From a Terminal Variable to a Martingale
A martingale can be built by taking the conditional expectation of a future quantity using the information available at each time. As information accumulates, the conditional value may change, but its conditional average at an earlier time equals the value already assigned at that earlier time. This construction is useful both for finite information structures and for processes observed continuously.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where \((\mathcal{F}_t)\) is an increasing family of sigma-fields. The sigma-field \(\mathcal{F}_t\) represents the information available at time \(t\). Fix a finite horizon \(T\), and let \(Y\) be an integrable, \(\mathcal{F}_T\)-measurable random variable; integrable means \(\mathbb{E}[|Y|]<\infty\). The conditional expectation \(\mathbb{E}[Y\mid\mathcal{F}_t]\) is an \(\mathcal{F}_t\)-measurable random variable whose integral over each event in \(\mathcal{F}_t\) agrees with that of \(Y\).
Proof. By the definition of conditional expectation, \(M_t\) is \(\mathcal{F}_t\)-measurable. The conditional-expectation inequality \(|\mathbb{E}[Y\mid\mathcal{F}_t]|\le\mathbb{E}[|Y|\mid\mathcal{F}_t]\) gives \(\mathbb{E}[|M_t|]\le\mathbb{E}[|Y|]<\infty\), so \(M_t\) is integrable. Now fix \(s\le t\). For any event \(A\in\mathcal{F}_s\), we also have \(A\in\mathcal{F}_t\), and the defining property of conditional expectation gives \(\mathbb{E}[\mathbf{1}_A M_t]=\mathbb{E}[\mathbf{1}_A Y]=\mathbb{E}[\mathbf{1}_A M_s]\). Since \(M_s\) is \(\mathcal{F}_s\)-measurable and integrable, this equality for every \(A\in\mathcal{F}_s\) is precisely the defining property of \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\), up to almost-sure equality. Thus the process is a martingale. \(\square\)
The theorem makes no square-integrability assumption: \(\mathbb{E}[|Y|]<\infty\) suffices. It also makes no assertion that the martingale has continuous paths. In the examples, conditional expectations can be computed directly from the available information.
Finite-State Information Revealed in Stages
Worked Example: A Four-Outcome Information Tree
Let \(\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}\), with probability \(1/4\) assigned to each outcome. Set \(T=2\), let \(\mathcal{F}_t\) be trivial for \(0\le t<1\), let \(\mathcal{F}_t\) be generated by the two groups \(G_1=\{\omega_1,\omega_2\}\) and \(G_2=\{\omega_3,\omega_4\}\) for \(1\le t<2\), and let \(\mathcal{F}_t=\mathcal{F}\) for \(t\ge2\). Take \(Y(\omega_1)=2\), \(Y(\omega_2)=6\), \(Y(\omega_3)=-1\), and \(Y(\omega_4)=5\). Compute \(M_t=\mathbb{E}[Y\mid\mathcal{F}_t]\).
Before time \(1\), the only information set is all of \(\Omega\). Its probability is \(1\), and the mean of \(Y\) is \((2+6-1+5)/4=12/4=3\). Thus \(M_t=3\) for \(0\le t<1\).
Between times \(1\) and \(2\), the observed group determines the relevant conditional average. Each group has probability \(1/2\), and each of its two outcomes has conditional probability \(1/2\). On \(G_1\), \((2+6)/2=8/2=4\); on \(G_2\), \((-1+5)/2=4/2=2\). Therefore \(M_t=4\) on \(G_1\) and \(M_t=2\) on \(G_2\) for \(1\le t<2\). At and after time \(2\), the outcome is known, so \(M_t=Y\).
Check: Averaging the two possible values at time \(1\) gives \((4+2)/2=3\), the value before the group is revealed. Within \(G_1\), the average of the terminal values is \((2+6)/2=4\); within \(G_2\), it is \((-1+5)/2=2\). The conditional averages at both stages agree with the computed values.
This calculation is an instance of a general finite-partition formula. If the information sigma-field is generated by a finite partition with positive-probability atoms \(A_1,\ldots,A_m\), then on atom \(A_j\) the conditional expectation of \(Y\) is \(\mathbb{E}[Y\mathbf{1}_{A_j}]/\mathbb{P}(A_j)\). For finite-valued \(Y\), this is the probability-weighted average of its values on that atom.
Proof. The displayed random variable is \(\mathcal{G}\)-measurable because it is constant on each atom. Every event \(C\in\mathcal{G}\) is a union of atoms. For an atom \(A_j\), its integral is \(\mathbb{E}[\mathbf{1}_{A_j}Z]=\mathbb{E}[Y\mathbf{1}_{A_j}]\), by the definition of the constant value assigned on \(A_j\). Summing this identity over the atoms contained in \(C\) gives \(\mathbb{E}[\mathbf{1}_C Z]=\mathbb{E}[\mathbf{1}_C Y]\). These are exactly the defining properties of the conditional expectation. \(\square\)
If an atom has probability zero, the quotient in the formula is undefined, but its assigned value does not affect the conditional expectation up to almost-sure equality. The positive-probability assumption avoids that case in the proposition and in the calculation above.
A Martingale When an Indicator Is Revealed
Worked Example: A Hidden Event Becomes Known
Let \(\Omega=\{a,b\}\), with \(\mathbb{P}(\{a\})=2/5\) and \(\mathbb{P}(\{b\})=3/5\), and put \(A=\{a\}\). Take \(T=2\). Let \(\mathcal{F}_t\) be trivial for \(0\le t<1\) and \(\mathcal{F}_t=\sigma(A)\) for \(1\le t\le2\). For the terminal variable \(Y=\mathbf{1}_A\), compute its conditional-expectation martingale.
