The L2 Brownian Martingale Representation Theorem

From a Terminal Payoff to a Martingale

Fix a finite horizon $T>0$. In the Brownian natural filtration, information available by time $t$ is generated by the Brownian path up to that time. The $L^2$ Brownian martingale representation theorem says that any square-integrable terminal random variable can be built from its mean and an Ito integral against that path. Thus, it describes not only a terminal variable but also the entire conditional-expectation martingale leading to it.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. On $[0,T]$, assume $(\mathcal{F}_t)$ is the usual augmentation of the natural filtration generated by $B$, and $\mathcal{F}=\mathcal{F}_T$ up to completion. In particular, no additional randomness is included in the terminal sigma-field. Write $\mathcal{L}^2$ for the predictable processes $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. These processes are admissible Ito integrands; in particular, they are progressively measurable.

Definition: A random variable $X$ is square-integrable if $\mathbb{E}[X^2]<\infty$. Its conditional-expectation martingale is $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$, $0\le t\le T$. A representation of $X$ is an equality $X=\mathbb{E}[X]+\int_0^T H_s\,dB_s$ almost surely, for some $H\in\mathcal{L}^2$.

The Ito isometry, established in the earlier material on square-integrable predictable integrands, states that $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. For $H\in\mathcal{L}^2$, the Ito integral is a square-integrable martingale: its increments have conditional expectation zero, so $\mathbb{E}[\int_0^T H_s\,dB_s\mid\mathcal{F}_t]=\int_0^t H_s\,dB_s$. This property follows first for elementary predictable integrands from independent Brownian increments, and then for all $H\in\mathcal{L}^2$ by the Ito-isometry limit.

Theorem: Let $X\in L^2(\Omega,\mathcal{F}_T,\mathbb{P})$ and suppose $(\mathcal{F}_t)$ is the usual augmentation of the natural filtration of $B$ on $[0,T]$. There exists a predictable $H\in\mathcal{L}^2$ such that, for every $t\in[0,T]$, $$ \mathbb{E}[X\mid\mathcal{F}_t]=\mathbb{E}[X]+\int_0^t H_s\,dB_s $$ almost surely. The integrand is unique up to equality outside a set of zero measure for $d\mathbb{P}\,dt$ on $\Omega\times[0,T]$.

Why the Representation Space Is Closed

Separate the theorem into two tasks: show that represented terminal variables form a closed subspace of $L^2$, then show that a dense family of terminal variables has representations. The constant term is essential: every Ito integral here has mean zero, whereas $X$ need not.

Proposition: The set $\mathcal{R}=\{c+\int_0^T H_s\,dB_s:c\in\mathbb{R},\,H\in\mathcal{L}^2\}$ is a closed subspace of $L^2(\mathcal{F}_T)$.

Proof. Linearity of the Ito integral makes $\mathcal{R}$ a subspace. Let $Y_n=c_n+\int_0^T H_s^{(n)}\,dB_s$ be a Cauchy sequence in $L^2$. Ito integrals have mean zero, so $\mathbb{E}[Y_n]=c_n$. Also, the Ito isometry gives $$ \mathbb{E}[(Y_n-Y_m)^2]=(c_n-c_m)^2+\mathbb{E}\int_0^T(H_s^{(n)}-H_s^{(m)})^2\,ds. $$ Both terms are nonnegative. Since the left side tends to zero as $n,m\to\infty$, $(c_n)$ is Cauchy in $\mathbb{R}$, and $(H^{(n)})$ is Cauchy in the $L^2(d\mathbb{P}\,dt)$ norm. Let their limits be $c$ and $H$, respectively. The predictable $L^2$ integrands are complete in this norm, so $H\in\mathcal{L}^2$. The Ito isometry now implies $Y_n\to c+\int_0^T H_s\,dB_s$ in $L^2$. Thus the $L^2$ limit lies in $\mathcal{R}$, proving closedness. $\square$

Proof of the Representation Theorem

We first consider a smooth function of finitely many Brownian observations. Suppose $0<t_1<\cdots<t_n\le T$ and $X=f(B_{t_1},\ldots,B_{t_n})$, where $f$ is smooth with compact support. On each time interval between successive observation times, the conditional expectation of $X$, given the observations already made and the current value $B_t$, is obtained by integrating $f$ over the remaining independent Gaussian increments. This gives a smooth function of the already observed values, $t$, and $B_t$. As a function of $t$ and $B_t$, it satisfies the backward heat equation $\partial_t u+\tfrac12\partial_{xx}u=0$: this follows by differentiating the Gaussian transition density, or equivalently the Gaussian convolution formula.

