Using Past Information to Weight Martingale Increments
A martingale increment has conditional mean zero given the information available just before it occurs. If we multiply that increment by a bounded coefficient chosen using only that same past information, the conditional mean remains zero. This simple operation is useful for forming new martingales from existing ones, including strategies that adjust their weights as information arrives. The timing of the coefficient is essential: a weight chosen after seeing the increment need not preserve the martingale property.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of information sets. Let \((M_n)_{n\ge0}\) be an integrable martingale adapted to this filtration: each \(M_n\) is integrable and \(\mathcal{F}_n\)-measurable, and \(\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]=M_{n-1}\) for \(n\ge1\). Its increment at time \(n\) is \(\Delta M_n=M_n-M_{n-1}\). The martingale condition gives \(\mathbb{E}[\Delta M_n\mid\mathcal{F}_{n-1}]=0\) almost surely.
Predictability specifies when the coefficient is chosen. It may be random and depend on the entire past, provided it is measurable with respect to \(\mathcal{F}_{n-1}\). Boundedness supplies a convenient sufficient condition for integrability: assume that each \(H_n\) is essentially bounded, with a common finite bound \(C\), so \(|H_n|\le C\) almost surely. The transformed partial sums, starting at zero, are
The Conditional Multiplication Rule
The key calculation is that a bounded coefficient measurable with respect to the conditioning information can be taken outside a conditional expectation. We state and prove the version needed here.
Proof. Let \(|Z|\le C\) almost surely. Then \(\mathbb{E}|ZX|\le C\mathbb{E}|X|<\infty\), so \(ZX\) is integrable. First suppose \(Z\) is a simple \(\mathcal{G}\)-measurable random variable, written \(Z=\sum_{r=1}^m c_r\mathbf{1}_{A_r}\) for disjoint \(A_r\in\mathcal{G}\). The defining integral property of conditional expectation gives \(\mathbb{E}[\mathbf{1}_{A_r}X\mid\mathcal{G}]=\mathbf{1}_{A_r}\mathbb{E}[X\mid\mathcal{G}]\), and linearity gives the claimed identity for \(Z\). For a general bounded \(Z\), choose simple \(\mathcal{G}\)-measurable \(Z_j\) with \(|Z_j-Z|\le 1/j\) almost surely. Then \(\mathbb{E}|(Z_j-Z)X|\le \mathbb{E}|X|/j\), and \(\mathbb{E}|(Z_j-Z)\mathbb{E}[X\mid\mathcal{G}]|\le \mathbb{E}|X|/j\), using \(\mathbb{E}|\mathbb{E}[X\mid\mathcal{G}]|\le\mathbb{E}|X|\). Thus both sides of the identity for \(Z_j\) converge in \(L^1\) to the corresponding sides for \(Z\), proving the identity almost surely. \(\square\)
Proof. Fix \(n\ge1\). Since \(H_{n-1}\) is \(\mathcal{F}_{n-1}\)-measurable and \(\Delta M_n\) is \(\mathcal{F}_n\)-measurable, their product is \(\mathcal{F}_n\)-measurable. Also, \(\mathbb{E}|H_{n-1}\Delta M_n|\le C\mathbb{E}|\Delta M_n|<\infty\). Therefore each transformed increment is integrable and adapted. The lemma and the martingale-difference condition give \(\mathbb{E}[H_{n-1}\Delta M_n\mid\mathcal{F}_{n-1}]=H_{n-1}\mathbb{E}[\Delta M_n\mid\mathcal{F}_{n-1}]=0\) almost surely. The sum defining \(Y_n\) is finite, so it is integrable and \(\mathcal{F}_n\)-measurable. Finally, \(Y_n=Y_{n-1}+H_{n-1}\Delta M_n\), hence \(\mathbb{E}[Y_n\mid\mathcal{F}_{n-1}]=Y_{n-1}\). This is the martingale condition, and \(Y_0=0\) is integrable. \(\square\)
For a finite time horizon, it is enough that each coefficient used on that horizon is bounded; a common bound is convenient when defining the transform for all \(n\). Boundedness is sufficient, not necessary. With unbounded coefficients, one must separately establish integrability of the products and the validity of taking the coefficient outside the conditional expectation.
Two-Step Calculations
Worked Example: Different Weights for Two Fair Increments
Let \(\Omega=\{(a,b):a,b\in\{-1,1\}\}\), with probability \(1/4\) assigned to each outcome. Let \(\xi_1=a\), \(\xi_2=b\), let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) contain the information in \(\xi_1\), and let \(\mathcal{F}_2\) contain both flips. Set \(M_0=0\), \(M_1=\xi_1\), and \(M_2=\xi_1+\xi_2\). Define \(H_0=2\) and \(H_1=\xi_1\). Find the transformed process and check its martingale property.
The first increment is \(\Delta M_1=\xi_1\). The second is \(\Delta M_2=\xi_2\), whose conditional mean given \(\mathcal{F}_1\) is zero because either value of \(\xi_2\) is equally likely after either value of \(\xi_1\). Also \(\mathbb{E}\xi_1=0\), so \(M\) is a martingale. The coefficients are predictable: \(H_0\) is constant and \(H_1=\xi_1\) is known at time \(1\). They are bounded by \(2\). Thus \(Y_1=2\xi_1\) and \(Y_2=2\xi_1+\xi_1\xi_2\).
Check by outcomes. If \(\xi_1=1\), then \(Y_1=2\); the values of \(Y_2\) for \(\xi_2=1,-1\) are \(3,1\), with average \((3+1)/2=2\). If \(\xi_1=-1\), then \(Y_1=-2\); the corresponding \(Y_2\) values are \(-3,-1\), with average \((-3-1)/2=-2\). Hence \(\mathbb{E}[Y_2\mid\mathcal{F}_1]=Y_1\). Also \(\mathbb{E}Y_1=(2+(-2))/2=0=Y_0\), verifying the martingale condition at both steps.