Before the event is revealed, \(\mathcal{F}_t\) contains no information distinguishing \(a\) from \(b\). The conditional mean of the indicator is its probability: \(\mathbb{E}[Y\mid\mathcal{F}_t]=\mathbb{P}(A)=2/5\) for \(t<1\). At time \(1\), the event is known, so \(Y\) is \(\mathcal{F}_1\)-measurable and \(M_t=Y=\mathbf{1}_A\) for \(1\le t\le2\).
Check: The value just before the revelation has expectation \(2/5\). The revealed value has expectation \((2/5)\cdot1+(3/5)\cdot0=2/5\). Conditioning the revealed value on the information available before time \(1\) therefore returns \(2/5\), as required. Both atoms have positive probability, so the conditional calculation is defined on each one.
The martingale here is not constant: it moves from \(2/5\) to either \(0\) or \(1\). The defining property does not say that future values equal present values outcome by outcome. It says that their conditional expectation, given the present information, equals the present value.
A Brownian Threshold Example
Now suppose \(B_t\) is a standard Brownian motion on the filtered probability space, with its completed natural filtration \(\mathcal{F}_t=\sigma(B_u:0\le u\le t)\) (completed by null sets). Fix \(T>0\), and consider \(Y=\mathbf{1}_{\{B_T>0\}}\). This is integrable and \(\mathcal{F}_T\)-measurable. Write \(\Phi\) for the standard normal cumulative distribution function.
Worked Example: Conditional Probability of a Positive Terminal Value
Compute \(M_t=\mathbb{E}[\mathbf{1}_{\{B_T>0\}}\mid\mathcal{F}_t]\) for \(0\le t\le T\).
For \(t<T\), the Brownian increment \(B_T-B_t\) is independent of \(\mathcal{F}_t\) and has distribution \(N(0,T-t)\). Conditional on \(\mathcal{F}_t\), write \(B_t=x\). Then \[ \mathbb{P}(B_T>0\mid\mathcal{F}_t) =\mathbb{P}(B_T-B_t>-B_t\mid\mathcal{F}_t) =\Phi\!\left(\frac{B_t}{\sqrt{T-t}}\right). \] The last equality follows by standardizing the centered normal increment and using the symmetry of its distribution. At \(t=T\), the terminal event is already known, so \(M_T=\mathbf{1}_{\{B_T>0\}}\). Thus the process is given by the displayed conditional probability for \(t<T\), with this separate terminal value.
Check: At \(t=0\), \(B_0=0\), so \(M_0=\Phi(0)=1/2\), which equals \(\mathbb{P}(B_T>0)\) by symmetry. For \(0<t<T\), an independent check of the mean is to take an independent standard normal variable \(Z\): conditional on \(B_t\), \(\Phi(B_t/\sqrt{T-t})=\mathbb{P}(B_t+\sqrt{T-t}Z>0\mid B_t)\). The variable \(B_t+\sqrt{T-t}Z\) has distribution \(N(0,T)\), so \(\mathbb{E}[M_t]=1/2\), matching \(\mathbb{E}[Y]\). At \(t=T\), \(\mathbb{E}[M_T]=\mathbb{P}(B_T>0)=1/2\) as well.
The terminal variable is bounded by \(0\) and \(1\), so every \(M_t\) is integrable. The conditional-expectation theorem proves the martingale identity for every deterministic \(s\le t\le T\); the computation above makes its values explicit. No stochastic integral is needed for this construction.
Common Mistakes and Why the Construction Matters
A common error is to use the eventual value before the information that determines it has arrived. In the revealed-event example, assigning \(M_t=\mathbf{1}_A\) for \(t<1\) is wrong: it is not measurable with respect to the trivial \(\mathcal{F}_t\), and it is not the conditional expectation. The correct value is \(2/5\) for both outcomes. For instance, on the event \(A\), the proposed value is \(1\), while the conditional expectation before revelation is \(2/5\); on \(A^c\), the proposed value is \(0\), while the conditional expectation is still \(2/5\). The correction is to average the terminal value over outcomes that remain possible under the current information.
Conditional-expectation martingales appear whenever a future quantity is repeatedly revalued as information arrives. In a finite model, the calculation reduces to conditional averages on information atoms. In the Brownian example, independent increments turn the conditional expectation into a normal tail probability. In each case, the martingale identity expresses consistency of these successive conditional valuations; it does not imply that the realized value stays fixed.
Check Your Understanding
For each question, state the information available at the time being considered and check that the conditional expectation is defined.
- In the four-outcome example, what is \(\mathbb{E}[Y\mid\mathcal{F}_t]\) on \(G_2\) for \(1\le t<2\), and what is its unconditional mean at that stage?
- If the revealed event in the two-outcome example had probability \(p\) with \(0<p<1\), what would the martingale be before and after revelation?
- Why does integrability of \(Y\) imply integrability of each \(\mathbb{E}[Y\mid\mathcal{F}_t]\)?
- For the Brownian threshold example, why is the conditional probability at \(t=0\) equal to \(1/2\)?
- Explain why setting the revealed indicator equal to its eventual value before time \(1\) fails the definition of a martingale in that example.