Why Stopping Preserves a Martingale
A martingale has increments whose conditional mean, given the information available just before each increment, is zero. If a process is stopped, it keeps an increment precisely when the process has not stopped before that increment occurs. The key fact is that this decision is already determined by the available information. Thus stopping retains the martingale’s zero-conditional-mean increment structure.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sigma-fields describing the information available over time. A discrete-time process \((M_n)_{n\ge0}\) is a martingale with respect to this filtration if each \(M_n\) is integrable and \(\mathcal{F}_n\)-measurable, and \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) almost surely for every \(n\ge1\). Define its increment by \(\Delta M_k=M_k-M_{k-1}\). The martingale condition is equivalently \(\mathbb{E}[\Delta M_k\mid\mathcal{F}_{k-1}]=0\) almost surely.
Although \(\tau\) may equal infinity, \(n\wedge\tau\) is finite at every fixed \(n\). Consequently, this result concerns the process evaluated at each finite time; it does not by itself make a claim about the random variable \(M_\tau\) when \(\tau\) can be infinite.
The Increment Identity
Proof. Fix an outcome \(\omega\) and a finite \(n\). For each \(k\in\{1,\ldots,n\}\), the indicator \(\mathbf{1}_{\{\tau(\omega)\ge k\}}\) equals \(1\) exactly when \(k\le n\wedge\tau(\omega)\). Thus the sum on the right includes the increments from \(1\) through \(n\wedge\tau(\omega)\), and no others. Telescoping those increments gives $$ M_0(\omega)+\sum_{k=1}^{n\wedge\tau(\omega)} \bigl(M_k(\omega)-M_{k-1}(\omega)\bigr) =M_{n\wedge\tau(\omega)}(\omega). $$ If \(\tau(\omega)=0\), the sum has no terms and both sides equal \(M_0(\omega)\). If \(\tau(\omega)\ge n\), including \(\tau(\omega)=\infty\), the sum runs through \(n\) and gives \(M_n(\omega)\). These cases establish the identity for every outcome for which the process values are defined, and hence almost surely. \(\square\)
The indicator in the identity records whether the process is still running just before step \(k\). The stopping-time property makes this decision measurable using \(\mathcal{F}_{k-1}\), since $$ \{\tau\ge k\}=\{\tau\le k-1\}^{c}\in\mathcal{F}_{k-1},\qquad k\ge1. $$ This measurability is what allows the martingale increment condition to be used.
Worked Example: A Stop Chosen After the First Toss
Let \((\xi_n)_{n\ge1}\) be independent fair coin increments taking values \(+1\) and \(-1\), on their product probability space. Let \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\), with \(\mathcal{F}_0\) trivial, and set \(M_0=0\), \(M_n=\sum_{k=1}^n\xi_k\). Each increment has conditional mean zero given \(\mathcal{F}_{n-1}\), so \((M_n)\) is a martingale. Stop at \(\tau=1\) if \(\xi_1=+1\), and at \(\tau=2\) if \(\xi_1=-1\).
This is a stopping time: \(\{\tau\le0\}=\varnothing\), \(\{\tau\le1\}=\{\xi_1=+1\}\in\mathcal{F}_1\), and \(\{\tau\le2\}=\Omega\). The identity at \(n=2\) gives $$ M_{2\wedge\tau}=\xi_1+\mathbf{1}_{\{\xi_1=-1\}}\xi_2. $$ Indeed, \(\tau\ge1\) always, while \(\tau\ge2\) exactly when \(\xi_1=-1\).
The four equally likely pairs \((\xi_1,\xi_2)\) give the following stopped values: \((+1,+1)\) gives \(1\), \((+1,-1)\) gives \(1\), \((-1,+1)\) gives \(0\), and \((-1,-1)\) gives \(-2\). Therefore $$ \mathbb{E}[M_{2\wedge\tau}] =\tfrac14(1+1+0-2)=0. $$ Also \(M_{1\wedge\tau}=M_1=\xi_1\), whose expectation is \(\tfrac12(1)+\tfrac12(-1)=0\). More strongly, conditional on \(\mathcal{F}_1\), if \(\xi_1=+1\) the stopped value at time \(2\) is \(1\); if \(\xi_1=-1\), its conditional mean is \(\tfrac12(0)+\tfrac12(-2)=-1\). These are exactly the two possible values of \(M_{1\wedge\tau}\).
Stopping Preserves the Martingale Property
Proof. Write \(Y_n=M_{n\wedge\tau}\). First, \(Y_n\) is \(\mathcal{F}_n\)-measurable: the increment identity expresses it as \(M_0\) plus a finite sum, and for \(1\le k\le n\), both \(\mathbf{1}_{\{\tau\ge k\}}\) and \(\Delta M_k\) are \(\mathcal{F}_n\)-measurable. It is integrable because that same finite sum consists of integrable terms: \(\mathbf{1}_{\{\tau\ge k\}}\) is bounded by \(1\), and \(\Delta M_k\) is integrable.
For \(n\ge1\), the pathwise identity for consecutive times gives $$ Y_n-Y_{n-1}=\mathbf{1}_{\{\tau\ge n\}}\Delta M_n. $$ The event \(\{\tau\ge n\}\) belongs to \(\mathcal{F}_{n-1}\). Taking conditional expectation and using the martingale increment condition therefore yields $$ \mathbb{E}[Y_n-Y_{n-1}\mid\mathcal{F}_{n-1}] =\mathbf{1}_{\{\tau\ge n\}}\mathbb{E}[\Delta M_n\mid\mathcal{F}_{n-1}] =0 $$ almost surely. Since \(Y_{n-1}\) is \(\mathcal{F}_{n-1}\)-measurable, this is equivalent to \(\mathbb{E}[Y_n\mid\mathcal{F}_{n-1}]=Y_{n-1}\) almost surely. Thus \(Y\) is adapted, integrable at every time, and satisfies the martingale condition. \(\square\)
As a direct consequence, for each fixed finite \(n\), \(\mathbb{E}[M_{n\wedge\tau}]=\mathbb{E}[M_0]\). This follows by taking expectations in the martingale condition for \(Y\) at each of the finitely many steps from \(1\) through \(n\). It is a statement about the stopped process at a deterministic finite time \(n\), not a claim that an unbounded stopping time can simply be substituted into an expectation.
