Why One-Step Increments Are Enough
A martingale is defined by comparing the conditional expectation of a process at a future time with its current value. In discrete time, it is enough to make this comparison one step at a time: the relevant quantity is the increment \(X_{n+1}-X_n\), conditioned on the information available at time \(n\). This reformulation is useful because increments are often simpler to calculate than the conditional expectation of the entire next value.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a filtration \((\mathcal{F}_n)_{n\ge0}\), meaning that \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\subseteq\mathcal{F}\). Let \((X_n)_{n\ge0}\) be a real-valued process with state space \(S=\mathbb{R}\). The process is adapted if \(X_n\) is \(\mathcal{F}_n\)-measurable for every \(n\), and it is integrable if \(\mathbb{E}[|X_n|]<\infty\) for every \(n\). Equalities involving conditional expectations are understood to hold almost surely.
The criterion below is an equivalent formulation of this definition. It does not require independent increments, a particular state space beyond the one stated, or any assumption about how the process is distributed. It does require integrability, adaptation, and conditioning on the filtration immediately before the increment.
Proof. Fix \(n\ge0\). Since \(X_{n+1}\) and \(X_n\) are integrable, their difference is integrable, so the conditional expectation of \(X_{n+1}-X_n\) is defined. If \((X_n)\) is a martingale, then \(X_n\) is \(\mathcal{F}_n\)-measurable and hence \(\mathbb{E}[X_n\mid\mathcal{F}_n]=X_n\). Linearity of conditional expectation gives \(\mathbb{E}[X_{n+1}-X_n\mid\mathcal{F}_n]=\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]-\mathbb{E}[X_n\mid\mathcal{F}_n]=X_n-X_n=0\). Conversely, suppose this conditional increment is zero. Again using \(\mathcal{F}_n\)-measurability of \(X_n\) and linearity, \(\mathbb{E}[X_{n+1}\mid\mathcal{F}_n]=\mathbb{E}[X_{n+1}-X_n\mid\mathcal{F}_n]+\mathbb{E}[X_n\mid\mathcal{F}_n]=0+X_n\). This is the martingale condition at time \(n\). Since the argument applies for every \(n\), the process is a martingale. \(\square\)
Thus the criterion is a direct equivalence, not an approximation: checking every one-step conditional increment is exactly checking the martingale condition. The word conditional matters. A mean-zero increment on average over the whole probability space need not have mean zero after the information in \(\mathcal{F}_n\) is taken into account.
From One Step to Several Steps
Proof. Fix \(n\). For \(m=n\), the identity follows because \(X_n\) is \(\mathcal{F}_n\)-measurable. Suppose it holds for some \(m=k\ge n\). The tower property of conditional expectation and the martingale condition at time \(k\) give \(\mathbb{E}[X_{k+1}\mid\mathcal{F}_n]=\mathbb{E}[\mathbb{E}[X_{k+1}\mid\mathcal{F}_k]\mid\mathcal{F}_n]=\mathbb{E}[X_k\mid\mathcal{F}_n]=X_n\), where \(\mathcal{F}_n\subseteq\mathcal{F}_k\) permits the tower property. Induction proves the identity for all \(m\ge n\). \(\square\)
This proposition says that, given the information at time \(n\), the conditional expectation of any later value is the current value. It follows from the one-step condition; it is not necessary to check every pair of times separately. A related immediate consequence is that a martingale has constant expectation: taking expectations in the defining conditional expectation gives \(\mathbb{E}[X_{n+1}]=\mathbb{E}[X_n]\). The reverse implication does not hold, as the second example shows.
Worked Example: A Martingale with Dependent Increments
Let \(\Omega=\{(r,v):r,v\in\{-1,1\}\}\), with each of its four outcomes having probability \(\tfrac14\). Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\), let \(\mathcal{F}_1\) reveal \(R(r,v)=r\), and let \(\mathcal{F}_n\) for \(n\ge2\) reveal both \(R\) and \(V(r,v)=v\). Define \(X_0=0\), \(X_1=R\), and \(X_2=R+V\) when \(R=1\), while \(X_2=R+2V\) when \(R=-1\). For \(n\ge2\), set \(X_n=X_2\). This is an adapted, integrable process with state space \(\mathbb{R}\).
