Why Linear Combinations Preserve Martingales
Martingales encode the idea that, given the information currently available, the expected next value is the present value. When several processes share the same information, a fixed weighted sum preserves that balance: conditional expectation distributes over the sum, and each process contributes zero conditional drift. This simple closure property is useful both for constructing new martingales and for removing a fluctuation shared by several processes.
Throughout, work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), meaning that \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\subseteq\mathcal{F}\). The processes considered here are real-valued, so their state space is \(S=\mathbb{R}\). A process \((M_n)_{n\ge0}\) is adapted if \(M_n\) is \(\mathcal{F}_n\)-measurable for every \(n\).
The common filtration matters because the conditional expectations in a proposed sum must be taken with respect to the same information. The coefficients in the result below are fixed real numbers, not random variables and not quantities selected after observing the process.
Proof. For each \(n\), every \(M_n^{(\ell)}\) is \(\mathcal{F}_n\)-measurable, so their finite linear combination \(Z_n\) is also \(\mathcal{F}_n\)-measurable. Moreover, the triangle inequality gives \(\mathbb{E}|Z_n|\le\sum_{\ell=1}^r |c_\ell|\mathbb{E}|M_n^{(\ell)}|<\infty\). By linearity of conditional expectation for a finite sum of integrable random variables, and because each coefficient is deterministic, we have \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=\sum_{\ell=1}^r c_\ell\mathbb{E}[M_{n+1}^{(\ell)}\mid\mathcal{F}_n]=\sum_{\ell=1}^r c_\ell M_n^{(\ell)}=Z_n\) almost surely. Thus adaptation, integrability, and the conditional-mean identity all hold, which proves the claim. \(\square\)
The same argument can be read in terms of increments. This is often the most direct way to find coefficients: write the increments of the processes and choose constants to eliminate a component. The resulting process is still adapted and integrable because it is a finite linear combination; its conditional increment has mean zero because each component does.
Proof. Subtracting \(Z_n=\sum_\ell c_\ell M_n^{(\ell)}\) from \(Z_{n+1}=\sum_\ell c_\ell M_{n+1}^{(\ell)}\) gives the increment identity. Taking conditional expectation and using the martingale condition for each \(M^{(\ell)}\) gives \(\mathbb{E}[\Delta Z_{n+1}\mid\mathcal{F}_n]=\sum_\ell c_\ell\cdot0=0\) almost surely. For the expectation statement, the defining identity and the tower property give \(\mathbb{E}Z_{n+1}=\mathbb{E}[\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]]=\mathbb{E}Z_n\). Iterating this equality from time \(0\) to time \(n\) proves \(\mathbb{E}Z_n=\mathbb{E}Z_0\). \(\square\)
Worked Example: Removing a Shared Second-Stage Fluctuation
Let \(\Omega=\{(u,v):u,v\in\{-1,1\}\}\), with probability \(1/4\) for each outcome. Let \(U(u,v)=u\) and \(V(u,v)=v\). Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_1=\sigma(U)\), and \(\mathcal{F}_n=\sigma(U,V)\) for \(n\ge2\). Define \(M_0=N_0=0\), \(M_1=N_1=U\), \(M_2=U+V\), \(N_2=U-V\), and keep both processes constant after time \(2\).
Each variable is bounded and adapted. At time \(0\), \(\mathbb{E}M_1=\mathbb{E}N_1=\mathbb{E}U=0\). Since \(U,V\) are independent and each has mean zero, \(\mathbb{E}M_2=\mathbb{E}U+\mathbb{E}V=0\) and \(\mathbb{E}N_2=\mathbb{E}U-\mathbb{E}V=0\). Given \(\mathcal{F}_1\), \(U\) is known and \(V\) has conditional mean zero, so \(\mathbb{E}[M_2\mid\mathcal{F}_1]=U\) and \(\mathbb{E}[N_2\mid\mathcal{F}_1]=U\). After time \(2\), both increments are zero. These computations verify that both processes are martingales.
Take their sum. At times \(0,1,2\), \(M_n+N_n\) equals \(0,2U,2U\), respectively, because \((U+V)+(U-V)=2U\). The \(V\)-fluctuation cancels at the second step. The value at time \(1\) has expectation \(2\mathbb{E}U=0\), and the value at time \(2\), conditional on \(\mathcal{F}_1\), has expectation \(\mathbb{E}[2U\mid\mathcal{F}_1]=2U\). Thus the combined process is a martingale and is constant from time \(1\) onward.
Check: For \(U=1\), the pair \((M_2,N_2)\) is \((2,0)\) when \(V=1\) and \((0,2)\) when \(V=-1\); either way \(M_2+N_2=2\). For \(U=-1\), the corresponding sums are \(-2\). This agrees with \(2U\), the value at time \(1\).
Choosing Coefficients to Cancel a Conditional Drift
Sometimes the processes being combined are not themselves martingales. In that situation, the closure theorem does not apply directly. A fixed linear combination can nevertheless be a martingale if the conditional expected increments cancel. The test is exact: for an adapted integrable candidate \(Z\), verify \(\mathbb{E}[\Delta Z_{n+1}\mid\mathcal{F}_n]=0\) at every step. This is the conditional-increment criterion for a martingale.
Worked Example: Opposing One-Step Drifts
Let \(\Omega=\{-1,1\}\), with each outcome having probability \(1/2\), and write \(R(\omega)=\omega\). Take \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(R)\) for \(n\ge1\). Define \(X_0=Y_0=0\), \(X_n=1+R\) and \(Y_n=-1+R\) for all \(n\ge1\). These processes are adapted and integrable, but their first conditional increments are not mean zero: \(\mathbb{E}[X_1-X_0\mid\mathcal{F}_0]=1\) and \(\mathbb{E}[Y_1-Y_0\mid\mathcal{F}_0]=-1\).
