Why Bounded Stopping Times Need an Approximation Argument
At deterministic times, the martingale property says that the future has conditional mean equal to the present. For stopping times taking only finitely many values, the same identity follows by adding martingale increments over the steps between the two stopping times. The continuous-time theorem extends this argument to arbitrary bounded stopping times. The key is to approximate them from above by finite-grid stopping times. Right-continuity gives convergence of the sampled process, while uniform integrability justifies convergence in expectation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), with a right-continuous filtration, carrying a Brownian motion \(B_t\). Brownian motion is not needed for this theorem, and no stochastic integrals are used here. A martingale \(M=(M_t)_{t\ge0}\) means an adapted process with \(\mathbb{E}|M_t|<\infty\) for every \(t\), and \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely whenever \(s\le t\). Assume throughout that \(M\) has right-continuous paths. A stopping time \(R\) is bounded if \(R\le K\) almost surely for some finite deterministic \(K\).
For an adapted right-continuous process, its value at a stopping time is measurable with respect to the sigma-field at that stopping time. We use this standard measurability fact for \(M_S\); it follows by approximating \(S\) by finite-valued stopping times and using the right-continuity of the filtration and of the paths. The proof below establishes the theorem’s conditional expectation identity and integrability directly.
The Finite-Grid Identity
Proof. For each \(k=1,\ldots,N\), the increment \(X_k-X_{k-1}\) is \(\mathcal{G}_k\)-measurable and has conditional expectation zero given \(\mathcal{G}_{k-1}\). The pathwise telescoping identity is $$ X_V-X_U=\sum_{k=1}^{N}\mathbf{1}_{\{U<k\le V\}}(X_k-X_{k-1}). $$ Fix \(A\in\mathcal{G}_U\). The event \(A\cap\{U<k\le V\}\) belongs to \(\mathcal{G}_{k-1}\): indeed, \(A\cap\{U\le k-1\}\in\mathcal{G}_{k-1}\) by the definition of \(\mathcal{G}_U\), and \(\{V\ge k\}\in\mathcal{G}_{k-1}\) because \(V\) is a stopping time. Therefore each term has mean zero after multiplication by \(\mathbf{1}_A\), and summing the finitely many terms gives \(\mathbb{E}[\mathbf{1}_A(X_V-X_U)]=0\). Since \(X_U\) is \(\mathcal{G}_U\)-measurable, this integral identity for every \(A\in\mathcal{G}_U\) is exactly \(\mathbb{E}[X_V\mid\mathcal{G}_U]=X_U\). \(\square\)
Rounding Stopping Times Upward
Choose an integer \(L\) strictly larger than a common deterministic bound for \(S\) and \(T\). For \(n\ge1\), let \(h_n=2^{-n}\) and define $$ S_n=h_n\left\lceil\frac{S}{h_n}\right\rceil,\qquad T_n=h_n\left\lceil\frac{T}{h_n}\right\rceil. $$ These are finite-valued stopping times on the grid \(\{0,h_n,2h_n,\ldots,L\}\). For example, if \(kh_n\le t<(k+1)h_n\), then \(\{S_n\le t\}=\{S\le kh_n\}\in\mathcal{F}_{kh_n}\subseteq\mathcal{F}_t\). The same argument applies to \(T_n\). Rounding by the same increasing function preserves order, so \(S_n\le T_n\). Also \(S_n\downarrow S\) and \(T_n\downarrow T\).
The grid lemma applied to \(M_{kh_n}\) gives $$ \mathbb{E}[M_{T_n}\mid\mathcal{F}_{S_n}]=M_{S_n}. $$ That identity alone does not permit passage to the limit: almost-sure convergence of random variables does not in general imply convergence of their expectations. We next provide the needed integrability control.
