Worked Problems on Brownian Martingale Representation

Set-up: The Filtration Determines the Question

Martingale representation is not a property of a random variable alone. It depends on the information available in the filtration. In the completed, right-continuous natural filtration of Brownian motion, a square-integrable terminal variable can be represented by an Ito integral. In a larger filtration, even one in which the same process remains Brownian motion, additional information can produce martingales that no Brownian Ito integral can represent.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. On a finite horizon $[0,T]$, all stochastic integrals below are Ito integrals with progressively measurable integrands $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. They are not Stratonovich integrals; no Stratonovich correction is being applied. The Ito multiplication rules include $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

Theorem: In the completed, right-continuous natural filtration of $B_t$, if $X$ is $\mathcal{F}_T$-measurable and $\mathbb{E}[X^2]<\infty$, then there is a progressively measurable $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ such that $$ X=\mathbb{E}[X]+\int_0^T H_s\,dB_s. $$ The integrand is unique up to equality almost everywhere for the product measure $ds\,d\mathbb{P}$. Equivalently, the martingale $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$ satisfies $M_t=\mathbb{E}[X]+\int_0^t H_s\,dB_s$. This is the Brownian martingale representation theorem; its proof is part of the earlier material on martingale representation.

The initial value matters: the theorem as stated uses a Brownian natural filtration, whose initial sigma-field is trivial up to null sets. If $\mathcal{F}_0$ contains nontrivial information, a representation for a martingale generally has the form $M_t=M_0+\int_0^tH_s\,dB_s$, not necessarily $\mathbb{E}[X]+\int_0^tH_s\,dB_s$.

Worked Representations from Conditional Expectations

Worked Example: A Product at Two Different Times

Use the natural filtration of $B_t$ on $[0,2]$ to represent $X=B_1B_2$. First check square integrability: write $B_2=B_1+(B_2-B_1)$. The increment $B_2-B_1$ is independent of $B_1$, has mean zero and variance $1$, so $$ \mathbb{E}[X^2]=\mathbb{E}[B_1^2B_2^2] =\mathbb{E}[B_1^4]+\mathbb{E}[B_1^2]\mathbb{E}[(B_2-B_1)^2] =3+1=4. $$

For $0\le t\le1$, condition first on $\mathcal{F}_1$: $\mathbb{E}[B_1B_2\mid\mathcal{F}_1]=B_1^2$. Then $B_1=B_t+(B_1-B_t)$, where the remaining increment is independent of $\mathcal{F}_t$, has mean zero and variance $1-t$. Thus $$ M_t=\mathbb{E}[B_1^2\mid\mathcal{F}_t]=B_t^2+(1-t). $$ For $1\le t\le2$, $B_1$ is already known and $\mathbb{E}[B_2\mid\mathcal{F}_t]=B_t$, so $M_t=B_1B_t$.

To find the integrand before time $1$, apply Ito's formula to $f(x)=x^2$, for which $f'(x)=2x$ and $f''(x)=2$. The correction term is $\tfrac12 f''(B_t)\,dt=dt$, giving $d(B_t^2)=2B_t\,dB_t+dt$. Therefore $dM_t=2B_t\,dB_t$ for $t<1$. After time $1$, $B_1$ is fixed and $dM_t=B_1\,dB_t$. The representation is $$ B_1B_2=1+\int_0^1 2B_s\,dB_s+\int_1^2 B_1\,dB_s. $$

Check by covariation: the coefficient of $dB_t$ in $dM_t$ is $2B_t$ before time $1$ and $B_1$ after it. Hence $[M,B]_t=\int_0^{t\wedge1}2B_s\,ds+\mathbf{1}_{\{t>1\}}B_1(t-1)$, which has precisely those densities on the two intervals. Also, $\mathbb{E}[X]=\mathbb{E}[B_1^2]=1$. The integrand is square-integrable since $$ \mathbb{E}\int_0^1(2B_s)^2\,ds=4\int_0^1s\,ds=2,\qquad \mathbb{E}\int_1^2B_1^2\,ds=1. $$ The total is $3$. Independently, $\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2=4-1=3$, as required by the Ito isometry.

