From Mean-Zero Increments to a Second-Moment Identity
A martingale has no predictable drift: conditional on the information available at the previous time, its next increment has mean zero. That fact has a stronger consequence than keeping the expected value constant. It also makes increments at different times orthogonal in \(L^2\), so when a martingale is written as its initial value plus its accumulated increments, the cross terms in its second moment vanish. The resulting identity expresses its second moment as an initial contribution plus a sum of nonnegative increment variances.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sub-\(\sigma\)-algebras of \(\mathcal{F}\). A random variable \(X\) is square-integrable if \(\mathbb{E}[X^2]<\infty\). For \(k\ge1\), define the increment \(\Delta M_k=M_k-M_{k-1}\).
The last assertion follows by subtracting \(M_{n-1}\) from the conditional-expectation equation, since \(M_{n-1}\) is \(\mathcal{F}_{n-1}\)-measurable. The decomposition below holds for every finite time \(n\). It does not require a limiting argument or an assumption that the martingale converges.
Why Distinct Increments Are Orthogonal
For square-integrable random variables \(U,V\), orthogonality means \(\mathbb{E}[UV]=0\); equivalently, their \(L^2\) inner product is zero. In this setting an earlier increment is already known by the time a later increment is taken, while the later increment has conditional mean zero.
Proof. Fix \(1\le i<j\). The variable \(\Delta M_i\) is \(\mathcal{F}_{j-1}\)-measurable, because \(M_i\) and \(M_{i-1}\) are measurable with respect to \(\mathcal{F}_i\), and \(\mathcal{F}_i\subseteq\mathcal{F}_{j-1}\). Both \(\Delta M_i\) and \(\Delta M_j\) are square-integrable, so their product is integrable by Cauchy–Schwarz. Conditional expectation gives \(\mathbb{E}[\Delta M_j\mid\mathcal{F}_{j-1}]=0\). Thus \(\mathbb{E}[\Delta M_i\Delta M_j]=0\). For an unbounded \(\mathcal{F}_{j-1}\)-measurable multiplier, this conditional-expectation step follows first for its bounded truncations and then by \(L^1\) convergence of the products, again by Cauchy–Schwarz. The same argument applies to \(M_0\), which is \(\mathcal{F}_{j-1}\)-measurable, and proves \(\mathbb{E}[M_0\Delta M_j]=0\). \(\square\)
Each increment has mean zero as well: taking expectations of its conditional mean gives \(\mathbb{E}[\Delta M_j]=0\). Consequently \(\operatorname{Var}(\Delta M_j)=\mathbb{E}[(\Delta M_j)^2]\). The next result puts these facts together.
Proof. Fix \(n\ge1\). Telescoping the increments gives \(M_n=M_0+\sum_{k=1}^n\Delta M_k\). Squaring and taking expectations yields the sum of the initial square, the increment squares, and twice each cross-product expectation. Every cross term between \(M_0\) and an increment has expectation zero by the lemma. Every cross term between distinct increments also has expectation zero by the lemma. Therefore \(\mathbb{E}[M_n^2]=\mathbb{E}[M_0^2]+\sum_{k=1}^n\mathbb{E}[(\Delta M_k)^2]\). Each increment has mean zero, so its variance equals its second moment. Also \(\mathbb{E}[M_n]=\mathbb{E}[M_0]+\sum_{k=1}^n\mathbb{E}[\Delta M_k]=\mathbb{E}[M_0]\). Subtracting \((\mathbb{E}[M_0])^2=(\mathbb{E}[M_n])^2\) from the second-moment identity gives the variance identity, since \(\operatorname{Var}(\Delta M_k)=\mathbb{E}[(\Delta M_k)^2]\). For \(n=0\), each sum is empty and the statements reduce to identities. \(\square\)
The result separates the contribution present at time zero from the variability added at each later step. If \(M_0=0\) almost surely, the second moment is exactly the sum of the increment variances. If \(M_0\) is random, its second moment must be retained; the variance formula instead retains its variance.
Worked Applications
Worked Example: A Three-Step Symmetric Random Walk
Let \(\Omega=\{-1,1\}^3\), with each of its eight outcomes having probability \(1/8\). Write the coordinate variables as \(R_1,R_2,R_3\), and let \(\mathcal{F}_0\) be the trivial \(\sigma\)-algebra and \(\mathcal{F}_k=\sigma(R_1,\ldots,R_k)\) for \(k=1,2,3\). Define \(M_0=0\) and \(M_k=\sum_{j=1}^kR_j\). Each \(R_k\) is equally likely to be \(-1\) or \(1\), independently of the earlier coordinates, so \(\mathbb{E}[R_k\mid\mathcal{F}_{k-1}]=0\). Thus \(M_k\) is a square-integrable martingale and \(\Delta M_k=R_k\).
For every \(k\), \(R_k^2=1\), so \(\mathbb{E}[(\Delta M_k)^2]=1\). The decomposition gives \(\mathbb{E}[M_3^2]=0+1+1+1=3\).
Check by the distribution. Among the eight equally likely sign triples, two have sum \(3\) or \(-3\), and six have sum \(1\) or \(-1\). Hence \(\mathbb{E}[M_3^2]=\tfrac{2}{8}\cdot9+\tfrac{6}{8}\cdot1=\tfrac{18}{8}+\tfrac{6}{8}=3\), agreeing with the decomposition.
