Building a Martingale by Multiplication
For sums of independent centered random variables, each new increment has conditional mean zero given the past. There is a multiplicative counterpart: if each new factor is independent of the factors already observed and has mean one, multiplying by it preserves the current value in conditional expectation. This produces a martingale even when the factors are not identically distributed, and even when some factors can be negative.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(Y_1,Y_2,\ldots\) be independent, integrable random variables with \(\mathbb{E}[Y_k]=1\) for every \(k\ge1\). Define the filtration by \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\) for \(n\ge1\). Thus, at time \(n\), the information consists of the first \(n\) factors. Define the empty product to be \(1\), and set \(M_0=1\) and \(M_n=\prod_{k=1}^nY_k\) for \(n\ge1\).
The product is adapted because \(M_n\) is a function of \(Y_1,\ldots,Y_n\), all of which are known at time \(n\). The substantive points are integrability and the conditional expectation identity.
Proof. Fix \(n\ge1\). Independence and integrability give \(\mathbb{E}[|M_n|]=\mathbb{E}[\prod_{k=1}^n|Y_k|]=\prod_{k=1}^n\mathbb{E}[|Y_k|]<\infty\). Thus \(M_n\) is integrable; \(M_0=1\) is integrable as well. Independence also gives \(\mathbb{E}[M_n]=\prod_{k=1}^n\mathbb{E}[Y_k]=1\), and the same expectation holds for \(n=0\) by the empty-product definition.
For the conditional assertion, the product \(Z_{m,n}=\prod_{k=m+1}^nY_k\) is defined to be \(1\) when \(m=n\). When \(m<n\), it is integrable and independent of \(\mathcal{F}_m=\sigma(Y_1,\ldots,Y_m)\), because the factors are independent. Its expectation is \(\prod_{k=m+1}^n\mathbb{E}[Y_k]=1\). Since \(M_n=M_mZ_{m,n}\), conditioning on \(\mathcal{F}_m\) gives \(\mathbb{E}[M_n\mid\mathcal{F}_m]=M_m\mathbb{E}[Z_{m,n}\mid\mathcal{F}_m]=M_m\). For \(m=n\), the identity is \(\mathbb{E}[M_n\mid\mathcal{F}_n]=M_n\). Taking \(m=n\) in the martingale setup, or taking \(m=n\) and the future index to be \(n+1\), the one-step identity is \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\). Adaptedness and integrability have already been checked, so the process is a martingale. \(\square\)
The one-step calculation is especially direct: \(M_{n+1}=M_nY_{n+1}\), and \(Y_{n+1}\) is independent of the information \(\mathcal{F}_n\) with mean one. Independence is what permits the new factor to be replaced by its mean when conditioning on the past.
Worked Products
Worked Example: Two Identically Distributed Factors
Let \(Y_1,Y_2,\ldots\) be independent, with each factor equal to \(\tfrac12\) or \(\tfrac32\), each with probability \(\tfrac12\). The mean of each factor is \(\mathbb{E}[Y_k]=\tfrac12(\tfrac12)+\tfrac12(\tfrac32)=\tfrac14+\tfrac34=1\). Find the distribution of \(M_2\), its expectation, and its conditional update from time \(1\) to time \(2\).
The four ordered pairs of factor values each have probability \(\tfrac14\). Their products are \(\tfrac14,\tfrac34,\tfrac34,\tfrac94\). Therefore \(M_2=\tfrac14\) with probability \(\tfrac14\), \(M_2=\tfrac34\) with probability \(\tfrac12\), and \(M_2=\tfrac94\) with probability \(\tfrac14\). Its expectation is \(\tfrac14\cdot\tfrac14+\tfrac12\cdot\tfrac34+\tfrac14\cdot\tfrac94=\tfrac1{16}+\tfrac38+\tfrac9{16}=1\).
