Orthogonal Martingales and Failure of Brownian Representation

Brownian Integrals and the Information in a Filtration

A Brownian martingale representation is a statement about both a random variable and the filtration in which it is observed. In the completed natural filtration of a Brownian motion, every square-integrable martingale can be represented by an integral with respect to that Brownian motion. If the filtration contains additional random information, the same conclusion can fail: Brownian integrals may not capture martingales driven by that information.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space satisfying the usual conditions, and let $B_t$ be a standard Brownian motion with respect to this filtration. For a fixed $T>0$, an admissible integrand is a progressively measurable process $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this class is denoted by $\mathcal{L}^2$. The Ito isometry gives $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$, and the integral is a square-integrable martingale.

Definition: A square-integrable martingale $N_t$ is orthogonal to $B_t$ if its predictable covariation with $B_t$ satisfies $\langle N,B\rangle_t=0$ for every $t$. Equivalently, $[N,B]_t=0$ when $B$ is continuous. In this case $N$ is also orthogonal in $L^2(\mathbb{P})$ to every Brownian integral $\int_0^T H_s\,dB_s$ with $H\in\mathcal{L}^2$.

Here $\langle N,B\rangle$ denotes predictable covariation, while $[N,B]$ denotes quadratic covariation. Orthogonality is stronger than merely having zero covariance at one time: it concerns the martingale’s entire interaction with Brownian motion. A local martingale is only a martingale after suitable stopping; the decomposition below concerns true square-integrable martingales and does not automatically extend to arbitrary local martingales.

The Orthogonal-Residual Decomposition

Theorem: Let $X\in L^2(\mathcal{F}_T)$, and put $X_0=\mathbb{E}[X\mid\mathcal{F}_0]$. There exist a progressively measurable $H\in\mathcal{L}^2$ and a square-integrable martingale $N_t$, with $N_0=0$, such that $$ X=X_0+\int_0^T H_s\,dB_s+N_T. $$ The residual $N$ is orthogonal to $B$, and hence to every Brownian stochastic integral. The representing integrand $H$ and residual terminal value $N_T$ are unique up to equality in their respective $L^2$ spaces. In particular, $X$ has a Brownian integral representation after its $\mathcal{F}_0$-measurable part is removed if and only if $N_T=0$ almost surely.

Proof. Consider the Hilbert space $L^2(\mathcal{F}_T)$, with inner product $\mathbb{E}[UV]$, and the subspace $$ \mathcal{K}=\left\{\int_0^T H_s\,dB_s:H\in\mathcal{L}^2\right\}. $$ The Ito isometry implies that $\mathcal{K}$ is closed: if a sequence of such integrals converges in $L^2$, the isometry makes the corresponding integrands a Cauchy sequence in $\mathcal{L}^2$; its limit is again an admissible integrand and its integral is the given $L^2$ limit. Every element of $\mathcal{K}$ has conditional expectation zero given $\mathcal{F}_0$, since it is the terminal value of a martingale starting at zero.

Apply the Hilbert-space projection theorem to $X-X_0$ and the closed subspace $\mathcal{K}$. It gives a unique $I\in\mathcal{K}$ such that $R=X-X_0-I$ is orthogonal to $\mathcal{K}$. Write $I=\int_0^T H_s\,dB_s$, and set $N_t=\mathbb{E}[R\mid\mathcal{F}_t]$, using the right-continuous martingale version. Because $R$ is orthogonal to every $\mathcal{F}_0$-measurable square-integrable variable, $\mathbb{E}[R\mid\mathcal{F}_0]=0$, so $N_0=0$. The construction gives the stated decomposition and orthogonality to all Brownian integrals.