Worked Example: A Martingale from Conditional Expectations
Let \(\Omega=\{a,b,c,d\}\) with probability \(1/4\) on each outcome. Define \(Y(a)=0\), \(Y(b)=2\), \(Y(c)=2\), and \(Y(d)=4\). Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) distinguish the blocks \(\{a,b\}\) and \(\{c,d\}\), and let \(\mathcal{F}_2\) contain all subsets of \(\Omega\). Set \(M_0=\mathbb{E}[Y\mid\mathcal{F}_0]\), \(M_1=\mathbb{E}[Y\mid\mathcal{F}_1]\), and \(M_2=Y\). These are conditional expectations of the same square-integrable variable, and their conditional means at successive times equal the preceding values; hence they form a square-integrable martingale.
The four values of \(Y\) sum to \(0+2+2+4=8\), so \(M_0=8/4=2\). On \(\{a,b\}\), the conditional mean is \((0+2)/2=1\); on \(\{c,d\}\), it is \((2+4)/2=3\). Therefore \(M_1\) takes values \(1,1,3,3\) on \(a,b,c,d\), respectively. The first increment takes values \(-1,-1,1,1\), so \(\mathbb{E}[(\Delta M_1)^2]=1\). The second increment \(Y-M_1\) takes values \(-1,1,-1,1\), so \(\mathbb{E}[(\Delta M_2)^2]=1\).
The identity gives \(\mathbb{E}[M_2^2]=2^2+1+1=6\). Directly, \(\mathbb{E}[Y^2]=(0^2+2^2+2^2+4^2)/4=(0+4+4+16)/4=6\). Also, \(\operatorname{Var}(Y)=6-2^2=2\), matching \(\operatorname{Var}(M_0)+\operatorname{Var}(\Delta M_1)+\operatorname{Var}(\Delta M_2)=0+1+1=2\).
Worked Example: A Random Initial Value
Let \(\Omega=\{-1,1\}^3\) with the uniform distribution, and denote its independent coordinate variables by \(Z,R,S\). Take \(\mathcal{F}_0=\sigma(Z)\), \(\mathcal{F}_1=\sigma(Z,R)\), and \(\mathcal{F}_2=\sigma(Z,R,S)\). Define \(M_0=Z\), \(M_1=Z+R\), and \(M_2=Z+R+2S\). Since \(R\) is independent of \(Z\) and has mean zero, \(\mathbb{E}[M_1\mid\mathcal{F}_0]=Z=M_0\). Since \(S\) is independent of \(\mathcal{F}_1\) and has mean zero, \(\mathbb{E}[M_2\mid\mathcal{F}_1]=Z+R=M_1\). This is a square-integrable martingale.
Here \(\mathbb{E}[M_0^2]=\mathbb{E}[Z^2]=1\), \(\Delta M_1=R\) has second moment \(1\), and \(\Delta M_2=2S\) has second moment \(4\). Thus \(\mathbb{E}[M_2^2]=1+1+4=6\).
Check directly. Conditional on \(Z,R\), put \(u=Z+R\). Averaging over the two equally likely values of \(S\) gives \(\mathbb{E}[(u+2S)^2\mid Z,R]=\big((u+2)^2+(u-2)^2\big)/2=u^2+4\). The four equally likely pairs \((Z,R)\) give \(u=-2,0,0,2\), so \(\mathbb{E}[u^2]=(4+0+0+4)/4=2\). Therefore \(\mathbb{E}[M_2^2]=2+4=6\), as required. This example also shows why the initial second moment cannot be dropped: the increment terms alone total \(1+4=5\), not \(6\).
Conditions, Uses, and a Common Error
Square-integrability matters because the claimed second moments and increment variances must be finite, and the cross-products must be integrable. The martingale condition matters because it supplies the conditional mean-zero property that removes cross terms. The decomposition is a finite-time identity under these hypotheses; it does not by itself assert that an infinite sum converges or that a martingale has a limit.
A useful reading of the formula is as an accounting identity: each step adds a nonnegative quantity to the second moment. In particular, \(\mathbb{E}[M_n^2]\ge\mathbb{E}[M_0^2]\). The variance version says that the spread around the common mean cannot decrease across these martingale times. These statements concern second moments and variances, not pathwise monotonicity: individual sample paths can move up or down.
Common mistake: applying the identity without the martingale condition. Let \(M_0=0\), \(M_1=1\), and \(M_2=2\) deterministically. The increments are both \(1\), so the claimed right-hand side would be \(0+1+1=2\), while \(\mathbb{E}[M_2^2]=4\). The failure is not an arithmetic issue: this process is not a martingale, since \(\mathbb{E}[M_1\mid\mathcal{F}_0]=1\ne0=M_0\). Its cross term is nonzero: expanding \(M_2^2=(\Delta M_1+\Delta M_2)^2\) gives \(1+1+2(1)(1)=4\). By contrast, for a martingale the conditional mean-zero property makes that cross term vanish in expectation.
When using the decomposition, first identify the filtration and verify the martingale condition. Then compute the initial second moment and each increment’s second moment. If the starting value is deterministic, its variance is zero, but its second moment is generally not zero; the two versions of the identity should not be confused.
Check Your Understanding
Use the martingale condition and the finite-time identities to answer the following questions.
- For a square-integrable martingale, explain why \(\mathbb{E}[\Delta M_k]=0\) and why this makes \(\mathbb{E}[(\Delta M_k)^2]=\operatorname{Var}(\Delta M_k)\).
- If \(M_0=3\) deterministically and the first three increment variances are \(2\), \(0\), and \(5\), find \(\mathbb{E}[M_3^2]\) and \(\operatorname{Var}(M_3)\).
- In the conditional-expectation example, identify the two possible values of \(M_1\) and compute the conditional mean of \(M_2\) on each block of \(\mathcal{F}_1\).
- Why does the proof require the earlier increment \(\Delta M_i\) to be measurable with respect to \(\mathcal{F}_{j-1}\) when \(i<j\)?
- Give a deterministic process whose increments have nonzero cross products, and explain which martingale condition fails.