Given \(\mathcal{F}_1\), the value of \(Y_1=M_1\) is known, while \(Y_2\) is still equally likely to be either factor value. Hence \(\mathbb{E}[M_2\mid\mathcal{F}_1]=M_1\mathbb{E}[Y_2\mid\mathcal{F}_1]=M_1\cdot1=M_1\). In particular, if \(Y_1=\tfrac12\), the two possible values of \(M_2\) are \(\tfrac14,\tfrac34\), whose average is \(\tfrac12\). If \(Y_1=\tfrac32\), they are \(\tfrac34,\tfrac94\), whose average is \(\tfrac32\).
Check: The probabilities sum to \(\tfrac14+\tfrac12+\tfrac14=1\), and the computed mean is \(1\), as the theorem requires. The two conditional averages equal the corresponding observed values of \(M_1\).
Worked Example: Nonidentical Factors and a Zero Product
Let \(Y_1\) equal \(0\) or \(2\) with equal probabilities, and let \(Y_2\), independently, equal \(\tfrac12\) or \(\tfrac32\) with equal probabilities. Both factors have mean one: \(\mathbb{E}[Y_1]=\tfrac12(0)+\tfrac12(2)=1\), and \(\mathbb{E}[Y_2]=\tfrac12(\tfrac12)+\tfrac12(\tfrac32)=\tfrac14+\tfrac34=1\). Compute the possible values of \(M_2\) and its conditional expectation given \(\mathcal{F}_1\).
Each of the four pairs of factor values has probability \(\tfrac14\). The resulting products are \(0\cdot\tfrac12=0\), \(0\cdot\tfrac32=0\), \(2\cdot\tfrac12=1\), and \(2\cdot\tfrac32=3\). Thus \(M_2\) is \(0\) with probability \(\tfrac12\), \(1\) with probability \(\tfrac14\), and \(3\) with probability \(\tfrac14\). Its expectation is \(0\cdot\tfrac12+1\cdot\tfrac14+3\cdot\tfrac14=1\).
If \(Y_1=0\), then \(M_1=0\) and \(M_2=0\) for either value of \(Y_2\), so the conditional average is \(0\). If \(Y_1=2\), then the two possible values of \(M_2\) are \(1\) and \(3\), with conditional average \(\tfrac12(1)+\tfrac12(3)=2=M_1\). Therefore \(\mathbb{E}[M_2\mid\mathcal{F}_1]=M_1\) in both cases.
Check: The unconditional mean from the product distribution is \(1\), and the two conditional means are \(0\) and \(2\), exactly the two possible values of \(M_1\). A factor may be zero; the martingale argument does not require division by any factor.
Worked Example: A Signed Factor
Let \(Y_1\) equal \(-1\) or \(3\) with equal probabilities, and let \(Y_2\), independently, equal \(0\) or \(2\) with equal probabilities. The means are \(\mathbb{E}[Y_1]=\tfrac12(-1)+\tfrac12(3)=1\) and \(\mathbb{E}[Y_2]=\tfrac12(0)+\tfrac12(2)=1\). Find the distribution of \(M_2\) and verify the conditional update.
The four pairs each have probability \(\tfrac14\), and the products are \((-1)(0)=0\), \((-1)(2)=-2\), \((3)(0)=0\), and \((3)(2)=6\). Consequently \(M_2=0\) with probability \(\tfrac12\), \(M_2=-2\) with probability \(\tfrac14\), and \(M_2=6\) with probability \(\tfrac14\). Its expectation is \(0\cdot\tfrac12+(-2)\cdot\tfrac14+6\cdot\tfrac14=0-\tfrac12+\tfrac32=1\).
When \(Y_1=-1\), the two possible next products are \(0\) and \(-2\), whose conditional average is \(-1=M_1\). When \(Y_1=3\), the two possible next products are \(0\) and \(6\), whose conditional average is \(3=M_1\). Thus the conditional update holds even though the factors and products need not be nonnegative.
Check: The probabilities sum to \(\tfrac12+\tfrac14+\tfrac14=1\); the expectation is \(1\); and the conditional averages equal \(-1\) and \(3\), respectively. No positivity assumption was used.