At time \(0\), the increment is \(\Delta X_1=R\). Its conditional expectation given the trivial \(\mathcal{F}_0\) is its ordinary expectation: \(\mathbb{E}[R\mid\mathcal{F}_0]=\tfrac12(1)+\tfrac12(-1)=0\).
At time \(1\), condition on the value of \(R\), which is exactly the information in \(\mathcal{F}_1\). If \(R=1\), then \(V\) has conditional probabilities \(\tfrac12\) for each of \(-1\) and \(1\), so \(\mathbb{E}[\Delta X_2\mid R=1]=\tfrac12(-1)+\tfrac12(1)=0\). If \(R=-1\), the increment is \(2V\), and \(\mathbb{E}[\Delta X_2\mid R=-1]=\tfrac12(-2)+\tfrac12(2)=0\). Therefore \(\mathbb{E}[\Delta X_2\mid\mathcal{F}_1]=0\). For \(n\ge2\), \(X_{n+1}=X_n\), so each later increment is zero. The criterion proves that \((X_n)\) is a martingale.
Check by listing the outcomes. In the order \((R,V)=(1,1),(1,-1),(-1,1),(-1,-1)\), the values of \(X_2\) are \(2,0,1,-3\). Given \(R=1\), their first two values average to \(1=X_1\); given \(R=-1\), their last two values average to \(-1=X_1\). Also \(\mathbb{E}[X_2]=\tfrac14(2+0+1-3)=0=\mathbb{E}[X_0]\), consistent with the martingale property. The increments are dependent in magnitude on \(R\), but their conditional means are zero.
Worked Example: Unconditional Mean Zero Is Not Enough
Let \(\Omega=\{-1,1\}\), with probability \(\tfrac12\) on each outcome, and write \(Z(\omega)=\omega\). Take \(\mathcal{F}_0\) to be trivial and let \(\mathcal{F}_n=\sigma(Z)\) for every \(n\ge1\). Define \(X_0=0\), \(X_1=Z\), \(X_2=2Z\), and \(X_n=X_2\) for \(n\ge2\). The process is adapted and integrable.
Both first increments have unconditional mean zero: \(\mathbb{E}[\Delta X_1]=\mathbb{E}[Z]=\tfrac12(1)+\tfrac12(-1)=0\), and \(\mathbb{E}[\Delta X_2]=\mathbb{E}[Z]=0\). However, the filtration at time \(1\) already reveals \(Z\). Thus \(\mathbb{E}[\Delta X_2\mid\mathcal{F}_1]=Z\), which is not zero: it equals \(1\) when \(Z=1\) and \(-1\) when \(Z=-1\). Equivalently, \(\mathbb{E}[X_2\mid\mathcal{F}_1]=2Z\ne Z=X_1\) on both outcomes. The process is not a martingale.
Check: The expectations of \(X_0,X_1,X_2\) are all \(0\), but the conditional martingale equation at time \(1\) fails. Constant expectations alone do not establish the martingale property.
Conditioning on a Revealed Regime
A useful application of the criterion is a sequence of payoffs whose conditional mean is set to zero after observing the current information. The probability of a positive payoff may depend on what has already been revealed; what matters is that the payoff is centered using that conditional probability, not an overall average.
Worked Example: Centering a Payoff by Its Conditional Probability
Let \(\Omega=\{1,\ldots,8\}\) have equal probabilities \(\tfrac18\). Define a regime \(R\) by \(R=1\) on outcomes \(1,2,3,4\) and \(R=-1\) on outcomes \(5,6,7,8\). Let \(I=1\) on outcomes \(1,2,5\) and \(I=0\) on the other outcomes. Set \(\mathcal{F}_0\) trivial, \(\mathcal{F}_1=\sigma(R)\), and \(\mathcal{F}_n=\sigma(R,I)\) for \(n\ge2\). Define \(X_0=0\), \(X_1=R\), and \(X_2=R+I-\tfrac12\) when \(R=1\), while \(X_2=R+I-\tfrac14\) when \(R=-1\). Keep \(X_n=X_2\) for \(n\ge2\). This finite-valued process is adapted and integrable.