Worked Example: A Brownian Threshold Event

In the natural Brownian filtration on $[0,1]$, represent $X=\mathbf{1}_{\{B_1>0\}}$. Since $B_1\sim N(0,1)$, $\mathbb{E}[X]=1/2$, and $X$ is square-integrable. For $t<1$, the conditional distribution of $B_1$ given $\mathcal{F}_t$ is normal with mean $B_t$ and variance $1-t$. Consequently, $$ M_t=\mathbb{P}(B_1>0\mid\mathcal{F}_t) =\Phi\left(\frac{B_t}{\sqrt{1-t}}\right). $$

To verify the Ito coefficient, put $u(t,x)=\Phi(x/\sqrt{1-t})$, $z=x/\sqrt{1-t}$, and let $\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}$. Its derivatives are $$ u_t=\frac{z\phi(z)}{2(1-t)},\qquad u_x=\frac{\phi(z)}{\sqrt{1-t}},\qquad u_{xx}=-\frac{z\phi(z)}{1-t}. $$ Ito's formula gives $du(t,B_t)=\big(u_t+\tfrac12u_{xx}\big)\,dt+u_x\,dB_t$. The explicit correction is $\tfrac12u_{xx}\,dt=-z\phi(z)\,dt/[2(1-t)]$; it cancels $u_t\,dt$. Thus $$ M_t=\frac12+\int_0^t\frac{\phi(B_s/\sqrt{1-s})}{\sqrt{1-s}}\,dB_s,\qquad 0\le t<1. $$

Here is an integrability check at the endpoint. For $r<1$, the Ito isometry gives $$ \mathbb{E}\int_0^r\left(\frac{\phi(B_s/\sqrt{1-s})}{\sqrt{1-s}}\right)^2ds =\mathbb{E}[(M_r-\tfrac12)^2]\le\frac14, $$ because $0\le M_r\le1$ and $\mathbb{E}[M_r]=1/2$. As $r\uparrow1$, $M_r$ converges almost surely to $X$ except on $\{B_1=0\}$, an event of probability zero; bounded convergence gives convergence in $L^2$. Monotone convergence for the integrals then gives expected squared-integrand integral $1/4$. Therefore the integrand is in the required class on $[0,1]$, and the representation extends to $t=1$.

Check at a specific point: at $t=1/2$ and $B_t=0$, the conditional probability is $\Phi(0)=1/2$, and the integrand is $\phi(0)/\sqrt{1/2}=1/\sqrt{\pi}$. These follow directly from the formulas. The terminal variance is $\mathbb{E}[(X-1/2)^2]=1/4$, matching the squared-integrand integral.

Covariation as a Diagnostic

For a represented continuous martingale, covariation with Brownian motion identifies the integrand. The next result makes this test precise.

Proposition: Suppose $H$ is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, and set $N_t=N_0+\int_0^tH_s\,dB_s$. Then, almost surely, $$ [N,B]_t=\int_0^tH_s\,ds,\qquad [N]_t=\int_0^tH_s^2\,ds. $$ In particular, the density of $[N,B]$ with respect to time, when it exists, must agree with the representing integrand almost everywhere.