Worked Example: A Stopping Time That May Be Infinite
Use the same fair coin walk \(M_n=\sum_{k=1}^n\xi_k\) and filtration as above. Define \(\tau=\inf\{k\ge1:\xi_k=+1\}\), with \(\inf\varnothing=\infty\). This is a stopping time because \(\{\tau\le n\}=\bigcup_{k=1}^n\{\xi_k=+1\}\in\mathcal{F}_n\); for \(n=0\) the event is empty.
At \(n=3\), the identity reads $$ M_{3\wedge\tau} =\xi_1+\mathbf{1}_{\{\xi_1=-1\}}\xi_2 +\mathbf{1}_{\{\xi_1=-1,\,\xi_2=-1\}}\xi_3. $$ The events \(\{\tau=1\}\), \(\{\tau=2\}\), \(\{\tau=3\}\), and \(\{\tau>3\}\) have probabilities \(\tfrac12,\tfrac14,\tfrac18,\tfrac18\), respectively, by independence and fairness. On these events the stopped values are \(1,0,-1,-3\), respectively. Their expectation is $$ \tfrac12(1)+\tfrac14(0)+\tfrac18(-1)+\tfrac18(-3) =\tfrac48-\tfrac18-\tfrac38=0=\mathbb{E}[M_0]. $$ The indicator terms also show why the process is stopped only after the increment at time \(\tau\) has been included: on \(\{\tau=3\}\), the third increment is retained. The theorem applies at every finite \(n\) without first requiring a finite bound on \(\tau\).
Why the Stopping-Time Condition Matters
The condition is more than a technical way to describe a random time. It ensures that the decision to retain the next increment uses only information already available. If the decision depends on future information, the indicator multiplying an increment need not be measurable at the preceding time, so the conditional-mean-zero argument can fail.
Worked Example: A Random Time That Is Not a Stopping Time
For the fair coin walk and filtration above, define \(\sigma=1\) if \(\xi_2=+1\), and \(\sigma=2\) if \(\xi_2=-1\). This is not a stopping time: \(\{\sigma\le1\}=\{\xi_2=+1\}\notin\mathcal{F}_1\), because at time \(1\) the second toss has not been observed.
The process stopped at \(\sigma\) has \(M_{1\wedge\sigma}=\xi_1\), since \(\sigma\ge1\) always. At time \(2\), if \(\xi_2=+1\) it has stopped at time \(1\), while if \(\xi_2=-1\) it includes the second increment. Hence $$ M_{2\wedge\sigma} =\mathbf{1}_{\{\xi_2=+1\}}\xi_1 +\mathbf{1}_{\{\xi_2=-1\}}(\xi_1+\xi_2). $$ Conditioning on \(\mathcal{F}_1\), the second toss is still fair, so $$ \mathbb{E}[M_{2\wedge\sigma}\mid\mathcal{F}_1] =\tfrac12\xi_1+\tfrac12(\xi_1-1) =\xi_1-\tfrac12. $$ This is not \(M_{1\wedge\sigma}=\xi_1\). For \(\xi_1=+1\), the conditional mean is \(\tfrac12\), rather than \(1\); for \(\xi_1=-1\), it is \(-\tfrac32\), rather than \(-1\). Thus this stopped process is not a martingale.
Common indexing mistake. In the increment formula, the correct indicator is \(\mathbf{1}_{\{\tau\ge k\}}\), not \(\mathbf{1}_{\{\tau>k\}}\). For the first example at \(n=1\), \(\tau\ge1\) always, so the correct formula gives \(M_{1\wedge\tau}=M_0+\xi_1=\xi_1\). Replacing the indicator by \(\mathbf{1}_{\{\tau>1\}}\) would give zero when \(\tau=1\), incorrectly deleting the increment that brings the process to its stopping time. The stopping decision for increment \(k\) is whether the process has stopped by time \(k-1\), which is why the event is \(\{\tau>k-1\}=\{\tau\ge k\}\).
The identity is useful because it turns stopping into a finite sum of original martingale increments, each multiplied by an indicator known just before that increment. It establishes the martingale property of the stopped process without any assumption that the stopping time is bounded. The exact conclusion is deliberately finite-time: for each deterministic \(n\), \(M_{n\wedge\tau}\) is integrable and obeys the martingale conditional expectation relation. That conclusion alone does not justify evaluating the original process at an unbounded random time.
Check Your Understanding
For each question, identify the information available at the relevant time and use the increment identity where appropriate.
- Why does \(\{\tau\ge k\}\in\mathcal{F}_{k-1}\) follow from the definition of a stopping time?
- In the increment representation, what does the term \(\mathbf{1}_{\{\tau\ge k\}}\Delta M_k\) mean on the event \(\{\tau=k\}\)?
- Which three properties must be checked to establish that \(M_{n\wedge\tau}\) is a martingale with respect to the original filtration?
- For the fair coin walk stopped at the first \(+1\) increment, list the probabilities and stopped values at time \(n=2\), and verify that their expected value is zero.
- For the random time \(\sigma\) depending on \(\xi_2\), which stopping-time condition fails, and what conditional expectation shows that the stopped process is not a martingale?