Three Separate Conditions
A martingale claim is not verified by checking that the process has mean zero, or even that its expectations are constant. The definition imposes three distinct requirements at every time: integrability, adaptation to the information available, and a conditional-mean identity. These conditions serve different purposes. Integrability makes the relevant expectations well-defined; adaptation says the current value is known at the current time; and the conditional identity says that, given the current information, the expected next value is the current value.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(S=\mathbb{R}\), and let \((X_n)_{n\ge0}\) be a discrete-time process with filtration \((\mathcal{F}_n)_{n\ge0}\). A filtration is an increasing sequence of sigma-fields: \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). The process is adapted if \(X_n\) is \(\mathcal{F}_n\)-measurable for every \(n\). It is integrable at time \(n\) if \(\mathbb{E}|X_n|<\infty\).
The equality is an almost-sure equality of random variables, not merely an equality of their averages. Since \(X_n\) is integrable and \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]=X_n\), the definition ensures integrability at each time; in the increment formulation, integrability of the process must still be checked separately. The filtration is part of the claim: changing what is known at time \(n\) can change the conditional expectation.
Proof. Assume first that \((X_n)\) is a martingale. Conditional expectation is linear for integrable random variables, so \(\mathbb{E}[X_{n+1}-X_n\mid\mathcal{F}_n]=\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]-\mathbb{E}[X_n\mid\mathcal{F}_n]\). Because \(X_n\) is \(\mathcal{F}_n\)-measurable, its conditional expectation given \(\mathcal{F}_n\) is \(X_n\); the martingale identity makes the first term \(X_n\) as well. The difference is zero almost surely. Conversely, suppose the conditional increment is zero. The same measurability gives \(\mathbb{E}[X_n\mid\mathcal{F}_n]=X_n\), so \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]=\mathbb{E}[X_n+(X_{n+1}-X_n)\mid\mathcal{F}_n]=X_n+0=X_n\). Adaptation and integrability were assumed, so all the martingale conditions hold. \(\square\)
This criterion organizes a check, but does not eliminate any of the three conditions. In particular, a conditional-mean computation cannot repair a failure of adaptation, and conditional expectations of non-integrable variables are not available under the definition used here.
Worked Example: Symmetry Does Not Supply Integrability
Let \(\Omega=\{(n,+),(n,-):n\ge1\}\), with \(\mathbb{P}((n,+))=\mathbb{P}((n,-))=\frac{1}{2n(n+1)}\). Set \(X(n,+)=n\), \(X(n,-)=-n\), and use \(\mathcal{F}_n=\sigma(X)\) for every time \(n\ge0\). Consider the constant process \(M_n=X\).
First check that this is a probability distribution. The total mass is \(\sum_{n\ge1}1/[n(n+1)]\). Since \(1/[n(n+1)]=1/n-1/(n+1)\), the partial sum through \(N\) is \(1-1/(N+1)\), which tends to \(1\). Each \(M_n\) is adapted, because \(M_n=X\) is \(\sigma(X)\)-measurable.
However, \(\mathbb{E}[|M_n|]=\sum_{n\ge1}2n/[2n(n+1)]=\sum_{n\ge1}1/(n+1)=\infty\). More precisely, both \(\mathbb{E}[M_n^+]\) and \(\mathbb{E}[M_n^-]\) equal \(\frac12\sum_{n\ge1}1/(n+1)=\infty\), so \(\mathbb{E}[M_n]\) is undefined. Symmetry makes each finite symmetric truncation have mean zero, but it does not make the untruncated random variable integrable or give it expectation zero.
Check: The probabilities sum to one and adaptation holds, but \(\mathbb{E}|M_n|=\infty\). Thus \((M_n)\) is not a martingale. In particular, writing \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) is not a valid martingale check: the required integrability is absent.
Adaptation Is About Current Information
Adaptation does not assert that a process has a particular average. It asserts that the value at time \(n\) can be determined from \(\mathcal{F}_n\). A process may have integrable values and may pass some conditional-mean calculations, yet fail to be adapted at a particular time. Because adaptation is required at every time, one failure is enough to rule out a martingale.
Worked Example: A Conditional-Mean Check Cannot Replace Adaptation
Take \(\Omega=\{-1,1\}\) with probability \(1/2\) at each point, and write \(Z(\omega)=\omega\). Let \(\mathcal{F}_0=\mathcal{F}_1=\{\varnothing,\Omega\}\), and let \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge2\). Define \(M_0=0\) and \(M_n=Z\) for \(n\ge1\). This is a filtration, and every \(M_n\) is integrable.
At time zero, \(\mathbb{E}[M_1\mid\mathcal{F}_0]=\mathbb{E}[Z]=\frac12(1)+\frac12(-1)=0=M_0\). At time one, \(\mathbb{E}[M_2\mid\mathcal{F}_1]=\mathbb{E}[Z]=0\), but this is not equal to \(M_1=Z\) almost surely: it equals \(0\) at both outcomes, whereas \(Z\) is either \(1\) or \(-1\). For \(n\ge2\), \(M_n\) is known from \(\mathcal{F}_n=\sigma(Z)\), so \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=Z=M_n\).