Proof. Apply the finite-grid lemma with the later time fixed at \(L\). It gives \(M_{S_n}=\mathbb{E}[M_L\mid\mathcal{F}_{S_n}]\), and likewise \(M_{T_n}=\mathbb{E}[M_L\mid\mathcal{F}_{T_n}]\). For any integrable \(X\), the family \(\{\mathbb{E}[X\mid\mathcal{G}]\}\), over all sub-sigma-fields \(\mathcal{G}\), is uniformly integrable. To verify this, write \(X=X^{(K)}+R^{(K)}\), where \(X^{(K)}=X\mathbf{1}_{\{|X|\le K\}}\) and \(R^{(K)}=X\mathbf{1}_{\{|X|>K\}}\). Put \(Y=\mathbb{E}[X\mid\mathcal{G}]\), \(A=\mathbb{E}[X^{(K)}\mid\mathcal{G}]\), and \(D=\mathbb{E}[R^{(K)}\mid\mathcal{G}]\). Then \(|A|\le K\), \(Y=A+D\), and, for \(R>2K\), the event \(E=\{|Y|>R\}\) implies \(|D|>R-K>R/2\). Consequently \(\mathbb{P}(E)\le 2\mathbb{E}|D|/R\le2\mathbb{E}|R^{(K)}|/R\), and $$ \mathbb{E}[|Y|\mathbf{1}_E]\le K\mathbb{P}(E)+\mathbb{E}|D| \le\left(\frac{2K}{R}+1\right)\mathbb{E}|R^{(K)}|. $$ First choose \(K\) so that \(\mathbb{E}|R^{(K)}|\) is small, which is possible since \(X\) is integrable, and then take any \(R>2K\). This proves the claimed uniform tail bound. Taking \(X=M_L\) proves uniform integrability of both approximation families. \(\square\)
Proof of the Bounded Optional Sampling Theorem
Proof. Since \(S_n\downarrow S\) and \(T_n\downarrow T\), right-continuity of the paths gives \(M_{S_n}\to M_S\) and \(M_{T_n}\to M_T\) almost surely. The uniform integrability lemma implies convergence in \(L^1\). For completeness, this implication follows by truncating the variables at a fixed level: the truncated variables are bounded and converge almost surely, so their difference converges in \(L^1\) by dominated convergence; the expectations of the discarded tails are uniformly small by uniform integrability. The limit has integrable tails by Fatou’s lemma, so the truncation level can then tend to infinity. Thus \(M_S,M_T\) are integrable and both approximation sequences converge in \(L^1\).
Now fix any \(A\in\mathcal{F}_S\). Because \(S\le S_n\), the sigma-fields satisfy \(\mathcal{F}_S\subseteq\mathcal{F}_{S_n}\). The finite-grid identity therefore gives $$ \mathbb{E}[\mathbf{1}_A M_{T_n}] =\mathbb{E}[\mathbf{1}_A M_{S_n}]. $$ The \(L^1\) convergence allows passage to the limit on both sides, yielding \(\mathbb{E}[\mathbf{1}_A M_T]=\mathbb{E}[\mathbf{1}_A M_S]\). Since \(M_S\) is \(\mathcal{F}_S\)-measurable and integrable, this identity for every \(A\in\mathcal{F}_S\) is the defining property of conditional expectation. Hence \(\mathbb{E}[M_T\mid\mathcal{F}_S]=M_S\) almost surely. \(\square\)
Worked Applications and Checks
Worked Example: A Stop Chosen from a Four-Outcome Martingale
Let \(\Omega=\{a_1,a_2,a_3,a_4\}\), with probability \(1/4\) for each outcome. Let \(Z\) take values \(2,2,2,-2\), respectively. Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1=\sigma(\{a_1,a_2\},\{a_3,a_4\})\), and let \(\mathcal{F}_2\) contain all subsets. Define \(M_t=\mathbb{E}[Z\mid\mathcal{F}_t]\) for \(t=0,1,2\), extending the filtration and process constantly between these times. Take \(T=2\), and let \(S=1\) on \(\{a_1,a_2\}\) and \(S=2\) on \(\{a_3,a_4\}\).