Worked Example: Predictable Weights with Conditional Increments
Use the same four equally likely outcomes \((a,b)\in\{-1,1\}^2\) and the same filtration. This time let \(\Delta M_1=a\). Define \(\Delta M_2=2b\) when \(a=1\), and \(\Delta M_2=b\) when \(a=-1\), and set \(M_0=0\), \(M_1=\Delta M_1\), \(M_2=M_1+\Delta M_2\). Take \(H_0=1\) and \(H_1=2\) when \(a=1\), \(H_1=\tfrac12\) when \(a=-1\).
At time \(1\), \(\mathbb{E}[\Delta M_1]=((1)+(-1))/2=0\). Given either value of \(a\), the two values of \(b\) remain equally likely. Thus the conditional mean of \(\Delta M_2\) is \((2+(-2))/2=0\) if \(a=1\), and \((1+(-1))/2=0\) if \(a=-1\). These increments define a martingale. The coefficient \(H_1\) depends only on \(a\), so it is predictable, and \(|H_1|\le2\).
The transformed values are \(Y_1=a\) and \(Y_2=a+H_1\Delta M_2\). When \(a=1\), \(Y_2=1+2(2b)=1+4b\), giving \(5\) for \(b=1\) and \(-3\) for \(b=-1\). Their average is \((5-3)/2=1=Y_1\). When \(a=-1\), \(Y_2=-1+\tfrac12b\), giving \(-\tfrac12\) for \(b=1\) and \(-\tfrac32\) for \(b=-1\). Their average is \((-\tfrac12-\tfrac32)/2=-1=Y_1\). Finally, \(\mathbb{E}Y_1=((1)+(-1))/2=0=Y_0\). The conditional checks verify the transformed process at both steps.
Worked Example: Stop Weighting After a Loss
Let \(\xi_1,\xi_2\) be two independent fair signs, each taking \(1\) or \(-1\) with probability \(1/2\), and use the filtration that reveals \(\xi_1\) at time \(1\) and both signs at time \(2\). Set \(M_0=0\), \(M_1=\xi_1\), \(M_2=\xi_1+\xi_2\). Choose \(H_0=1\) and \(H_1=\mathbf{1}_{\{\xi_1=1\}}\). This coefficient makes the second weight zero if the first increment was negative.
Here \(Y_1=\xi_1\) and \(Y_2=\xi_1+\mathbf{1}_{\{\xi_1=1\}}\xi_2\). The four outcomes \((\xi_1,\xi_2)\) give \(Y_2\) values \(2,0,-1,-1\), respectively: the first two have \(\xi_1=1\), and the last two have \(\xi_1=-1\). Conditional on \(\xi_1=1\), the average is \((2+0)/2=1=Y_1\); conditional on \(\xi_1=-1\), it is \((-1-1)/2=-1=Y_1\). At time zero, \(\mathbb{E}Y_1=(1+(-1))/2=0=Y_0\). Thus this history-dependent weighting still gives a martingale. Its coefficient is bounded by \(1\) and is known before \(\xi_2\) is observed.
Why Timing Matters
The theorem does not say that every random weighting preserves the martingale property. In the fair-sign setup of the first example, consider the tempting but invalid choice \(H_1=\mathbf{1}_{\{\xi_2=1\}}\). This coefficient uses the second flip itself, so it is not \(\mathcal{F}_1\)-measurable. The weighted second increment is \(H_1\Delta M_2=\mathbf{1}_{\{\xi_2=1\}}\xi_2=\mathbf{1}_{\{\xi_2=1\}}\). Its conditional expectation given \(\mathcal{F}_1\) is \(1/2\), not zero. The resulting process therefore has a positive conditional drift at time \(2\).
Common pitfall. A wrong calculation is to write \(\mathbb{E}[H_1\Delta M_2\mid\mathcal{F}_1]=H_1\mathbb{E}[\Delta M_2\mid\mathcal{F}_1]=0\) for that invalid choice. The conditional multiplication rule requires \(H_1\) to be \(\mathcal{F}_1\)-measurable. Since this \(H_1\) depends on \(\xi_2\), the rule cannot be applied; direct calculation gives \(1/2\). Always check the information available when the coefficient is selected before pulling it outside a conditional expectation.
Predictable transforms allow past information to determine the size or sign of each upcoming martingale increment without changing its conditional mean-zero property. The conclusion is about the martingale property, not about independence of the transformed increments: coefficients can depend on the past, and the resulting increments can therefore be dependent. The essential checks are that the original increments are conditionally centered, the coefficient is measurable at the preceding time, and the products are integrable.
Check Your Understanding
For each question, identify the relevant conditioning information and verify the conditional-mean calculation.
- If \(H_{n-1}=3\) is constant and \(\mathbb{E}[\Delta M_n\mid\mathcal{F}_{n-1}]=0\), what is \(\mathbb{E}[H_{n-1}\Delta M_n\mid\mathcal{F}_{n-1}]\)?
- In the first worked example, compute \(Y_2\) on the outcome \((\xi_1,\xi_2)=(-1,1)\).
- In the second worked example, why is \(H_1\) predictable even though its value is random?
- For the stopping-weight example, calculate the conditional average of \(Y_2\) given \(\xi_1=-1\).
- In the pitfall calculation, what is the conditional expectation of \(\mathbf{1}_{\{\xi_2=1\}}\xi_2\) given \(\mathcal{F}_1\), and why can the coefficient not be pulled out?