathbb P(A)+\mathbb E[|X|\mathbf 1_{\{|X|>K/2\}}]\). But \(K\mathbb P(A)\le\mathbb E[|Z|\mathbf 1_A]\), so \(\mathbb E[|Z|\mathbf 1_A]\le2\mathbb E[|X|\mathbf 1_{\{|X|>K/2\}}]\). The right-hand side tends to zero independently of \(\mathcal G\), proving uniform integrability.
For the second assertion, Fatou's lemma applied to the tail indicators shows that \(Y\) has tails tending to zero: for each \(K>0\), \(\mathbb E[|Y|\mathbf 1_{\{|Y|>2K\}}]\le\liminf_n\mathbb E[|Y_n|\mathbf 1_{\{|Y_n|>K\}}]\). Thus \(Y\) is integrable and its tails vanish. For \(L>0\), let \(\phi_L(x)=\max(-L,\min(x,L))\). The function is bounded and continuous, so almost sure convergence and bounded convergence give \(\mathbb E|\phi_L(Y_n)-\phi_L(Y)|\to0\). Also, \(|x-\phi_L(x)|\le |x|\mathbf 1_{\{|x|>L\}}\). The triangle inequality therefore bounds \(\mathbb E|Y_n-Y|\) by this truncated difference plus the two tail expectations. First choose \(L\) large using uniform integrability and the tail bound for \(Y\), then let \(n\to\infty\). This proves convergence in \(L^1\), and hence convergence of expectations. \(\square\)
Proof of the theorem. If \(T=0\), then \(\tau=0\) almost surely and the assertion is immediate. Suppose \(T>0\). For each integer \(n\ge0\), use the grid \(t_{n,k}=kT/2^n\), \(0\le k\le2^n\), and let \(\tau_n\) be the first grid point at or after \(\tau\). Thus \(\tau\le\tau_n\le T\), \(\tau_n-\tau\le T/2^n\), and the nested grids imply \(\tau_n\downarrow\tau\). Each \(\tau_n\) is a stopping time: for \(t\) between consecutive grid points, \(\{\tau_n\le t\}\) is the event that \(\tau\) is at or before the preceding grid point, which belongs to the filtration at that grid point and hence to \(\mathcal F_t\).
Fix \(n\), write \(K_n=2^n\tau_n/T\), and let \(\mathcal G_k=\mathcal F_{t_{n,k}}\). The sampled process \(M_{t_{n,k}}\) is a discrete-time martingale, and \(K_n\) is a bounded discrete-time stopping index. The bounded discrete-time optional stopping identity gives \(\mathbb E[M_{\tau_n}]=\mathbb E[M_0]\). We need more than this identity: we need to pass to the limit in its left-hand side.
For that purpose, define \(\mathcal G_{K_n}\) to consist of events \(A\) such that \(A\cap\{K_n=k\}\in\mathcal G_k\) for every \(k\). Since \(M_{t_{n,k}}=\mathbb E[M_T\mid\mathcal G_k]\), for every \(A\in\mathcal G_{K_n}\) we have \(\mathbb E[\mathbf 1_A M_T]=\sum_k\mathbb E[\mathbf 1_{A\cap\{K_n=k\}}M_T]=\sum_k\mathbb E[\mathbf 1_{A\cap\{K_n=k\}}M_{t_{n,k}}]=\mathbb E[\mathbf 1_A M_{\tau_n}]\). Also \(M_{\tau_n}\) is \(\mathcal G_{K_n}\)-measurable, since it has the value \(M_{t_{n,k}}\) on \(\{K_n=k\}\). Consequently \(M_{\tau_n}=\mathbb E[M_T\mid\mathcal G_{K_n}]\) almost surely. The lemma shows that \((M_{\tau_n})\) is uniformly integrable.
Because \(\tau_n\downarrow\tau\) and \(M\) has right-continuous paths, \(M_{\tau_n}\to M_\tau\) almost surely. The lemma now gives \(\mathbb E|M_{\tau_n}-M_\tau|\to0\). In particular \(M_\tau\) is integrable, and taking limits in \(\mathbb E[M_{\tau_n}]=\mathbb E[M_0]\) proves \(\mathbb E[M_\tau]=\mathbb E[M_0]\). Applying this equality separately to two stopping times bounded by \(T\) proves the final assertion. \(\square\)
Worked Examples
Worked Example: Stopping a Finite-State Continuous-Time Chain
Let \(S=\{0,1,2\}\), and let \((X_t)\) be a continuous-time Markov chain started at \(X_0=1\), with generator \(Q\) in the state order \(0,1,2\). States 0 and 2 are absorbing; from state 1 the chain jumps to each of them at rate 1. Find the expectation of \(X_\tau\), where \(\tau\) is the first jump time capped at 1.
The generator and row-sum check are \(Q=\begin{bmatrix}0&0&0\\1&-2&1\\0&0&0\end{bmatrix}\), with row sums \(0\), \(1-2+1=0\), and \(0\). From state 1, the holding time is exponential with rate 2, and conditional on a jump, either absorbing state is chosen with probability \(1/2\). Write \(H\) for the first jump time and set \(\tau=\min(H,1)\). This is a stopping time and \(0\le\tau\le1\).
The coordinate process \(M_t=X_t\) is a martingale. Indeed, starting from state 1, after an interval of length \(u\) the probabilities of still being at 1 and of having reached each absorbing state are respectively \(e^{-2u}\), \((1-e^{-2u})/2\), and \((1-e^{-2u})/2\). Its expected state is therefore \(e^{-2u}+2(1-e^{-2u})/2=1\). Starting at 0 or 2 leaves the state fixed. The Markov property then gives \(\mathbb E[X_t\mid\mathcal F_s]=X_s\).