First compute the unconditional mean: $$ \mathbb{E}[Z]=\frac{2+2+2-2}{4}=1. $$ Thus \(M_0=1\). On \(\{a_1,a_2\}\), the conditional mean is \((2+2)/2=2\); on \(\{a_3,a_4\}\), it is \((2-2)/2=0\). Hence \(M_1\) equals \(2\) on the first pair and \(0\) on the second. At time \(2\), \(M_2=Z\). The stopping rule is valid because the decision to stop at time \(1\) is made on an event in \(\mathcal{F}_1\). The sigma-field \(\mathcal{F}_S\) has atoms \(\{a_1,a_2\},\{a_3\},\{a_4\}\). On these atoms, the conditional means of \(M_T=Z\) are \(2,2,-2\), respectively. These are exactly the corresponding values of \(M_S\). As a check, \(\mathbb{E}[M_S]=(2+2+2-2)/4=1=\mathbb{E}[M_T]\).
Worked Example: Deterministic Sampling in a Two-Stage Information Flow
On four equally likely outcomes, let \(X\) take values \(4,0,2,-2\), respectively. At time \(0\), reveal nothing; at time \(1/2\), reveal whether the outcome is among the first two or the last two; at time \(1\), reveal the outcome. Let \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\), with the information held constant between these times. Set \(S=1/2\) and \(T=1\).
The unconditional mean is \((4+0+2-2)/4=1\), so \(M_0=1\). At time \(1/2\), the two group means are \((4+0)/2=2\) and \((2-2)/2=0\). Therefore \(M_S\) equals \(2\) on the first group and \(0\) on the second. Since \(\mathcal{F}_S\) distinguishes precisely those groups, \(\mathbb{E}[M_T\mid\mathcal{F}_S]\) has the same two values. The unconditional check is \(\mathbb{E}[M_S]=(2+2+0+0)/4=1\), equal to \(\mathbb{E}[M_T]=(4+0+2-2)/4=1\).
Worked Example: Checking the Grid Approximation for a Jump Martingale
Let \(X\) equal \(3\) or \(-3\), each with probability \(1/2\). Let \(\mathcal{F}_t\) be trivial for \(t<1/2\) and reveal \(X\) for \(t\ge1/2\). Define \(M_t=0\) for \(t<1/2\) and \(M_t=X\) for \(t\ge1/2\). This is a right-continuous martingale: before the reveal the conditional mean of \(X\) is \((3-3)/2=0\), and after the reveal the process is constant. Take \(S=1/3\) and \(T=5/6\).
The original sampled values are \(M_S=0\) and \(M_T=X\). Since \(\mathcal{F}_S\) is trivial, \(\mathbb{E}[M_T\mid\mathcal{F}_S]=\mathbb{E}[X]=(3-3)/2=0=M_S\). On the grid with spacing \(1/16\), rounding upward gives \(S_1=6/16=3/8\) and \(T_1=14/16=7/8\). Thus \(M_{S_1}=0\) and \(M_{T_1}=X\), and the grid identity also gives \(\mathbb{E}[M_{T_1}\mid\mathcal{F}_{S_1}]=0=M_{S_1}\). As the mesh tends to zero, the rounded times approach \(S,T\); they stay on the same sides of the jump time \(1/2\), so the sampled values remain \(0\) and \(X\), confirming the limiting identity directly.
Common Mistake: Replacing Conditional Expectation by the Overall Mean
In the four-outcome example, it would be wrong to conclude that \(\mathbb{E}[M_T\mid\mathcal{F}_S]\) is the constant \(\mathbb{E}[M_T]=1\). The sigma-field \(\mathcal{F}_S\) contains information: on \(\{a_1,a_2\}\), the conditional mean is \(2\), not \(1\). Optional sampling says the conditional mean given the information at \(S\) equals \(M_S\); it does not say that this conditional mean is constant. Another error is to pass from almost-sure convergence of \(M_{T_n}\) to convergence of expectations without checking integrability. The proof needs uniform integrability precisely to justify that passage.
Check Your Understanding
Use the hypotheses and proof steps above to answer the following questions.
- Why does rounding \(S\) and \(T\) upward on the same grid preserve the order \(S_n\le T_n\)?
- For the four-outcome example, list the atoms of \(\mathcal{F}_S\) and compute the conditional mean of \(M_T\) on each.
- Which step in the proof uses right-continuity of the martingale paths?
- Why is almost-sure convergence alone insufficient to pass from the finite-grid identity to the bounded optional sampling identity?
- In the jump-martingale example, what fails in the stated calculation if \(S\) is changed to \(2/3\) while \(T=5/6\)?