What the Martingale Condition Requires
A martingale formalizes the idea that, given the information currently available, the expected future value is the present value. This statement is about conditional expectation, not merely about averages over the whole probability space. The distinction matters: a process can have the same expectation at every time and still change in a systematically predictable way once its past is known.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where \(\mathcal{F}_s\subseteq\mathcal{F}_t\subseteq\mathcal{F}\) whenever \(s\le t\). The filtration represents the information available over time. When Brownian motion is used in an example, \(B_t\) denotes a standard Brownian motion adapted to the filtration, with increments after time \(s\) independent of \(\mathcal{F}_s\). No stochastic integrals are needed here.
Each condition has a separate role. Adaptedness says that the present value is available from the present information. Integrability makes the conditional expectation well-defined as an integrable random variable. The final condition says that no information in \(\mathcal{F}_s\) changes the conditional mean of the future value from \(M_s\). The definition includes the case \(s=t\), where the conditional-expectation identity follows from \(\mathcal{F}_t\)-measurability and integrability.
The definition is relative to a filtration. A process may be a martingale with respect to one filtration and fail to be a martingale with respect to a larger one, because the larger filtration may reveal information about its future. The expectation and conditional expectations here are all under \(\mathbb{P}\); in a change-of-measure setting the measure must be specified, since martingale properties can depend on it.
Equivalent Ways to Test the Conditional Identity
Conditional expectation can be characterized by averaging over events in the conditioning sigma-field. In particular, if \(X\) is integrable and \(Y\) is \(\mathcal{G}\)-measurable and integrable, then \(\mathbb{E}[X\mid\mathcal{G}]=Y\) almost surely exactly when \(\mathbb{E}[\mathbf{1}_A X]=\mathbb{E}[\mathbf{1}_A Y]\) for every \(A\in\mathcal{G}\). This is the defining characterization of conditional expectation. It yields useful tests for the martingale property.
- \(\mathbb{E}[\mathbf{1}_A M_t]=\mathbb{E}[\mathbf{1}_A M_s]\) for every \(A\in\mathcal{F}_s\);
- \(\mathbb{E}[Y M_t]=\mathbb{E}[Y M_s]\) for every bounded \(\mathcal{F}_s\)-measurable random variable \(Y\).
Proof. First suppose \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) almost surely. For any \(A\in\mathcal{F}_s\), the defining event property of conditional expectation gives \(\mathbb{E}[\mathbf{1}_A M_t]=\mathbb{E}[\mathbf{1}_A M_s]\). For any bounded \(\mathcal{F}_s\)-measurable \(Y\), the conditional-expectation averaging property gives \(\mathbb{E}[Y M_t]=\mathbb{E}[Y\,\mathbb{E}[M_t\mid\mathcal{F}_s]]=\mathbb{E}[Y M_s]\). These expectations exist because \(Y\) is bounded and both \(M_t,M_s\) are integrable. Conversely, if the event identity holds for every \(A\in\mathcal{F}_s\), then \(M_s\) is \(\mathcal{F}_s\)-measurable and integrable, and it has the same integrals as \(M_t\) over every event in \(\mathcal{F}_s\). By the defining characterization and almost-sure uniqueness of conditional expectation, \(M_s=\mathbb{E}[M_t\mid\mathcal{F}_s]\) almost surely. The bounded-variable test implies the event test by choosing \(Y=\mathbf{1}_A\), so all three statements are equivalent. \(\square\)
The bounded-variable version makes the information interpretation explicit: multiplying a future value by any bounded quantity measurable at time \(s\) does not change its expectation if \(M_t\) is replaced by \(M_s\). The event version is often convenient for direct verification, while the conditional-expectation form is the definition used most often.
A Basic Consequence: Constant Expectations
Proof. Since \(M_t\) is integrable, its conditional expectation given \(\mathcal{F}_s\) has expectation \(\mathbb{E}[M_t]\), by the unconditional-mean property of conditional expectation. The martingale identity therefore gives \(\mathbb{E}[M_t]=\mathbb{E}[\mathbb{E}[M_t\mid\mathcal{F}_s]]=\mathbb{E}[M_s]\). Taking \(s=0\) gives the second assertion. \(\square\)
This consequence is necessary but not sufficient. It checks only the average over \(\Omega\), whereas the martingale identity checks averages conditional on every event known at time \(s\). The third worked example makes that difference concrete.
Worked Example: Brownian Motion Is a Martingale
Let \(B\) be standard Brownian motion with respect to the filtration \((\mathcal{F}_t)\). Verify the three martingale conditions.
Adaptedness. Brownian motion is assumed adapted, so \(B_t\) is \(\mathcal{F}_t\)-measurable.
Integrability. Since \(B_t\) has distribution \(N(0,t)\), it has finite first absolute moment. For \(t>0\), \(\mathbb{E}[|B_t|]=\sqrt{2t/\pi}<\infty\), and at \(t=0\) the expectation is \(0\).
Conditional identity. For \(s\le t\), write \(B_t=B_s+(B_t-B_s)\). The increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has mean zero, including the degenerate case \(s=t\), when the increment is zero. Thus \(\mathbb{E}[B_t-B_s\mid\mathcal{F}_s]=0\), and \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\).
Check: Taking expectations gives \(\mathbb{E}[B_t]=0=\mathbb{E}[B_s]\). This confirms the constant-expectation consequence; the independent-increment calculation verifies the stronger conditional identity.
