The Orthogonal-Complement Proof of Representation

Representation Through Orthogonality

Fix a finite horizon $T>0$. The $L^2$ Brownian martingale representation theorem says that a square-integrable terminal variable in a Brownian filtration can be written as a constant plus an Ito integral. The orthogonal-complement proof makes the key idea visible: if a terminal variable is orthogonal to constants and to every such integral, then it is orthogonal to a dense family of exponential martingales, and hence must be zero.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. For this result, assume that $(\mathcal{F}_t)$ is the completed, right-continuous natural filtration generated by $B$, and consider variables in $L^2(\Omega,\mathcal{F}_T,\mathbb{P})$. All stochastic integrals below are Ito integrals. Their integrands $H$ are progressively measurable and satisfy $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this class is denoted by $\mathcal{L}^2$.

Definition: Let $\mathcal{S}$ be the set of terminal variables of the form $c+\int_0^T H_s\,dB_s$, where $c\in\mathbb{R}$ and $H\in\mathcal{L}^2$. Orthogonality in $L^2$ means $\mathbb{E}[XY]=0$ for every element $Y$ of the specified set.

The Ito isometry gives $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. Such an integral is square-integrable and has mean zero. More generally, its process $M_t=\int_0^t H_s\,dB_s$ is a martingale: its increments have conditional expectation zero, first for simple adapted integrands by the Brownian increment property and then for all $H\in\mathcal{L}^2$ by $L^2$ approximation. Its second moment is finite by the isometry. These are true martingales, not merely local martingales.

Why the Subspace Is Closed

Proposition: The set $\mathcal{S}$ is a closed linear subspace of $L^2(\Omega,\mathcal{F}_T,\mathbb{P})$.

Proof. Linearity follows from linearity of the Ito integral. To show closedness, let $c_n+\int_0^T H_s^n\,dB_s$ be a Cauchy sequence in $L^2$. Since each stochastic integral has mean zero, constants are orthogonal to these integrals. Thus, for every $n,m$, the square of the distance between two terms is

|c_n-c_m|^2+\mathbb{E}\int_0^T(H_s^n-H_s^m)^2\,ds.

Both terms are nonnegative, so the Cauchy property implies that $(c_n)$ is Cauchy in $\mathbb{R}$, and $(H^n)$ is Cauchy in the $L^2(\mathbb{P}\times dt)$ norm. The space of progressively measurable square-integrable integrands is complete in this norm, so $c_n\to c$ and $H^n\to H$ for some $c\in\mathbb{R}$ and $H\in\mathcal{L}^2$. By the Ito isometry, $\int_0^T H_s^n\,dB_s\to\int_0^T H_s\,dB_s$ in $L^2$. Therefore the original sequence converges to $c+\int_0^T H_s\,dB_s\in\mathcal{S}$, proving that $\mathcal{S}$ is closed. $\square$

Exponential Martingales Lie in the Subspace

Take a deterministic step function $h:[0,T]\to\mathbb{R}$, and define, for $0\le t\le T$,

Z_t(h)=\exp\left(\int_0^t h(s)\,dB_s-\frac12\int_0^t h(s)^2\,ds\right).

This is a true martingale. Indeed, for $s\le t$, the ratio $Z_t(h)/Z_s(h)$ is the exponential of the integral over $[s,t]$ minus half its variance. That future integral is independent of $\mathcal{F}_s$, is normal with variance $\int_s^t h(u)^2\,du$, and its centered exponential has expectation one. Consequently $\mathbb{E}[Z_t(h)\mid\mathcal{F}_s]=Z_s(h)$. It is integrable; in fact, the normal moment-generating function gives $\mathbb{E}[Z_t(h)^2]=\exp(\int_0^t h(s)^2\,ds)<\infty$.

Lemma: For every deterministic step function $h$, $Z_T(h)\in\mathcal{S}$.

Proof. On any interval where $h$ is constant, set $X_t=\int_0^t h(s)\,dB_s$ and $A_t=\int_0^t h(s)^2\,ds$. Apply Ito's formula to $f(t,x)=\exp(x-\tfrac12 A_t)$ with $x=X_t$. The derivatives are $f_x=f$, $f_{xx}=f$, and $f_t=-\tfrac12 h(t)^2 f$ on that interval. Since $dX_t=h(t)\,dB_t$ and $d[X]_t=h(t)^2dt$, the formula, including its second-derivative correction, is

dZ_t(h)=-\frac12h(t)^2Z_t(h)\,dt+h(t)Z_t(h)\,dB_t +\frac12h(t)^2Z_t(h)\,dt =h(t)Z_t(h)\,dB_t.