To identify that orthogonality with zero predictable covariation, use the standard cross-isometry for square-integrable martingales and the Kunita–Watanabe inequality. The latter implies that $d\langle N,B\rangle_t=a_t\,dt$ for a predictable $a$ with $\mathbb{E}\int_0^T a_t^2\,dt<\infty$. The cross-isometry gives $$ 0=\mathbb{E}\left[N_T\int_0^T a_s\,dB_s\right] =\mathbb{E}\int_0^T a_s\,d\langle N,B\rangle_s =\mathbb{E}\int_0^T a_s^2\,ds. $$ Thus $a=0$ almost everywhere for $dt\,d\mathbb{P}$, and $\langle N,B\rangle=0$. The cross-isometry and Kunita–Watanabe inequality are standard martingale results; their technical proofs are not included here. Conversely, zero predictable covariation implies orthogonality to every Brownian integral by the cross-isometry. Finally, the orthogonal projection onto a closed subspace is unique, and the Ito isometry makes its integrand unique in $\mathcal{L}^2$. The residual vanishes exactly when $X-X_0$ belongs to $\mathcal{K}$. $\square$

The theorem characterizes the obstruction: a nonzero residual is precisely the part of the terminal variable not captured by Brownian stochastic integrals. The result does not say that a residual must be independent of $B$. It says that the residual martingale is orthogonal to Brownian motion in the covariation sense.

Independent Brownian Noise

Worked Example: A Second Brownian Motion Cannot Be Represented

Let $B_t$ and $W_t$ be independent standard Brownian motions, and let $\mathcal{F}_t$ be the completed filtration they jointly generate. The process $B_t$ remains Brownian in this larger filtration. On $[0,1]$, show that $W_1$ cannot be written as a Brownian integral with respect to $B$.

Independence gives $\mathbb{E}[W_1]=0$ and $\mathbb{E}[W_1^2]=1$. Also, $\mathbb{E}[W_1\mid\mathcal{F}_t]=W_t$: the increment $W_1-W_t$ is independent of the joint information up to time $t$ and has mean zero. Thus $W_t$ is a square-integrable martingale. Independent Brownian motions have $[W,B]_t=0$, so this martingale is orthogonal to $B$.

If $W_1=\int_0^1 H_s\,dB_s$ for some $H\in\mathcal{L}^2$, orthogonality would give $$ 1=\mathbb{E}[W_1^2] =\mathbb{E}\left[W_1\int_0^1 H_s\,dB_s\right] =0, $$ a contradiction. In the orthogonal-residual decomposition, the Brownian integral component is zero and $N_1=W_1$. Check: the residual variance is $\mathbb{E}[W_1^2]=1$, so it is not zero almost surely and representation fails.

Independent Jump Noise

Worked Example: A Compensated Poisson Residual

Let $J_t$ be a Poisson process of rate $2$, independent of $B_t$, and let $\mathcal{F}_t$ be the completed filtration jointly generated by $B$ and $J$. Define $M_t=J_t-2t$. Show that $M_1$ is not a Brownian integral with respect to $B$.

The independent Poisson increments have mean $2(t-s)$, so $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $s\le t$. Also, $\mathbb{E}[M_t]=0$ and $\mathbb{E}[M_t^2]=2t$, the variance formula for a Poisson random variable of mean $2t$. Hence $M_t$ is a square-integrable martingale. Its jumps occur at the jump times of $J$, while $B$ is continuous; independence and the zero drift of $M$ give $[M,B]_t=0$.

At $t=1$, $\mathbb{E}[M_1^2]=2$. For any $H\in\mathcal{L}^2$, orthogonality gives $\mathbb{E}[M_1\int_0^1H_s\,dB_s]=0$. If $M_1$ equalled that integral, then $$ 2=\mathbb{E}[M_1^2] =\mathbb{E}\left[M_1\int_0^1 H_s\,dB_s\right] =0, $$ which is impossible. Check: the variance $2$ is the Poisson variance at parameter $2$; it confirms that the residual terminal value is nonzero.

