Why Condition at the Earlier Stopping Time?
A martingale has no expected drift when one conditions on the information currently available. If it is observed at two bounded stopping times \(\sigma\le\tau\), the same principle says that, given the information available at \(\sigma\), the expected value at \(\tau\) is the value already observed at \(\sigma\). The conditioning sigma-field is essential: it specifies exactly what is known at the earlier random time.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sigma-fields representing the information available by time \(n\). An integrable process \((M_n)_{n\ge0}\) is a martingale with respect to this filtration if \(M_n\) is \(\mathcal{F}_n\)-measurable and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely for every \(n\).
For bounded \(\sigma\), the stopped value \(M_\sigma\) is integrable and \(\mathcal{F}_\sigma\)-measurable. For example, if \(\sigma\le N\), then \(M_\sigma=\sum_{k=0}^N M_k\mathbf{1}_{\{\sigma=k\}}\), which is integrable as a finite sum. Its measurability at \(\sigma\) follows because on \(\{\sigma\le n\}\) it is assembled from the \(\mathcal{F}_n\)-measurable values \(M_k\), \(k\le n\); for \(n\ge N\), the whole stopped value is \(\mathcal{F}_n\)-measurable.
The ordering is only assumed almost surely, so the increment identity used in the proof holds almost surely, not necessarily on every outcome. That is all the expectation argument requires. The deterministic bound ensures that only finitely many martingale increments are involved.
Proof. Both stopped values are integrable by boundedness. To verify the asserted conditional expectation, it is enough to show that for every \(A\in\mathcal{F}_\sigma\), \(\mathbb{E}[\mathbf{1}_A M_\tau]=\mathbb{E}[\mathbf{1}_A M_\sigma]\). For each outcome on which \(\sigma\le\tau\), the increments from \(\sigma\) up to, but not including, \(\tau\) give $$ M_\tau-M_\sigma=\sum_{n=0}^{N-1}\mathbf{1}_{\{\sigma\le n<\tau\}}(M_{n+1}-M_n). $$ This identity therefore holds almost surely. Fix \(A\in\mathcal{F}_\sigma\). For each \(n<N\), the event \(A\cap\{\sigma\le n\}\) belongs to \(\mathcal{F}_n\) by the definition of \(\mathcal{F}_\sigma\); also \(\{\tau>n\}\in\mathcal{F}_n\) because \(\tau\) is a stopping time. Hence \(D_n=A\cap\{\sigma\le n<\tau\}\in\mathcal{F}_n\). The martingale property and the defining property of conditional expectation now give $$ \mathbb{E}\!\left[\mathbf{1}_{D_n}(M_{n+1}-M_n)\right] =\mathbb{E}\!\left[\mathbf{1}_{D_n}\mathbb{E}[M_{n+1}-M_n\mid\mathcal{F}_n]\right]=0. $$ Multiplying the almost-sure increment identity by \(\mathbf{1}_A\), taking expectations, and summing these zero terms yields \(\mathbb{E}[\mathbf{1}_A(M_\tau-M_\sigma)]=0\). This holds for every \(A\in\mathcal{F}_\sigma\), and \(M_\sigma\) is integrable and \(\mathcal{F}_\sigma\)-measurable. These are precisely the defining conditions for \(\mathbb{E}[M_\tau\mid\mathcal{F}_\sigma]=M_\sigma\) almost surely. \(\square\)
Taking expectations in the conditional identity gives \(\mathbb{E}[M_\tau]=\mathbb{E}[M_\sigma]\). The conditional statement is stronger: it says the equality still holds after accounting for every event determined by the information at \(\sigma\). The following finite examples compute that conditional expectation directly.