Apply Ito’s formula on each such interval. For the function $u$, the derivatives are $u_t$, $u_x$, and $u_{xx}$; the formula is $$ du(t,B_t)=\left(u_t(t,B_t)+\tfrac12u_{xx}(t,B_t)\right)dt+u_x(t,B_t)\,dB_t=u_x(t,B_t)\,dB_t. $$ The displayed $\tfrac12u_{xx}\,dt$ correction is what cancels the time derivative. At an observation time the conditional-expectation process has matching left and right limits: immediately before that time, the current Brownian value tends to the newly observed value, and immediately after it, that same value is included among the observations. Adding the interval representations therefore gives $X=\mathbb{E}[X]+\int_0^T H_s\,dB_s$. The integrand is the corresponding $u_x$ on each interval; it is predictable up to choices at finitely many deterministic times, which do not affect the integral. It belongs to $\mathcal{L}^2$, since the derivatives are bounded for the smooth compactly supported payoff on a finite horizon.

These smooth cylinder variables are dense in $L^2(\mathcal{F}_T)$. To see why, take the increasing dyadic grids on $[0,T]$ and let $\mathcal{G}_n$ be the sigma-field generated by Brownian values on the $n$th grid. Path continuity implies that the union of these sigma-fields generates $\mathcal{F}_T$, up to completion. The standard $L^2$ conditional-expectation convergence theorem gives $\mathbb{E}[X\mid\mathcal{G}_n]\to X$ in $L^2$. Each such conditional expectation is a function of a finite-dimensional Gaussian vector. Smooth compactly supported functions are dense in $L^2$ of a finite-dimensional Gaussian law: one may first truncate to a bounded region, approximate there by continuous functions, and then smooth by convolution. Consequently, finite linear combinations of the smooth cylinder variables approximate $X$ in $L^2$.

Every smooth cylinder variable has a representation, and sums of represented variables remain represented. Hence $\mathcal{R}$ contains a dense subset of $L^2(\mathcal{F}_T)$. The closedness proposition implies $\mathcal{R}=L^2(\mathcal{F}_T)$, proving existence. For any representation $X=c+\int_0^T H_s\,dB_s$, taking expectations gives $c=\mathbb{E}[X]$, and the martingale property of the Ito integral gives the stated conditional-expectation formula at every $t$. Finally, if $H$ and $\widetilde H$ both represent $X$, subtraction and the Ito isometry yield $$ \mathbb{E}\int_0^T(H_s-\widetilde H_s)^2\,ds =\mathbb{E}\left[\left(\int_0^T(H_s-\widetilde H_s)\,dB_s\right)^2\right]=0. $$ A nonnegative function with integral zero vanishes almost everywhere for $d\mathbb{P}\,dt$. Thus the integrands agree up to product-measure equality. $\square$

Worked Representations

Worked Example: A Linear Terminal Payoff

Take $T=1$ and $X=B_1$. Since $B_t$ is measurable with respect to $\mathcal{F}_t$, and $B_1-B_t$ is independent of $\mathcal{F}_t$ with mean zero, $\mathbb{E}[B_1\mid\mathcal{F}_t]=B_t$. Also $\mathbb{E}[B_1]=0$. The candidate integrand is $H_s=1$, so $$ \mathbb{E}[X]+\int_0^t H_s\,dB_s=0+\int_0^t 1\,dB_s=B_t. $$ It is admissible because $\mathbb{E}\int_0^1 1^2\,ds=1$. At $t=1$, the expression equals $X$, as required. As a check, the Ito isometry gives $\mathbb{E}[(\int_0^1 1\,dB_s)^2]=1=\mathbb{E}[B_1^2]$.