There is probability \(e^{-2}\) of no jump by time 1. The probability of a jump to either specified absorbing state by time 1 is \((1-e^{-2})/2\). Thus \(X_\tau=1\) if there is no jump, \(X_\tau=0\) after a jump to 0, and \(X_\tau=2\) after a jump to 2. Hence \(\mathbb E[X_\tau]=e^{-2}\cdot1+\frac{1-e^{-2}}2\cdot0+\frac{1-e^{-2}}2\cdot2=1\). Check: the probabilities sum to \(e^{-2}+(1-e^{-2})/2+(1-e^{-2})/2=1\), and the computed expectation equals \(X_0=1\).
Worked Example: A Capped First Arrival of a Poisson Process
Let \((N_t)\) be a Poisson process of rate 1 with its natural filtration, and let \(M_t=N_t-t\). If \(S_1\) is its first arrival time, compute \(\mathbb E[M_\tau]\) for \(\tau=\min(S_1,1)\).
The compensated Poisson process \(M_t\) is a martingale, and \(\tau\) is a stopping time bounded by 1. The probability of no arrival by time 1 is \(e^{-1}\); on that event \(M_\tau=-1\). The first-arrival density on \(0\le s\le1\) is \(e^{-s}\), and if the arrival occurs at \(s\), then \(M_\tau=1-s\). Therefore \(\mathbb E[M_\tau]=\int_0^1(1-s)e^{-s}\,ds-e^{-1}\).
Evaluate the integral term by term: \(\int_0^1e^{-s}\,ds=1-e^{-1}\), while integration by parts gives \(\int_0^1se^{-s}\,ds=1-2e^{-1}\). Their difference is \(e^{-1}\), so \(\mathbb E[M_\tau]=e^{-1}-e^{-1}=0=M_0\). Check: the first-arrival density has total mass \(1-e^{-1}\) on \([0,1]\), and the remaining probability \(e^{-1}\) is exactly the no-arrival case included separately.
Worked Example: A Stopping Rule Based on Revealed Information
Let \(\Omega=\{-1,1\}\), with each point having probability \(1/2\), and write \(Y(\omega)=\omega\). Use the filtration that is trivial for \(t<1\) and equals \(\sigma(Y)\) for \(t\ge1\). Define \(M_t=0\) for \(t<1\) and \(M_t=Y\) for \(t\ge1\). Show that the bounded stopping time \(\tau=1\) when \(Y=1\), and \(\tau=2\) when \(Y=-1\), preserves expectation.
The process is adapted and right-continuous. Before time 1 its conditional future mean is \(\mathbb E[Y]=0=M_t\); from time 1 onward \(Y\) is known and the process stays equal to \(Y\). Thus it is a martingale. For \(t<1\), \(\{\tau\le t\}\) is empty; for \(1\le t<2\), it is \(\{Y=1\}\in\mathcal F_t\); and for \(t\ge2\), it is all of \(\Omega\). Hence \(\tau\) is a stopping time bounded by 2.
On \(Y=1\), \(M_\tau=M_1=1\); on \(Y=-1\), \(M_\tau=M_2=-1\). Therefore \(\mathbb E[M_\tau]=\frac12(1)+\frac12(-1)=0=\mathbb E[M_0]\). Check: the two possible stopped values have equal probabilities and opposite signs, so their mean is zero.
Why Both Hypotheses Matter
Right-continuity and boundedness address different parts of the proof. Right-continuity gives almost sure convergence when the rounded times decrease to \(\tau\). Uniform integrability, obtained here from the integrable terminal value \(M_T\), turns that convergence into convergence of expectations. Without the stopping-time condition, even a bounded choice of time can select outcomes using information that was not yet available.
Common pitfall: treating any bounded random time as a stopping time. In the information-revelation example, define instead \(\sigma=0\) when \(Y=1\) and \(\sigma=1\) when \(Y=-1\). This time is bounded, but \(\{\sigma\le0\}=\{Y=1\}\notin\mathcal F_0\), so it is not a stopping time. The incorrect calculation would apply optional stopping and claim \(\mathbb E[M_\sigma]=\mathbb E[M_0]=0\). In fact, \(M_\sigma=0\) when \(Y=1\) and \(M_\sigma=-1\) when \(Y=-1\), giving \(\mathbb E[M_\sigma]=-\tfrac12\). Boundedness alone is not enough.
The grid proof is useful whenever a continuous-time stopping rule can be approximated from above and the process has right-continuous paths. The limit step should never be omitted: almost sure convergence of stopped values does not by itself justify convergence of their expectations. Here the conditional-expectation representation supplies the necessary uniform integrability on every grid.
Check Your Understanding
For each question, identify which hypothesis or step of bounded optional stopping is being used.
- Why does rounding \(\tau\) up to the next point of a nested grid give \(\tau_n\downarrow\tau\), rather than merely \(\tau_n\to\tau\)?
- In the proof, why can \(M_{\tau_n}\) be written as a conditional expectation of \(M_T\)?
- For the finite-state chain, verify directly that the generator rows sum to zero and recompute \(\mathbb E[X_\tau]\).
- For the rate-1 Poisson example, what is the probability of no arrival by time 1, and what value does \(M_\tau\) take on that event?
- In the information-revelation example, which event shows that \(\sigma\) is not a stopping time, and what is \(\mathbb E[M_\sigma]\)?