Check: \(M_1\) is not \(\mathcal{F}_1\)-measurable, because the only events in \(\mathcal{F}_1\) are \(\varnothing\) and \(\Omega\), while \(\{M_1=1\}\) is a nonempty proper subset of \(\Omega\). Therefore the process is not adapted and is not a martingale. The conditional equation at time zero does hold; it does not establish adaptation or the martingale property at later times.
Unconditional Mean Is Not Conditional Mean
For an integrable martingale, taking expectations in the defining conditional identity shows that its expectation is constant over time. The converse does not hold: knowing \(\mathbb{E}[M_{n+1}]=\mathbb{E}[M_n]\) only checks an average over all outcomes. The martingale condition checks the expected next value separately, given each piece of information available now.
Proof. If \((M_n)\) is a martingale, its defining identity and the tower property of conditional expectation give \(\mathbb{E}[M_{n+1}]=\mathbb{E}[\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]]=\mathbb{E}[M_n]\). To show that the converse fails, take a fair sign \(Z\in\{-1,1\}\), let \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge1\), and define \(M_0=0\), \(M_1=Z\), \(M_2=2Z\). All three variables are integrable and the process is adapted. Their expectations are all zero, since \(\mathbb{E}[Z]=\frac12(1)+\frac12(-1)=0\). But \(\mathbb{E}[M_2\mid\mathcal{F}_1]=2Z\), which is not \(M_1=Z\): for \(Z=1\) the two values are \(2\) and \(1\), and for \(Z=-1\) they are \(-2\) and \(-1\). Thus the expectations are constant but the martingale identity fails. \(\square\)
Worked Example: Centered Increments That Are Not Fair Given the Past
Use the fair sign \(Z\), the filtration \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_n=\sigma(Z)\) for \(n\ge1\), and the process \(M_0=0\), \(M_1=Z\), \(M_2=2Z\). The first increment is \(M_1-M_0=Z\), with unconditional mean \(\frac12(1)+\frac12(-1)=0\). The second increment is \(M_2-M_1=Z\), also with unconditional mean zero.
The second increment is already known at time one. Therefore \(\mathbb{E}[M_2-M_1\mid\mathcal{F}_1]=\mathbb{E}[Z\mid\sigma(Z)]=Z\), not zero almost surely. At \(Z=1\) the conditional increment is \(1\), and at \(Z=-1\) it is \(-1\). In contrast, the first conditional increment is \(\mathbb{E}[Z\mid\mathcal{F}_0]=0\), because \(\mathcal{F}_0\) contains no information about \(Z\).
Common wrong calculation: “\(\mathbb{E}[M_2-M_1]=0\), so \(\mathbb{E}[M_2-M_1\mid\mathcal{F}_1]=0\).” The second equality does not follow from the first. The correction is to condition using the information in \(\mathcal{F}_1\): since \(Z\) is \(\mathcal{F}_1\)-measurable, its conditional expectation is \(Z\). Thus this increment has zero unconditional mean but nonzero conditional mean almost surely.
Check: The increment values are \(1\) and \(-1\), so their average is \(0\). Their conditional expectation given \(\mathcal{F}_1\) equals the increment itself, which is \(1\) or \(-1\), not \(0\). Hence the second martingale identity fails, despite equality of the unconditional means.
A Practical Diagnostic
When assessing a proposed martingale, check the conditions in an order that prevents one type of calculation from obscuring another. First identify the filtration and verify that each \(M_n\) is measurable with respect to \(\mathcal{F}_n\). Next establish \(\mathbb{E}|M_n|<\infty\) for every time under consideration. Only then compute \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]\), or equivalently the conditional mean of the increment. The identity must hold almost surely for each \(n\), not just after taking an ordinary expectation.
Some familiar calculations can provide useful checks but cannot replace these steps. For example, constant expectations are necessary for an integrable martingale, as proved above, but are not sufficient. Likewise, an increment with unconditional mean zero need not have conditional mean zero. Symmetry can establish a zero expectation only when the expectation exists; checking finite truncations does not establish integrability of a limiting variable. And even a correct conditional-mean calculation does not show that the process is adapted.
Verify \(\mathbb{E}|M_n|<\infty\) at each time. If this fails, the standard martingale definition does not apply.
Verify that \(M_n\) is \(\mathcal{F}_n\)-measurable. Information revealed only later cannot be used as if it were already available.
Compute \(\mathbb{E}[M_{n+1}-M_n\mid\mathcal{F}_n]\) and verify that it is zero almost surely at every step.
The distinction between unconditional and conditional means is the central safeguard. A martingale has no expected one-step gain after conditioning on the available past. That is stronger than having no expected gain when all outcomes are averaged together.
Check Your Understanding
For each question, distinguish what is known from what still has to be checked.
- A process has \(\mathbb{E}[M_n]=0\) for every \(n\). Which martingale conditions remain unverified?
- For the fair-sign process in the last example, compute \(\mathbb{E}[M_2-M_1]\) and \(\mathbb{E}[M_2-M_1\mid\mathcal{F}_1]\). Why are they different?
- In the adaptation example, which event shows that \(M_1\) is not \(\mathcal{F}_1\)-measurable?
- Why do symmetric finite truncations with mean zero not establish that the random variable in the integrability example has expectation zero?
- State the conditional-increment criterion, including the assumptions needed to use it.