Set \(Z_n=X_n+Y_n\). At time \(0\), \(Z_0=0\); for \(n\ge1\), \(Z_n=(1+R)+(-1+R)=2R\). Therefore \(\mathbb{E}[Z_1\mid\mathcal{F}_0]=2\mathbb{E}R=0=Z_0\). For \(n\ge1\), \(Z_{n+1}=Z_n\), so \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=Z_n\) as well. Thus \(Z\) is a martingale, although neither summand is one.
Check by outcomes: If \(R=1\), then \(X_1=2\), \(Y_1=0\), and \(Z_1=2\). If \(R=-1\), then \(X_1=0\), \(Y_1=-2\), and \(Z_1=-2\). The two values of \(Z_1\) have equal probability, so \(\mathbb{E}Z_1=(2+(-2))/2=0=Z_0\). The individual expectations are \(\mathbb{E}X_1=(2+0)/2=1\) and \(\mathbb{E}Y_1=(0+(-2))/2=-1\), confirming the opposing drifts and their cancellation.
This example illustrates a related calculation, not an extension of the closure theorem's assumptions: the theorem combines martingales, whereas here the summands have nonzero conditional drift. The general cancellation check is to compute each conditional expected increment, combine those quantities with the proposed coefficients, and verify that the result is zero almost surely at every time.
Worked Example: Solving for Coefficients in a Two-Stage Martingale
Use the four-outcome probability space and filtration from the first example. Define \(A_0=B_0=0\), \(A_1=U\), \(A_2=U+V\), \(B_1=2U\), and \(B_2=2U+3V\), keeping both processes constant after time \(2\). All variables are bounded and adapted. At time \(0\), \(\mathbb{E}A_1=\mathbb{E}U=0\) and \(\mathbb{E}B_1=2\mathbb{E}U=0\). At time \(1\), \(\mathbb{E}[A_2\mid\mathcal{F}_1]=U=A_1\) and \(\mathbb{E}[B_2\mid\mathcal{F}_1]=2U=B_1\), since \(\mathbb{E}[V\mid\mathcal{F}_1]=0\). Their later increments are zero, so both are martingales.
Consider \(Z_n=aA_n+bB_n\), with deterministic coefficients \(a,b\). Its second-stage increment is \(\Delta Z_2=aV+3bV=(a+3b)V\). To eliminate this increment, solve \(a+3b=0\). Taking \(a=3\) and \(b=-1\) gives \(Z=3A-B\). At time \(1\), \(Z_1=3U-2U=U\). At time \(2\), \(Z_2=3(U+V)-(2U+3V)=U\). Hence this combination removes the second-stage fluctuation completely.
Check: The first increment is \(Z_1-Z_0=U\), whose expectation conditional on the trivial \(\mathcal{F}_0\) is \(\mathbb{E}U=(1+(-1))/2=0\). The second increment is \(Z_2-Z_1=U-U=0\). All subsequent increments are also zero. Thus every conditional expected increment is zero, and the calculated combination is indeed a martingale.
Scope and a Common Mistake
The deterministic-coefficient restriction is substantive. A coefficient that depends on observed randomness cannot simply be pulled outside a conditional expectation. Even if a process is a martingale, multiplying it by an arbitrary random variable need not preserve the martingale property.
Common mistake: Let \(R\) take values \(-1\) and \(1\) with probability \(1/2\) each, let \(\mathcal{F}_0\) be trivial and \(\mathcal{F}_1=\sigma(R)\), and define \(M_0=0\), \(M_1=R\). Then \(M\) is a martingale over these two times because \(\mathbb{E}R=0\). A tempting but incorrect argument says that \(Z_1=H M_1\) must also have mean zero for any coefficient \(H\). Choose \(H=\mathbf{1}_{\{R=1\}}\). Direct computation gives \(Z_1=1\) when \(R=1\) and \(Z_1=0\) when \(R=-1\), so \(\mathbb{E}Z_1=(1+0)/2=1/2\), while \(Z_0=HM_0=0\). The martingale identity fails. The closure theorem applies to fixed real coefficients; random coefficients require separate hypotheses and arguments.
For fixed coefficients, the practical procedure is short: check that all component processes use the same filtration, identify the desired linear combination, and use conditional-expectation linearity. If the goal is cancellation, calculate the relevant increments before choosing coefficients. If the component processes are not martingales, check the combined conditional expected increments directly rather than applying the closure theorem.
Check Your Understanding
Use conditional expectations and increments to justify each answer.
- If \(M\) and \(N\) are martingales for the same filtration, what conditions on \(a\) and \(b\) ensure that \(aM_n+bN_n\) is covered by the finite linear-combination theorem?
- In the first worked example, calculate \(M_2-N_2\) and compare it with \(M_1-N_1\). Which component has been removed?
- For the processes in the coefficient-solving example, find a nonzero pair \((a,b)\) that cancels the second-stage increment \(aV+3bV\).
- In the opposing-drifts example, calculate \(\mathbb{E}[X_1+Y_1\mid\mathcal{F}_0]\). Why does this not imply that \(X\) and \(Y\) are martingales?
- For the random coefficient pitfall, compute \(\mathbb{E}[H M_1]\) and compare it with \(H M_0\). Which step in the deterministic-coefficient proof cannot be used?