From a Terminal Variable to Its Conditional Expectations
Suppose a square-integrable random variable \(X\) is observed at a fixed terminal time \(T\), while information arrives over time through a Brownian motion. The process \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\) records the best conditional expectation of \(X\) using information available by time \(t\). In a Brownian filtration, the martingale representation theorem says that this process can be written as its initial mean plus an Ito integral. Once the integrand is known, the representation gives both the conditional expectations and the increments of \(M\).
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Assume \((\mathcal{F}_t)\) is the usual augmentation of the filtration generated by \(B\), so the Brownian martingale representation theorem applies. Fix a finite horizon \(T\). An Ito integrand \(H\) belongs to \(\mathcal{L}^2\) on \([0,T]\) if it is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). All stochastic integrals below are Ito integrals under \(\mathbb{P}\).
The integral here is an Ito integral, not a Stratonovich integral. When the covariation is defined, conversion to Stratonovich uses \(\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t\); no such correction is part of the representation above. The Ito multiplication rules include \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\).
The Conditional-Expectation Representation
Proof. For each \(t\), split the terminal representation into integrals over \([0,t]\) and \((t,T]\): $$ X=\mathbb{E}[X]+\int_0^t H_s\,dB_s+\int_t^T H_s\,dB_s. $$ The Ito integral over \((t,T]\) has conditional expectation zero given \(\mathcal{F}_t\). To justify this fact, first take an adapted step integrand on \((t,T]\). Each integral term is a past-measurable coefficient multiplied by a Brownian increment independent of the past and of mean zero, so its conditional expectation is zero. General progressively measurable square-integrable integrands are obtained in the Ito construction as limits of such integrands in \(L^2\). The Ito isometry gives convergence of their integrals in \(L^2\), and conditional expectation is a contraction in \(L^2\); therefore the zero conditional-expectation identity passes to the limit. The integral up to \(t\) is \(\mathcal{F}_t\)-measurable, so conditioning the split identity gives the stated formula.
For \(0\le s\le t\le T\), the increment \(M_t-M_s=\int_s^t H_u\,dB_u\) has conditional mean zero given \(\mathcal{F}_s\), by the same argument. Thus \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\), which is the martingale property. Also, by Jensen's inequality, \(\mathbb{E}[M_t^2]\le\mathbb{E}[X^2]<\infty\); hence \(M_t\) is integrable at every time and \(M\) is a square-integrable martingale, not merely a local martingale. The Ito integral has a continuous version, giving the claimed continuous version of \(M\). At \(t=0\) the integral is zero, and at \(t=T\) the formula is the terminal representation. \(\square\)
Proof. The conditional-mean identity is the martingale-increment property just proved. For the second identity, let \(A\in\mathcal{F}_s\). The process \(\mathbf{1}_A H_u\) on \((s,t]\), extended by zero elsewhere, is progressively measurable and square-integrable. Since \(\mathbf{1}_A\) is \(\mathcal{F}_s\)-measurable, its integral over \((s,t]\) equals \(\mathbf{1}_A(M_t-M_s)\). Applying the Ito isometry gives $$ \mathbb{E}\big[\mathbf{1}_A(M_t-M_s)^2\big] =\mathbb{E}\left[\mathbf{1}_A\int_s^t H_u^2\,du\right]. $$ This equality holds for every \(A\in\mathcal{F}_s\), which is precisely the defining property of the asserted conditional expectation. Both sides are integrable because \(H\in\mathcal{L}^2\). \(\square\)
Taking expectations in the second identity recovers the Ito-isometry formula \(\mathbb{E}[(M_t-M_s)^2]=\mathbb{E}\int_s^t H_u^2\,du\). The conditional version gives more information: it describes the expected squared size of the increment using the information already available at time \(s\).
Worked Conditional-Expectation Calculations
Worked Example: A Deterministic Step Integrand
Let \(T=2\) and define \(H_s=1\) on \([0,1]\) and \(H_s=2\) on \((1,2]\). Set \(X=\int_0^2 H_s\,dB_s\). Find \(M_t=\mathbb{E}[X\mid\mathcal{F}_t]\), and compute the variance of the increment from time \(1\) to time \(3/2\).
The integrand is deterministic and square-integrable, since \(\mathbb{E}\int_0^2 H_s^2\,ds=1+4=5\). Splitting at time \(1\) gives $$ X=(B_1-B_0)+2(B_2-B_1)=B_1+2(B_2-B_1)=2B_2-B_1. $$ Each Brownian increment has mean zero, so \(\mathbb{E}[X]=0\). The representation theorem yields \(M_t=\int_0^t H_s\,dB_s\), hence $$ M_t= \begin{cases} B_t,&0\le t\le1,\\ B_1+2(B_t-B_1)=2B_t-B_1,&1<t\le2. \end{cases} $$ In particular, \(M_{3/2}-M_1=2(B_{3/2}-B_1)\). Its conditional mean given \(\mathcal{F}_1\) is zero, and its variance is \(4(3/2-1)=2\), since the increment is independent of \(\mathcal{F}_1\) and has variance \(1/2\).
Check: Independence of \(B_1\) and \(B_2-B_1\) gives \(\mathbb{E}[X^2]=1+4=5\). The full-integral isometry gives \(\mathbb{E}[M_2^2]=\int_0^1 1^2\,ds+\int_1^2 2^2\,ds=1+4=5\), in agreement.
Worked Example: An Integrand Chosen from Earlier Information
Let \(a=1/2\), and set $$ X=B_a+\mathbf{1}_{\{B_a>0\}}(B_1-B_a). $$ Find the conditional-expectation process and the conditional second moment of the increment from \(a\) to \(1\).