Worked Example: A Conditional-Probability Martingale
Flip a fair coin twice. Let \(A\) be the event that both flips are heads, and let \(\mathcal{F}_n\) record the first \(n\) flips. Define \(M_n=\mathbb{P}(A\mid\mathcal{F}_n)\) for \(n=0,1,2\). Take \(\sigma=1\), and let \(\tau=1\) if the first flip is tails and \(\tau=2\) if it is heads. Compute \(\mathbb{E}[M_\tau\mid\mathcal{F}_1]\).
The process is a martingale by the defining conditional-expectation identity. Its possible values over these times are \(0,\tfrac14,\tfrac12,1\): initially \(M_0=\tfrac14\); after a tail \(M_1=0\), and after a head \(M_1=\tfrac12\); at time 2, \(M_2=\mathbf{1}_A\). The time \(\tau\) is a stopping time because \(\{\tau\le1\}\) is the event that the first flip is tails, which belongs to \(\mathcal{F}_1\); also \(\sigma\le\tau\).
On the first-flip-tail atom, \(\tau=1\) and \(M_\tau=M_1=0\). On the first-flip-head atom, \(\tau=2\), and the second flip is still fair conditional on \(\mathcal{F}_1\). Thus \(M_\tau\) is 1 or 0 with conditional probabilities \(\tfrac12\) and \(\tfrac12\) on this atom. Consequently, its conditional expectation there is \(\tfrac12(1)+\tfrac12(0)=\tfrac12\). The two atoms give $$ \mathbb{E}[M_\tau\mid\mathcal{F}_1]= \begin{cases} 0,&\text{if the first flip is tails},\\ \tfrac12,&\text{if the first flip is heads}, \end{cases} =M_1=M_\sigma. $$ As an unconditional check, \(\mathbb{E}[M_\tau]=\tfrac12\cdot0+\tfrac12\cdot\tfrac12=\tfrac14\), while \(\mathbb{E}[M_\sigma]=\tfrac12\cdot0+\tfrac12\cdot\tfrac12=\tfrac14\).
Worked Example: A Random Walk with a Random Earlier Time
Let \(\Omega=\{-1,1\}^3\) assign probability \(\tfrac18\) to each three-increment path. Let \(\mathcal{F}_n\) reveal the first \(n\) increments, and let \(M_n\) be their sum, with \(M_0=0\). The increments are independent, equally likely to be \(1\) or \(-1\), so \(M_n\) is a martingale. Set \(\tau=3\); set \(\sigma=1\) if the first increment is \(1\), and \(\sigma=2\) otherwise. Find the conditional expectation of \(M_3\) given \(\mathcal{F}_\sigma\).
The choice of \(\sigma\) is a stopping time: \(\{\sigma\le1\}\) is the event that the first increment is \(1\), and \(\{\sigma\le2\}=\Omega\). Also \(0\le\sigma\le\tau=3\). The atoms of \(\mathcal{F}_\sigma\) are the first-increment-\(1\) event, and the two events specifying first increment \(-1\) and second increment \(1\) or \(-1\).
On the first atom, \(\sigma=1\) and \(M_\sigma=1\). The four equally likely terminal sums, as the last two increments vary, are \(3,1,1,-1\); their average is \((3+1+1-1)/4=1\). On the atom with first two increments \((-1,1)\), \(\sigma=2\), \(M_\sigma=0\), and the two terminal sums are \(1,-1\), with average \((1-1)/2=0\). On the atom with first two increments \((-1,-1)\), \(\sigma=2\), \(M_\sigma=-2\), and the terminal sums are \(-1,-3\), with average \((-1-3)/2=-2\). Therefore the conditional expectation equals \(M_\sigma\) on every atom. As a check, the unconditional mean is \(\tfrac12(1)+\tfrac14(0)+\tfrac14(-2)=0=\mathbb{E}[M_3]\).
