Martingales from Conditional Mean-One Multipliers

Why Conditional Mean One Matters

A product of independent mean-one random variables is a familiar way to construct a martingale. Independence is stronger than necessary. What matters at each step is that the new multiplier has conditional mean one given the information already available. This condition permits the current product to be carried through a conditional expectation, even when the multiplier's distribution depends on the past.

There is an important qualification: the product at each time must be integrable. In particular, integrability of each multiplier separately does not always imply integrability of their product if signed multipliers are allowed. We will state the theorem with this requirement, give a useful condition that guarantees it automatically, and see why the qualification matters.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), an increasing sequence of sigma-fields representing information through time \(n\). Suppose \(Y_{n+1}\) is \(\mathcal{F}_{n+1}\)-measurable, so the new multiplier is known by time \(n+1\). Define \(M_0=1\) and \(M_n=\prod_{k=1}^nY_k\) for \(n\ge1\). A process is a martingale with respect to this filtration if each \(M_n\) is integrable and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely.

The hypothesis is conditional: it must hold given the whole past information in \(\mathcal{F}_n\), not merely on average over the probability space. The separate product-integrability hypothesis ensures that the conditional expectation defining a martingale is well-defined. It is enough to check integrability of the products through every time under consideration.

Proof. The product \(M_n\) is \(\mathcal{F}_n\)-measurable because every factor \(Y_k\), \(k\le n\), is \(\mathcal{F}_k\)-measurable and \(\mathcal{F}_k\subseteq\mathcal{F}_n\). It is integrable by assumption. Fix \(n\). Since \(M_n\) is \(\mathcal{F}_n\)-measurable, the conditional expectation product rule gives \(\mathbb{E}[M_nY_{n+1}\mid\mathcal{F}_n]=M_n\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]\), provided the product is integrable; that product is \(M_{n+1}\), which is integrable by assumption. The conditional mean-one hypothesis therefore gives \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely. Adaptedness and integrability have already been checked, so this is the martingale property. \(\square\)

A Useful Integrability Guarantee

For nonnegative multipliers, one need not separately assume that each product is integrable. The conditional mean-one condition itself gives the needed integrability, by induction. This is especially useful for likelihood ratios, weights, and other nonnegative multiplicative constructions.

Proof. We use induction. The starting value \(M_0=1\) is nonnegative, integrable, and has expectation one. Suppose \(M_n\) is nonnegative and integrable. Because \(M_n\) is \(\mathcal{F}_n\)-measurable and both factors \(M_n,Y_{n+1}\) are nonnegative, the conditional expectation identity for nonnegative random variables gives \(\mathbb{E}[M_nY_{n+1}\mid\mathcal{F}_n]=M_n\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]=M_n\). This identity is valid even before integrability of the product is known, as an identity for nonnegative conditional expectations. Taking expectations gives \(\mathbb{E}[M_{n+1}]=\mathbb{E}[M_n]=1\), so \(M_{n+1}\) is integrable and nonnegative. Induction proves the claim at every finite time. The theorem now applies. \(\square\)

Worked Computations

Why the Conditions Cannot Be Weakened Casually

Unconditional means alone do not control the mean of a product. For a concrete wrong calculation, let \(U\) be a fair sign and take \(Y_1=Y_2=1+\tfrac12U\). Each factor has mean \(\tfrac12(\tfrac32)+\tfrac12(\tfrac12)=1\). It would be wrong to conclude that the product has mean one: direct computation gives \(\mathbb{E}[Y_1Y_2]=\tfrac12(\tfrac32)^2+\tfrac12(\tfrac12)^2=\tfrac12(\tfrac94+\tfrac14)=\tfrac54\). With \(\mathcal{F}_1=\sigma(U)\), the second factor is already known, so \(\mathbb{E}[Y_2\mid\mathcal{F}_1]=Y_2\), not one. The correction is to check the conditional mean given the actual past, rather than just the unconditional mean of each factor.

There is a separate issue with signed multipliers: even the conditional mean-one conditions do not by themselves guarantee integrability of the product. Here is an example showing why the theorem includes that hypothesis. Let \(K\) take positive integer values with \(\mathbb{P}(K=k)=Ck^{-3}\), where \(C=(\sum_{k=1}^{\infty}k^{-3})^{-1}\). Then \(\mathbb{E}[K]=C\sum_{k=1}^{\infty}k^{-2}\) is finite, whereas \(\mathbb{E}[K^2]=C\sum_{k=1}^{\infty}k^{-1}\) is infinite. Let \(\eta\) be an independent fair sign, take \(\mathcal{F}_0\) trivial, \(\mathcal{F}_1=\sigma(K)\), and \(\mathcal{F}_2=\sigma(K,\eta)\). Set \(Y_1=K/\mathbb{E}[K]\) and \(Y_2=1+K\eta\). Both multipliers are integrable; also \(\mathbb{E}[Y_1]=1\) and \(\mathbb{E}[Y_2\mid\mathcal{F}_1]=1\).

Nevertheless, \(M_2=Y_1Y_2\) is not integrable. Conditional on \(K=k\), the mean of \(|1+k\eta|\) is \(\tfrac12(|1+k|+|1-k|)=k\), since \(k\ge1\). It follows that \(\mathbb{E}[|M_2|]=\mathbb{E}[K^2]/\mathbb{E}[K]=\infty\). Thus \(M_2\) cannot be a martingale value. This example does not contradict the nonnegative corollary: \(Y_2\) can be negative, and nonnegativity was what allowed product integrability to follow by induction.

When checking a proposed product martingale, use the filtration that records the information available at each step. Verify that each new factor is measurable at the next time and has conditional mean one given the current information. Then check that the products are integrable, or establish nonnegativity so the corollary applies. These steps distinguish a genuine conditional-mean argument from an unjustified multiplication of unconditional means.