Write \(X=\int_0^1 H_s\,dB_s\) with \(H_s=1\) on \([0,a]\) and \(H_s=\mathbf{1}_{\{B_a>0\}}\) on \((a,1]\). The second value is known at time \(a\), so this is an adapted progressively measurable step integrand. Its square-integrability follows from $$ \mathbb{E}\int_0^1 H_s^2\,ds=\frac12+\frac12\,\mathbb{P}(B_a>0)=\frac12+\frac14=\frac34. $$ Here \(B_a\) has a centered, nondegenerate normal distribution, so \(\mathbb{P}(B_a>0)=1/2\). The representation gives $$ M_t= \begin{cases} B_t,&0\le t\le a,\\ B_a+\mathbf{1}_{\{B_a>0\}}(B_t-B_a),&a<t\le1. \end{cases} $$ For \(t\le a\), the still-unobserved increment \(B_1-B_a\) has conditional mean zero even after multiplication by \(\mathbf{1}_{\{B_a>0\}}\), because it is independent of \(\mathcal{F}_a\). For \(t>a\), the displayed expression follows directly by splitting the Ito integral at \(t\).
At the terminal time, \(M_1=X\). On the interval from \(a\) to \(1\), \(M_1-M_a=\mathbf{1}_{\{B_a>0\}}(B_1-B_a)\). Thus $$ \mathbb{E}[M_1-M_a\mid\mathcal{F}_a]=0,\qquad \mathbb{E}[(M_1-M_a)^2\mid\mathcal{F}_a] =\frac12\mathbf{1}_{\{B_a>0\}}. $$ The second equality uses the independence of \(B_1-B_a\) from \(\mathcal{F}_a\) and its variance \(1-a=1/2\).
Check: Taking expectations of the conditional second moment gives \((1/2)(1/2)=1/4\). The Ito isometry gives the same value for the increment: \(\mathbb{E}\int_a^1 H_s^2\,ds=(1/2)\mathbb{P}(B_a>0)=1/4\).
Worked Example: A Nonzero Initial Mean
For \(T=1\), let \(H_s=2\) on \([0,1/4]\) and \(H_s=-1\) on \((1/4,1]\). Set \(X=7+\int_0^1 H_s\,dB_s\). Find \(M_t\), its value at \(t=1/2\), and \(\mathbb{E}[(X-7)^2]\).
First compute the terminal variable: $$ X=7+2B_{1/4}-(B_1-B_{1/4})=7+3B_{1/4}-B_1. $$ The Ito integral has mean zero, so \(\mathbb{E}[X]=7\). The conditional-expectation representation is $$ M_t= \begin{cases} 7+2B_t,&0\le t\le1/4,\\ 7+2B_{1/4}-(B_t-B_{1/4})=7+3B_{1/4}-B_t,&1/4<t\le1. \end{cases} $$ Therefore \(M_{1/2}=7+3B_{1/4}-B_{1/2}\). The variance around the mean is, by the Ito isometry, $$ \mathbb{E}[(X-7)^2] =\int_0^{1/4} 2^2\,ds+\int_{1/4}^1 (-1)^2\,ds =4\left(\frac14\right)+\left(1-\frac14\right) =1+\frac34=\frac74. $$
Check: The increment \(3B_{1/4}-B_1\) can be rewritten as \(2B_{1/4}-(B_1-B_{1/4})\). These two terms are independent and centered, with variances \(4(1/4)=1\) and \(1-1/4=3/4\). Their summed variance is \(1+3/4=7/4\), as above.
Common Pitfall: Omitting the Conditional Mean
A common error is to identify \(M_t\) with only the integral accumulated by time \(t\). That works only when \(\mathbb{E}[X]=0\). In the last example, the incorrect calculation \(M_0=0\) contradicts the defining condition \(M_0=\mathbb{E}[X]=7\). The correction is to retain the constant mean: \(M_t=7+\int_0^t H_s\,dB_s\). At \(t=1\), this gives \(M_1=X\), as required.
There is also an important distinction between the unconditional mean-zero property of a future integral and independence of that integral from the past. With an adapted integrand, a future integral need not be independent of \(\mathcal{F}_t\). What is guaranteed for a square-integrable Ito integral is its conditional mean zero given \(\mathcal{F}_t\). The representation uses that conditional-mean property; it does not require independence.
| Quantity | Representation-based formula | Interpretation |
|---|---|---|
| Conditional expectation | \(M_t=\mathbb{E}[X]+\int_0^t H_s\,dB_s\) | Mean plus the integral accumulated by time \(t\) |
| Martingale increment | \(M_t-M_s=\int_s^t H_u\,dB_u\) | Conditional mean zero given \(\mathcal{F}_s\) |
| Conditional squared increment | \(\mathbb{E}[(M_t-M_s)^2\mid\mathcal{F}_s]=\mathbb{E}[\int_s^t H_u^2\,du\mid\mathcal{F}_s]\) | Expected increment size given the information at time \(s\) |
Check Your Understanding
Use the representation and the Ito isometry to justify your calculations.
- If \(X=c+\int_0^T H_s\,dB_s\) with deterministic \(c\), what are \(M_0\) and \(M_T\)?
- For \(0\le s\le t\le T\), why does \(\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\) not require the future integral to be independent of \(\mathcal{F}_s\)?
- If \(H_s=3\) on an interval \((s,t]\) of length \(1/4\), what is the unconditional second moment of \(M_t-M_s\)?
- In the adapted-integrand example, explain why \(\mathbf{1}_{\{B_{1/2}>0\}}\) can multiply the increment after time \(1/2\) while preserving adaptedness.
- For a square-integrable terminal variable, why is its conditional-expectation process a true martingale rather than only a local martingale?