Worked Example: Computing on Unequal-Size Information Atoms
Let \(\Omega=\{1,2,3,4,5,6\}\), with each outcome having probability \(\tfrac16\). Let \(\mathcal{F}_1\) distinguish \(B_1=\{1,2\}\) from \(B_2=\{3,4,5,6\}\), and let \(\mathcal{F}_2\) reveal the outcome. For \(C=\{1,2,3\}\), define \(M_n=\mathbb{P}(C\mid\mathcal{F}_n)\) for \(n=0,1,2\). Set \(\sigma=1\) and set \(\tau=2\) on \(B_1\), \(\tau=1\) on \(B_2\). Find \(\mathbb{E}[M_\tau\mid\mathcal{F}_1]\).
The initial value is \(M_0=\mathbb{P}(C)=\tfrac36=\tfrac12\). On \(B_1\), all two outcomes belong to \(C\), so \(M_1=2/2=1\); on \(B_2\), one of four outcomes belongs to \(C\), so \(M_1=1/4\). At time 2, \(M_2=\mathbf{1}_C\). The time \(\tau\) is a stopping time since \(\{\tau\le1\}=B_2\in\mathcal{F}_1\), and \(\sigma\le\tau\).
On \(B_1\), \(M_\tau=M_2=1\) at both outcomes, so its conditional average is \(1=M_1\). On \(B_2\), \(M_\tau=M_1=\tfrac14\) throughout, so its conditional average is \(\tfrac14=M_1\). Thus \(\mathbb{E}[M_\tau\mid\mathcal{F}_1]=M_1=M_\sigma\) on both atoms. The probabilities of the atoms are \(2/6=\tfrac13\) and \(4/6=\tfrac23\), so the unconditional check is \(\mathbb{E}[M_\tau]=\tfrac13(1)+\tfrac23(\tfrac14)=\tfrac12=\mathbb{E}[M_\sigma]\).
Why the Conditioning Field Matters
A frequent error is to replace the theorem’s conditioning field \(\mathcal{F}_\sigma\) with a later field, or to infer a conditional identity merely from equality of unconditional means. For a concrete distinction, use a fair one-step random walk with \(M_0=0\), \(M_1\in\{-1,1\}\), \(\sigma=0\), and \(\tau=1\). The times are bounded stopping times with \(\sigma\le\tau\). The correct identity is \(\mathbb{E}[M_1\mid\mathcal{F}_0]=M_0=0\), since \(\mathcal{F}_0\) contains no information about the increment. A wrong calculation would condition on \(\mathcal{F}_1\) and claim \(\mathbb{E}[M_1\mid\mathcal{F}_1]=M_0\); in fact \(\mathbb{E}[M_1\mid\mathcal{F}_1]=M_1\), which is not zero almost surely. The theorem specifies \(\mathcal{F}_\sigma\), not an arbitrary later sigma-field.
The boundedness and ordering hypotheses also have precise roles. The proof uses only finitely many increments because both times are bounded by \(N\). The ordering makes the sum of increments run forward from \(\sigma\) to \(\tau\); if the ordering holds only almost surely, that sum identity is correspondingly an almost-sure identity. One must not assert it on exceptional outcomes where \(\sigma>\tau\). Finally, \(\mathcal{F}_\sigma\) is not interchangeable with \(\mathcal{F}_n\) for a fixed \(n\): it records information at a random time.
Check Your Understanding
For each question, identify the conditioning information before calculating.
- In the two-coin example, what are the values of \(M_\sigma\) on the first-head and first-tail atoms?
- In the random-walk example, list the three atoms of \(\mathcal{F}_\sigma\) and the conditional average of \(M_3\) on each.
- Why is \(\{\tau\le1\}\in\mathcal{F}_1\) the relevant check for showing that a time taking values 1 and 2 is a stopping time?
- State the event \(D_n\) used in the proof and explain why it belongs to \(\mathcal{F}_n\) when \(A\in\mathcal{F}_\sigma\).
- For a fair one-step random walk with \(\sigma=0\) and \(\tau=1\), compare conditioning on \(\mathcal{F}_0\) with conditioning on \(\mathcal{F}_1\).