The first increment is \(\Delta X_1=R\). There are four outcomes in each regime, so \(\mathbb{E}[R\mid\mathcal{F}_0]=\tfrac12(1)+\tfrac12(-1)=0\). Given \(R=1\), exactly two of the four outcomes have \(I=1\), and hence \(\mathbb{E}[I\mid R=1]=\tfrac24=\tfrac12\). Given \(R=-1\), exactly one of the four outcomes has \(I=1\), and hence \(\mathbb{E}[I\mid R=-1]=\tfrac14\). Therefore the next increment has conditional mean \(\mathbb{E}[\Delta X_2\mid R=1]=\tfrac12-\tfrac12=0\) and \(\mathbb{E}[\Delta X_2\mid R=-1]=\tfrac14-\tfrac14=0\). All later increments are zero. By the criterion, \((X_n)\) is a martingale.
Check by conditional averages. When \(R=1\), \(X_2\) is \(\tfrac32\) on two outcomes and \(\tfrac12\) on two outcomes, so its conditional average is \(\tfrac24\cdot\tfrac32+\tfrac24\cdot\tfrac12=\tfrac34+\tfrac14=1=X_1\). When \(R=-1\), \(X_2=-\tfrac14\) on one outcome and \(-\tfrac54\) on three outcomes, so its conditional average is \(\tfrac14(-\tfrac14)+\tfrac34(-\tfrac54)=-\tfrac1{16}-\tfrac{15}{16}=-1=X_1\). The conditional averages agree with the criterion in both regimes.
A Reliable Test and a Common Mistake
For a proposed martingale, first identify the filtration: it specifies what is known at each time. Then verify integrability and adaptation, calculate \(X_{n+1}-X_n\), and condition that increment on \(\mathcal{F}_n\). If the result is zero almost surely at every step, the criterion proves the martingale property. If it is nonzero on an event of positive probability at even one step, the process is not a martingale with respect to that filtration.
Common mistake: replacing conditional mean zero with ordinary mean zero. In the second example, the incorrect calculation is \(\mathbb{E}[\Delta X_2]=\mathbb{E}[Z]=0\), followed by the claim that the second martingale condition holds. That calculation averages over both values of information already available at time \(1\). The required calculation is \(\mathbb{E}[\Delta X_2\mid\mathcal{F}_1]=Z\), which is nonzero on each outcome. Ordinary mean zero is a necessary consequence of the conditional criterion, by taking expectations, but it is not a substitute for conditioning.
The criterion is useful whenever the process is naturally described by successive changes: it turns a global martingale question into a sequence of conditional calculations. The conditioning information must match the filtration, and a zero conditional expectation is required at every step, not just on average across time.
Check Your Understanding
For each question, state which filtration is being used and distinguish conditional expectation from ordinary expectation.
- Suppose \((X_n)\) is integrable and adapted and \(\mathbb{E}[X_{n+1}-X_n\mid\mathcal{F}_n]=0\) for every \(n\). Which part of the martingale definition follows directly from adaptation, and which part follows from the increment condition?
- In the first worked example, what are the two possible values of \(\Delta X_2\) conditional on \(R=-1\), and what is their conditional mean?
- For the process \(X_0=0,X_1=Z,X_2=2Z\) in the second example, compute \(\mathbb{E}[\Delta X_2\mid\mathcal{F}_1]\) separately for \(Z=1\) and \(Z=-1\).
- In the third example, how many outcomes have \(I=1\) in each regime, and what conditional probability does each count give?
- Explain how the tower property gives \(\mathbb{E}[X_m\mid\mathcal{F}_n]=X_n\) from the one-step martingale equations when \(m>n\).