Proof. For an adapted step integrand $H_s=\sum_j H_j\mathbf{1}_{(t_j,t_{j+1}]}(s)$, with each $H_j$ measurable at time $t_j$, the integral on each interval is $H_j(B_t-B_{t_j})$. The Brownian quadratic-variation and covariation rules therefore give $[N,B]_t=\int_0^tH_s\,ds$ and $[N]_t=\int_0^tH_s^2\,ds$. For a general square-integrable progressively measurable $H$, approximate it in the norm $(\mathbb{E}\int_0^T H_s^2ds)^{1/2}$ by adapted step integrands. The Ito isometry gives convergence of the corresponding integrals in $L^2$; the defining covariation identities pass to the limit, yielding the displayed formulas. Finally, if two integrands represent the same martingale, the Ito isometry applied to their difference gives $\mathbb{E}\int_0^T(H_s-\widetilde H_s)^2ds=0$, so they agree $ds\,d\mathbb{P}$-almost everywhere. $\square$

This test is necessary for a representation, not by itself sufficient for every martingale in every filtration. In a Brownian natural filtration, the representation theorem supplies existence, and covariation identifies the integrand. In a larger filtration, martingales can contain components orthogonal to $B_t$; a zero covariation with $B_t$ does not imply that the martingale is constant.

When Brownian Representation Fails

Worked Example: New Information Creates a Nonrepresentable Martingale

Let the horizon be $T=1$. Let $U$, independent of the entire Brownian motion, take values $1$ and $-1$ with probability $1/2$ each. Use the filtration generated by $B_t$ and reveal $U$ at time $a=1/2$: before $a$, only Brownian information is available; at and after $a$, $U$ is also available. Complete the filtration. Brownian motion remains Brownian relative to this filtration because $U$ is independent of the Brownian path.

Consider $M_t=0$ for $t<a$ and $M_t=U$ for $t\ge a$. It is a martingale: before $a$, the conditional expectation of the unrevealed, mean-zero $U$ is $0$; at and after $a$, its value is known and remains $U$. It is square-integrable since $\mathbb{E}[M_t^2]\le\mathbb{E}[U^2]=1$, and $M_0=0$.

This martingale cannot equal $\int_0^tH_s\,dB_s$ for any progressively measurable $H$ with $\mathbb{E}\int_0^1H_s^2ds<\infty$. Such an Ito integral has continuous paths, while $M_t$ has a jump of size $U$, whose absolute value is $1$, at time $a$. Equivalently, its terminal variable $U$, with $\mathbb{E}[U]=0$, has no representation $U=\int_0^1H_s\,dB_s$ in this filtration. The failure is caused by the additional information, not by a defect in Brownian motion.

Check by covariation: the jump occurs while $B_t$ is continuous, so $[M,B]_t=0$; yet $[M]_t=0$ before $a$ and $[M]_t=U^2=1$ after $a$. This jump in quadratic variation is incompatible with the quadratic variation $\int_0^tH_s^2ds$ of a Brownian Ito integral, which is continuous. Covariation alone would not detect this failure: the path discontinuity and quadratic variation do.

Common Mistake: Using the Theorem Without Checking the Filtration

A wrong argument is: “$U$ is square-integrable and the space carries Brownian motion, so $U=\mathbb{E}[U]+\int_0^1H_s\,dB_s$.” This wrongly applies the Brownian representation theorem without its natural-filtration hypothesis. In the example, it would say $U=\int_0^1H_s\,dB_s$, which is impossible because the right side has continuous paths and the martingale revealing $U$ jumps. The correction is to check the filtration first. If the filtration is the Brownian natural filtration, use the theorem; if it includes additional information, test the proposed martingale’s paths and covariations and do not infer representation merely from square integrability.

Check Your Understanding

For each question, state the filtration and integrability assumptions you use.

For $X=B_1B_2$ in the Brownian natural filtration, what are the conditional-expectation formulas for $M_t$ on $0\le t\le1$ and $1\le t\le2$?
For the threshold-event representation, why must the Ito correction term cancel the time derivative before identifying the integrand?
If $N_t=N_0+\int_0^tH_s\,dB_s$, what does the time density of $[N,B]_t$ identify?
Why does zero covariation with $B_t$ not imply that a martingale in an enlarged filtration is constant?
What path property rules out representing the martingale that reveals an independent $U$ at a fixed positive time?