Applying this on each constant interval and joining the increments yields $Z_T(h)=1+\int_0^T h(s)Z_s(h)\,dB_s$. The integrand is progressively measurable, and

\mathbb{E}\int_0^T h(s)^2Z_s(h)^2\,ds =\int_0^T h(s)^2\exp\left(\int_0^s h(u)^2\,du\right)ds<\infty,

because $h$ is bounded on the finite interval. Thus the integral is in the required class and $Z_T(h)\in\mathcal{S}$. $\square$

The density input is the Brownian exponential-density theorem: for the completed natural filtration on $[0,T]$, the linear span of $Z_T(h)$, over deterministic step functions $h$, is dense in $L^2(\Omega,\mathcal{F}_T,\mathbb{P})$. This is the Gaussian-cylinder density result established in the earlier material on exponential martingales and density. The hypotheses matter: the conclusion is for the Brownian-generated terminal sigma-field, not an arbitrary larger sigma-field containing additional randomness.

Worked Example: A Step-Kernel Exponential

Let $T=1$, and let $h(s)=2$ for $0\le s\le\tfrac12$ and $h(s)=0$ for $\tfrac12<s\le1$. Compute its terminal exponential, mean, variance, and the squared-integrand norm in its Ito representation.

The stochastic integral is $\int_0^1 h(s)\,dB_s=2B_{1/2}$, and $\int_0^1h(s)^2\,ds=4\cdot\tfrac12=2$. Hence $Z_1(h)=\exp(2B_{1/2}-1)$. Since $B_{1/2}$ is normal with mean zero and variance $\tfrac12$, its moment-generating function gives $\mathbb{E}[Z_1(h)]=\exp(\tfrac12\cdot 2^2\cdot\tfrac12-1)=1$. Also $\mathbb{E}[Z_1(h)^2]=\exp(2)=e^2$, so $\operatorname{Var}(Z_1(h))=e^2-1$.

The Ito representation is $Z_1(h)=1+\int_0^{1/2}2Z_s(h)\,dB_s$. For $s\le\tfrac12$, $\mathbb{E}[Z_s(h)^2]=\exp(4s)$, so the isometry gives

\mathbb{E}\int_0^{1/2}4Z_s(h)^2\,ds =\int_0^{1/2}4e^{4s}\,ds =e^2-1.

Check: the squared-integrand norm equals $\operatorname{Var}(Z_1(h))$, as it must by the isometry applied to $Z_1(h)-1$. The mean is one, consistent with the martingale calculation.

The Orthogonal-Complement Argument

Theorem: For the Brownian filtration specified above, every $X\in L^2(\Omega,\mathcal{F}_T,\mathbb{P})$ has a representation $$ X=\mathbb{E}[X]+\int_0^T H_s\,dB_s $$ for some $H\in\mathcal{L}^2$. The integrand is unique up to equality almost everywhere with respect to $\mathbb{P}\times dt$.

Proof. First take any $Y\in\mathcal{S}^{\perp}$, the orthogonal complement of $\mathcal{S}$. By the lemma, $Z_T(h)\in\mathcal{S}$ for every deterministic step function $h$, so $\mathbb{E}[YZ_T(h)]=0$ for every such $h$. Linearity implies that $Y$ is orthogonal to the span of these terminal exponentials. That span is dense in $L^2(\mathcal{F}_T)$, so there is a sequence $V_n$ in the span with $V_n\to Y$ in $L^2$. Cauchy–Schwarz gives $\mathbb{E}[YV_n]\to\mathbb{E}[Y^2]$. Each inner product on the left is zero, and therefore $\mathbb{E}[Y^2]=0$, which means $Y=0$ almost surely. Hence $\mathcal{S}^{\perp}=\{0\}$.

For completeness, a closed subspace of a Hilbert space with zero orthogonal complement is the whole space. To see why, fix $X$ and choose a sequence $S_n\in\mathcal{S}$ whose distances to $X$ decrease to $d=\inf_{S\in\mathcal{S}}\|X-S\|_2$. The parallelogram identity and $(S_n+S_m)/2\in\mathcal{S}$ give

\|S_n-S_m\|_2^2 =2\|X-S_n\|_2^2+2\|X-S_m\|_2^2 -4\left\|X-\frac{S_n+S_m}{2}\right\|_2^2 \le 2\|X-S_n\|_2^2+2\|X-S_m\|_2^2-4d^2.

The right-hand side tends to zero, so closedness gives a limit $S\in\mathcal{S}$ attaining the distance $d$. For any $V\in\mathcal{S}$, minimizing $\|X-S-aV\|_2^2$ over real $a$ shows $\mathbb{E}[(X-S)V]=0$. Thus $X-S\in\mathcal{S}^{\perp}$, which is zero. It follows that $X=S$, so $\mathcal{S}$ is all of $L^2(\mathcal{F}_T)$.

