Doob Martingales from Terminal Random Variables

From a Terminal Variable to a Martingale

A martingale can be built from a single integrable random variable by asking what its expected value is, given the information currently available. As the information grows, these conditional expectations change, but their conditional average at the next time remains equal to the current value. This construction is useful whenever a quantity is ultimately determined but is revealed gradually.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((\mathcal{F}_n)_{n\ge 0}\) be a filtration: each \(\mathcal{F}_n\) is a sigma-field contained in \(\mathcal{F}\), and \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). Let \(Y\) be an integrable, real-valued random variable, meaning \(\mathbb{E}[|Y|]<\infty\). For each \(n\), the conditional expectation \(\mathbb{E}[Y\mid\mathcal{F}_n]\) is defined up to almost-sure equality. It is an integrable, \(\mathcal{F}_n\)-measurable random variable whose integral over each event in \(\mathcal{F}_n\) agrees with that of \(Y\).

Here “terminal” refers to the quantity \(Y\) whose conditional expectations are being taken; it does not by itself assert that \(Y\) is revealed at any particular finite time. The defining properties of conditional expectation include integrability and measurability, uniqueness up to almost-sure equality, and the tower property: if \(\mathcal{G}\subseteq\mathcal{H}\) are sigma-fields, then \(\mathbb{E}[\mathbb{E}[Y\mid\mathcal{H}]\mid\mathcal{G}]=\mathbb{E}[Y\mid\mathcal{G}]\) almost surely. We use these standard conditional-expectation properties below.

Proof. Fix \(n\ge0\). By the defining properties of conditional expectation, \(M_n\) is integrable and \(\mathcal{F}_n\)-measurable. Since the filtration is increasing, \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). Apply the tower property with \(\mathcal{G}=\mathcal{F}_n\) and \(\mathcal{H}=\mathcal{F}_{n+1}\): \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=\mathbb{E}[\mathbb{E}[Y\mid\mathcal{F}_{n+1}]\mid\mathcal{F}_n]=\mathbb{E}[Y\mid\mathcal{F}_n]=M_n\) almost surely. These are exactly the integrability, adaptedness, and conditional-expectation requirements for a martingale. \(\square\)

The argument requires the sigma-fields to be nested in the stated direction and \(Y\) to be integrable. It does not require independence, a finite state space, or that the filtration eventually reveal \(Y\). The conditional expectation is always defined up to almost-sure equality, so the martingale identity is also an almost-sure identity.

Consequences of the Construction

Two useful facts follow directly. First, the mean of each \(M_n\) is the mean of \(Y\). Second, the increments of the process have conditional mean zero. These statements are consequences of the construction, not extra assumptions about \(Y\).

Proof. Taking expectations in the defining property of conditional expectation gives \(\mathbb{E}[M_n]=\mathbb{E}[Y]\). For \(n\ge1\), the difference \(D_n\) is integrable because both \(M_n\) and \(M_{n-1}\) are integrable. By linearity of conditional expectation and the martingale identity just proved, \(\mathbb{E}[D_n\mid\mathcal{F}_{n-1}]=\mathbb{E}[M_n\mid\mathcal{F}_{n-1}]-M_{n-1}=0\) almost surely. Finally, if \(Y\) is \(\mathcal{F}_N\)-measurable and integrable, then \(Y\) itself satisfies the defining integral property of \(\mathbb{E}[Y\mid\mathcal{F}_N]\). Uniqueness of conditional expectation up to almost-sure equality gives \(M_N=Y\) almost surely. \(\square\)

The finite-time condition in the last statement matters. If \(Y\) is not \(\mathcal{F}_N\)-measurable, the conditional expectation at time \(N\) need not equal \(Y\). Nor does the martingale identity alone assert that \(M_n\) converges to \(Y\) as \(n\) increases. Such convergence requires additional hypotheses and is a separate question.

Computing the Conditional-Expectation Process

Common Pitfall: Averaging Over the Wrong Information

A conditional expectation given a sigma-field is computed within the events that the sigma-field can distinguish. It is not generally the unconditional mean repeated on every event. In the four-outcome example, the unconditional mean of \(Y\) is \(3\). A mistaken calculation might assign \(M_1=3\) on both blocks \(\{a,b\}\) and \(\{c,d\}\), by averaging over all four outcomes even after the block is known. But the relevant conditional averages are \((0+2)/2=1\) and \((4+6)/2=5\).

The mistaken values do not satisfy the defining property of conditional expectation: on \(\{a,b\}\), the integral of the proposed constant \(3\) is \(\tfrac14(3)+\tfrac14(3)=\tfrac32\), whereas the integral of \(Y\) is \(\tfrac14(0)+\tfrac14(2)=\tfrac12\). The correction is to condition on each event in the current sigma-field. The tower property then applies in the direction from finer information \(\mathcal{F}_{n+1}\) back to coarser information \(\mathcal{F}_n\), yielding \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\).

To construct a Doob martingale in a new problem, identify the integrable terminal variable and the filtration, then compute its conditional expectation at each time. The proof that the result is a martingale is always the same tower-property argument; the calculations that produce each conditional expectation depend on the particular information available at that time.