Second Moments and Spread
The mean of the product remains one, but the product need not stay close to one. A useful measure of its spread is available when the factors have finite second moments.
Proof. Squaring the finite product gives \(M_n^2=\prod_{k=1}^nY_k^2\). The variables \(Y_1^2,\ldots,Y_n^2\) are independent and integrable, so factorization of expectation yields \(\mathbb{E}[M_n^2]=\prod_{k=1}^n\mathbb{E}[Y_k^2]\). Since \(\mathbb{E}[M_n]=1\), the definition of variance gives \(\operatorname{Var}(M_n)=\mathbb{E}[M_n^2]-(\mathbb{E}[M_n])^2=\prod_{k=1}^n\mathbb{E}[Y_k^2]-1\). For \(n=0\), the empty product is \(1\), and \(M_0=1\) has variance \(0=1-1\). \(\square\)
In the first worked example, \(\mathbb{E}[Y_k^2]=\tfrac12(\tfrac14)+\tfrac12(\tfrac94)=\tfrac18+\tfrac98=\tfrac54\). The proposition gives \(\mathbb{E}[M_2^2]=(\tfrac54)^2=\tfrac{25}{16}\), and therefore \(\operatorname{Var}(M_2)=\tfrac{25}{16}-1=\tfrac9{16}\). Directly from its distribution, \(\mathbb{E}[M_2^2]=\tfrac14(\tfrac1{16})+\tfrac12(\tfrac9{16})+\tfrac14(\tfrac{81}{16})=\tfrac1{64}+\tfrac9{32}+\tfrac{81}{64}=\tfrac{100}{64}=\tfrac{25}{16}\), which verifies the formula.
Why Independence Matters
Mean-one marginals do not by themselves guarantee a martingale. The new factor must have conditional mean one given the information already observed. Independence ensures that condition for the natural filtration, but dependence can make the conditional mean different from one.
Common mistake. Suppose \(Y_1\) is \(0\) or \(2\), each with probability \(\tfrac12\), and set \(Y_2=Y_1\). Each factor has mean \(\tfrac12(0)+\tfrac12(2)=1\), but the factors are dependent. The incorrect calculation is \(\mathbb{E}[Y_1Y_2]=\mathbb{E}[Y_1]\mathbb{E}[Y_2]=1\); that factorization requires independence and is not valid here. In fact \(M_2=Y_1^2\), so \(M_2\) is \(0\) or \(4\), each with probability \(\tfrac12\), and \(\mathbb{E}[M_2]=0\cdot\tfrac12+4\cdot\tfrac12=2\). Moreover, given \(\mathcal{F}_1=\sigma(Y_1)\), \(Y_2\) is already known, and \(\mathbb{E}[M_2\mid\mathcal{F}_1]=Y_1^2\), which is \(0\) when \(Y_1=0\) and \(4\) when \(Y_1=2\). This is not \(M_1=Y_1\) on the event \(Y_1=2\).
The calculation identifies the right test: check \(\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]=1\) almost surely, not merely \(\mathbb{E}[Y_{n+1}]=1\). In this tutorial, independence from the factors already revealed supplies that conditional identity. In other settings, a multiplier may depend on the past and still preserve a martingale, but only if its conditional mean given that past is one.
Check Your Understanding
For each question, make explicit which information is available at the conditioning time.
- If independent factors have means one and \(M_0=1\), what is \(\mathbb{E}[M_4]\)? State the factorization that justifies your answer.
- Suppose \(Y_1\) and \(Y_2\) are independent, \(Y_1\) is \(1\) or \(3\) with equal probabilities, and \(Y_2\) is \(0\) or \(2\) with equal probabilities. Find the distribution of \(M_2\) and verify its expectation.
- Why does \(\mathbb{E}[Y_{n+1}]=1\) alone not establish \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\)?
- If all factors have finite second moments, write the formula for \(\mathbb{E}[M_n^2]\). Which assumption allows the product of expectations?
- Can the factors be negative or zero under the theorem's assumptions? Explain which part of the proof addresses integrability.