Worked Example: A Squared Brownian Payoff

For $T=1$, let $X=B_1^2$. Independence and the second moment of a normal increment give $\mathbb{E}[X\mid\mathcal{F}_t]=B_t^2+(1-t)$, and $\mathbb{E}[X]=1$. To verify the representation directly, set $f(t,x)=x^2+1-t$. Its derivatives are $f_t=-1$, $f_x=2x$, and $f_{xx}=2$. Ito’s formula, including its correction, gives $$ df(t,B_t)=\left(-1+\tfrac12\cdot2\right)dt+2B_t\,dB_t=2B_t\,dB_t. $$ Therefore $B_t^2+(1-t)=1+\int_0^t2B_s\,dB_s$, and at $t=1$ this is $B_1^2$. The integrand is admissible since $\mathbb{E}\int_0^1(2B_s)^2\,ds=4\int_0^1s\,ds=2$. As an independent check, $\operatorname{Var}(B_1^2)=\mathbb{E}[B_1^4]-1^2=3-1=2$, matching the Ito-isometry value.

Worked Example: An Indicator Payoff

Let $T=1$ and $X=\mathbf{1}_{\{B_1>0\}}$. For $t<1$, the conditional probability of $B_1>0$ is $M_t=\Phi(B_t/\sqrt{1-t})$. Write $z=x/\sqrt{1-t}$ and $f(t,x)=\Phi(z)$, where $\Phi$ and $\varphi$ are the standard normal distribution function and density. Then $f_x=\varphi(z)/\sqrt{1-t}$, $f_{xx}=-z\varphi(z)/(1-t)$, and $f_t=z\varphi(z)/(2(1-t))$. Ito’s formula gives $$ d f(t,B_t)=\left(\frac{z\varphi(z)}{2(1-t)}-\frac{z\varphi(z)}{2(1-t)}\right)dt+\frac{\varphi(z)}{\sqrt{1-t}}\,dB_t. $$ Thus $H_t=\varphi(B_t/\sqrt{1-t})/\sqrt{1-t}$ for $t<1$. As $t\uparrow1$, $M_t\to X$ almost surely because $\mathbb{P}(B_1=0)=0$, and also in $L^2$ because $0\le M_t\le1$. The isometry gives $\mathbb{E}\int_0^1H_t^2\,dt=\operatorname{Var}(X)=\tfrac12(1-\tfrac12)=\tfrac14$. Hence $H\in\mathcal{L}^2$, and $X=\tfrac12+\int_0^1H_t\,dB_t$. The mean $\tfrac12$ agrees with the symmetry of the centered normal law.

What the Assumptions Exclude

The natural-filtration assumption matters. If $\mathcal{F}_T$ contains an independent random variable not determined by the Brownian path, the theorem need not hold: an Ito integral against $B$ cannot encode that extra randomness. Likewise, square integrability is the condition that places the terminal payoff and its representing integrand in the $L^2$ framework of the Ito isometry.

Common mistake: applying the ordinary chain rule to $B_t^2$ gives the incorrect identity $B_1^2=2\int_0^1B_s\,dB_s$. Ito’s formula has $f(t,x)=x^2$, $f_t=0$, $f_x=2x$, and $f_{xx}=2$, so its correction is $\tfrac12f_{xx}\,dt=dt$. Therefore $B_1^2=2\int_0^1B_s\,dB_s+1$, with the constant matching $\mathbb{E}[B_1^2]=1$. The constant term in the representation cannot be dropped.

The theorem asserts uniqueness only up to equality almost everywhere for $d\mathbb{P}\,dt$, not equality at every time and every outcome. Altering an integrand at a single deterministic time, for example, changes neither its Ito integral nor its $L^2(d\mathbb{P}\,dt)$ class.

Check Your Understanding

Use the finite-horizon assumptions and the Ito isometry when answering.

Why does the representation include $\mathbb{E}[X]$ rather than only an Ito integral?
For $X=B_T$, identify the integrand and verify its $\mathcal{L}^2$ condition.
For the squared payoff, identify the Ito correction that cancels the time derivative of $B_t^2+(T-t)$.
How does the Ito isometry imply that two representing integrands agree $d\mathbb{P}\,dt$-almost everywhere?
Give an example of additional terminal information that would invalidate the natural-filtration assumption.