Worked Example: The Centered Square of Brownian Motion
For \(t\ge0\), define \(M_t=B_t^2-t\). Check that \(M\) is a martingale without using a stochastic differential.
Adaptedness and integrability. Since \(B_t\) is adapted, \(M_t\) is \(\mathcal{F}_t\)-measurable. Also \(\mathbb{E}[|M_t|]\le\mathbb{E}[B_t^2]+t=2t<\infty\), because \(\mathbb{E}[B_t^2]=t\).
Conditional identity. Let \(s\le t\) and write \(\Delta=B_t-B_s\). Then \(B_t=B_s+\Delta\), where \(\Delta\) is independent of \(\mathcal{F}_s\), \(\mathbb{E}[\Delta]=0\), and \(\mathbb{E}[\Delta^2]=t-s\). Expanding and conditioning, \(\mathbb{E}[B_t^2\mid\mathcal{F}_s]=B_s^2+2B_s\mathbb{E}[\Delta\mid\mathcal{F}_s]+\mathbb{E}[\Delta^2\mid\mathcal{F}_s]=B_s^2+(t-s)\). Consequently \(\mathbb{E}[M_t\mid\mathcal{F}_s]=B_s^2+(t-s)-t=B_s^2-s=M_s\).
Check: The mean is \(\mathbb{E}[M_t]=\mathbb{E}[B_t^2]-t=t-t=0\). The conditional calculation also works at \(s=t\), since then \(t-s=0\) and both sides equal \(M_t\).
A Finite-State Check at Every Pair of Times
Worked Example: A Martingale with Two Information Arrivals
Let \(\Omega=\{1,2,3,4\}\), give each outcome probability \(1/4\), and let \(U\) take values \(1,1,-1,-1\) on outcomes \(1,2,3,4\), respectively. Let \(V\) take values \(1,-1,1,-1\) on those outcomes. Then \(U\) and \(V\) are independent, each with mean zero. Use the filtration that is trivial for \(t<1\), reveals \(U\) for \(1\le t<2\), and reveals both \(U,V\) for \(t\ge2\). Define \(M_t=0\) for \(t<1\), \(M_t=U\) for \(1\le t<2\), and \(M_t=U+V\) for \(t\ge2\).
Every \(M_t\) is bounded by \(2\) in absolute value and is measurable with respect to the information available at time \(t\), so it is integrable and adapted. To check all pairs \(s\le t\), if both times lie in the same information interval, the process is unchanged, so its conditional expectation is itself. If \(s<1\le t<2\), then \(\mathbb{E}[M_t\mid\mathcal{F}_s]=\mathbb{E}[U]=0=M_s\). If \(s<1\) and \(t\ge2\), then \(\mathbb{E}[U+V]=0=M_s\). Finally, if \(1\le s<2\le t\), then \(U\) is known at time \(s\), while \(V\) is independent of \(U\) with mean zero. Hence \(\mathbb{E}[M_t\mid\mathcal{F}_s]=\mathbb{E}[U+V\mid U]=U=M_s\). These cases cover every pair \(s\le t\), including endpoints where the filtration changes.
Check: At each time the expectation is zero: before time \(1\), \(M_t=0\); between times \(1\) and \(2\), \(\mathbb{E}[U]=0\); and from time \(2\) onward, \(\mathbb{E}[U+V]=0+0=0\). The conditional check across the second information arrival is what establishes the martingale property, not the mean calculation alone.
Common Pitfall: Constant Means Are Not Enough
Here is a concrete counterexample on the same four equally likely outcomes. Keep the filtration from the preceding example, and define \(N_t=0\) for \(t<1\), \(N_t=U\) for \(1\le t<2\), and \(N_t=2U+V\) for \(t\ge2\). This process is adapted and integrable. Its expectation is zero at every time: \(\mathbb{E}[U]=0\) and \(\mathbb{E}[2U+V]=2\mathbb{E}[U]+\mathbb{E}[V]=0\).
Nevertheless, for \(1\le s<2\) and \(t\ge2\), the variable \(U\) is known at time \(s\), and \(V\) remains independent with mean zero. Therefore \(\mathbb{E}[N_t\mid\mathcal{F}_s]=2U\), which is not \(N_s=U\): since \(U\) is always either \(1\) or \(-1\), the two values differ on every outcome. The incorrect check “the expectation is zero at every time, so this is a martingale” has omitted the conditioning on past information. The correction is to test \(\mathbb{E}[N_t\mid\mathcal{F}_s]=N_s\) for every \(s\le t\), or an equivalent event or bounded-variable identity.
A martingale as defined here is a true martingale: each time value is integrable and the conditional identity holds. A local martingale is a different, weaker notion defined using a sequence of stopping times that localizes the process; a local martingale need not satisfy the true martingale identity without additional conditions. That distinction is important in stochastic calculus, but it does not weaken any condition in the definition above.
Check Your Understanding
For each question, distinguish unconditional expectations from conditional statements and state any integrability requirement you use.
- Which three conditions must an adapted process satisfy to be a martingale?
- Use the event test to restate \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for a fixed pair \(s\le t\).
- Why does integrability of \(M_t\) matter when defining \(\mathbb{E}[M_t\mid\mathcal{F}_s]\)?
- For Brownian motion, identify the conditional mean and conditional second moment of \(B_t-B_s\) given \(\mathcal{F}_s\).
- In the final example, compute \(\mathbb{E}[N_t\mid\mathcal{F}_s]\) when \(1\le s<2\le t\), and explain why its unconditional mean does not settle the martingale test.