Finally, write $X=c+\int_0^T H_s\,dB_s$. Taking expectations gives $c=\mathbb{E}[X]$, since the Ito integral has mean zero. If two integrands represent the same terminal variable, their difference has integral zero; the Ito isometry then gives $\mathbb{E}\int_0^T(H_s-\widetilde H_s)^2\,ds=0$, proving uniqueness up to $\mathbb{P}\times dt$-almost-everywhere equality. $\square$

Worked Example: Testing a Variable Against Exponentials

Let $T=1$, $Y=B_1$, and $Z_1(\lambda)=\exp(\lambda B_1-\tfrac12\lambda^2)$ for a constant $\lambda\in\mathbb{R}$. Compute $\mathbb{E}[YZ_1(\lambda)]$ and compare with the integral representation.

For a standard normal variable, $\mathbb{E}[e^{\lambda B_1}]=e^{\lambda^2/2}$. Differentiating this explicit moment-generating function gives $\mathbb{E}[B_1e^{\lambda B_1}]=\lambda e^{\lambda^2/2}$. Therefore

\mathbb{E}[YZ_1(\lambda)] =e^{-\lambda^2/2}\mathbb{E}[B_1e^{\lambda B_1}] =\lambda.

On the other hand, $B_1=\int_0^1 1\,dB_s$ and $Z_1(\lambda)=1+\int_0^1\lambda Z_s(\lambda)\,dB_s$. The Ito isometry’s inner-product identity gives $\mathbb{E}[YZ_1(\lambda)]=\mathbb{E}\int_0^1\lambda Z_s(\lambda)\,ds=\lambda$, since $\mathbb{E}[Z_s(\lambda)]=1$ at each $s$. Check: both methods give $\lambda$; at $\lambda=0$, the pairing is zero, as it must be because $\mathbb{E}[B_1]=0$.

Worked Example: A Centered Event Detected by an Exponential

Let $Y=\mathbf{1}_{\{B_1>0\}}-\tfrac12$. The event has probability $\tfrac12$, so $Y$ is centered and belongs to $L^2$. Compute its inner product with $Z_1(\lambda)$.

Write the standard normal density as $\varphi(x)=(2\pi)^{-1/2}e^{-x^2/2}$. Completing the square gives $e^{\lambda x-\lambda^2/2}\varphi(x)=\varphi(x-\lambda)$. Since $\mathbb{E}[Z_1(\lambda)]=1$,

\mathbb{E}[YZ_1(\lambda)] =\int_0^\infty \varphi(x-\lambda)\,dx-\frac12 =\Phi(\lambda)-\frac12.

For $\lambda>0$, this is strictly positive, so this centered variable is not orthogonal to the exponential family. At $\lambda=0$, it is zero because $\Phi(0)=\tfrac12$. Check: the calculation shows why testing only against constants is insufficient: centering checks only the constant direction, while the exponential family probes additional directions.

Common Pitfall and Why the Proof Matters

A common mistake is to treat “orthogonal to constants” as equivalent to “zero.” The incorrect calculation is: $\mathbb{E}[Y]=0$, therefore $Y=0$ almost surely. The conclusion does not follow. For the concrete centered variable $Y=B_1$, $\mathbb{E}[B_1]=0$, but $\mathbb{E}[B_1^2]=1$, so $B_1$ is not zero in $L^2$. The exponential test makes the missing information explicit: $\mathbb{E}[B_1Z_1(1)]=1$, not zero. In the proof, it is orthogonality to the entire dense family—not just to constants—that forces the residual to vanish.

The argument separates representation into two structural facts. The Ito isometry makes constants plus stochastic integrals a closed subspace; exponential martingales place a dense family inside that subspace. Orthogonal-complement reasoning then rules out any nonzero residual. This method is useful beyond this particular representation theorem whenever a candidate subspace is closed and a convenient dense family can be shown to lie in it.

Check Your Understanding

Use the assumptions and notation stated above.

Why are constants orthogonal to every square-integrable Brownian Ito integral?
For deterministic step $h$, what condition ensures $h(s)Z_s(h)$ is an admissible integrand in the representation of $Z_T(h)$?
Where does the density of deterministic-kernel exponential martingales enter the orthogonal-complement proof?
Why does the proof require the Brownian-generated terminal sigma-field rather than an arbitrary larger sigma-field?
If two integrands represent the same terminal variable, which identity implies they agree almost everywhere?