The Poisson example also distinguishes continuous Brownian noise from jump noise. A compensated Poisson process is a genuine martingale in the joint filtration, but its random jumps cannot be generated by integrating against the continuous process $B$. Brownian representation is therefore not a property of the probability space alone; it depends on whether the filtration’s information is fully captured by Brownian motion.

A Mixed Payoff and Its Two Components

Worked Example: Decomposing a Product of Independent Brownian Motions

In the joint filtration of independent Brownian motions $B$ and $W$, decompose $X=B_1W_1$ into a $B$-integral and a residual. For $f(x,y)=xy$, the derivatives are $f_x=y$, $f_y=x$, and $f_{xx}=f_{yy}=0$. The two-dimensional Ito formula, with $dB_t\,dW_t=0$, gives $$ d(B_tW_t)=W_t\,dB_t+B_t\,dW_t. $$

Integrating from $0$ to $1$ yields $$ B_1W_1=\int_0^1 W_s\,dB_s+\int_0^1 B_s\,dW_s. $$ The first integrand is admissible because $\mathbb{E}\int_0^1W_s^2\,ds=\int_0^1s\,ds=\tfrac12$. The second is also admissible because $\mathbb{E}\int_0^1B_s^2\,ds=\tfrac12$. Each integral is a square-integrable martingale with mean zero, by the Ito isometry and the martingale property of Ito integrals.

The residual $N_t=\int_0^tB_s\,dW_s$ is orthogonal to $B$, since the two Brownian coordinates have zero covariation. Its terminal variance, computed by the Ito isometry, is $$ \mathbb{E}[N_1^2]=\mathbb{E}\int_0^1B_s^2\,ds=\int_0^1s\,ds=\tfrac12. $$ As a separate check, independence gives $\mathbb{E}[X^2]=\mathbb{E}[B_1^2]\mathbb{E}[W_1^2]=1$. The two integrals are orthogonal in $L^2$, and each has variance $\tfrac12$, so their variances sum to $1$, as required. Since the residual variance is positive, $B_1W_1$ has no representation using only a $B$-integral.

What Orthogonality Does—and Does Not—Say

The orthogonal-residual decomposition is the Hilbert-space form of the representation question. If the residual is zero, the terminal variable is represented by a Brownian integral. If the residual is nonzero, no change of integrand can remove it: uniqueness of orthogonal projection separates the Brownian-generated component from the remainder. In particular, the examples above do not contradict Brownian representation in a Brownian filtration; they use filtrations enlarged by independent sources of randomness.

A common mistake is to argue that a martingale must be representable because it is adapted to a filtration containing $B$. Adaptedness only says that the process uses information available in the filtration. It does not say that all that information was generated by $B$. For the independent Brownian example, the incorrect step would be to assume that some $H$ satisfies $W_1=\int_0^1H_s\,dB_s$ merely because $W_1$ is adapted. The correction is to test orthogonality: every such integral has zero covariance with $W_1$, whereas $\mathbb{E}[W_1^2]=1$.

Orthogonality also does not mean that the residual is independent of the Brownian path in every model. Zero covariation is the relevant condition for this decomposition; independence is one sufficient way to obtain it in the examples. Finally, the square-integrable hypotheses matter: the theorem uses $L^2$ projection and the Ito isometry. A local martingale need not be a true martingale, and an $L^2$ decomposition theorem should not be applied to it without additional hypotheses.

Check Your Understanding

Use the orthogonal-residual criterion, and state any integrability assumption you need.

In a filtration jointly generated by independent Brownian motions $B$ and $W$, why can $W_T$ not equal $\int_0^T H_s\,dB_s$ when $T>0$?
For a rate-$\lambda$ Poisson process $J_t$, what are the mean and variance of $J_T-\lambda T$?
In the product example, compute the variance of $\int_0^T B_s\,dW_s$ for $T>0$.
What does a zero residual imply about representation, and what does a nonzero residual imply?
Why does adaptation to a filtration containing $B$ not by itself imply Brownian representation?