Martingale Representation for a Brownian Terminal Claim

From a Terminal Claim to a Martingale

A square-integrable terminal claim is not just a random payoff at time $T$. Its conditional expectation at each earlier time describes the information available then, and therefore forms a martingale. In a Brownian filtration, this martingale can be expressed as its initial value plus an Ito integral. The integrand records how the claim’s conditional value responds to Brownian risk.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carry a standard Brownian motion $B_t$, and fix $T<\infty$. Assume that $(\mathcal{F}_t)$ is the usual augmentation of the natural filtration generated by $B$, so that $\mathcal{F}_T$ contains no additional randomness beyond the Brownian path up to time $T$. A terminal claim $X$ is square-integrable if $X$ is $\mathcal{F}_T$-measurable and $\mathbb{E}[X^2]<\infty$.

Definition: Let $\mathcal{L}^2$ be the class of progressively measurable processes $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Brownian martingale representation of $X$ is an identity $$ X=\mathbb{E}[X]+\int_0^T H_s\,dB_s,\qquad H\in\mathcal{L}^2. $$ The Ito integral is defined on this class. Its integrand is unique up to equality almost everywhere for the product of time measure and $\mathbb{P}$.

The filtration assumption matters: if $\mathcal{F}_0$ contains an independent random variable, a claim depending on that variable cannot in general be represented using only an integral with respect to $B$. The theorem below is specifically about the Brownian filtration.

The Brownian Martingale Representation Theorem

Theorem: If $X$ is $\mathcal{F}_T$-measurable and $\mathbb{E}[X^2]<\infty$, then there exists $H\in\mathcal{L}^2$ such that, for every $t\in[0,T]$, $$ \mathbb{E}[X\mid\mathcal{F}_t]=\mathbb{E}[X]+\int_0^t H_s\,dB_s. $$ The process $H$ is unique up to equality almost everywhere for $ds\,d\mathbb{P}$.

Proof. Write $\mathcal{I}$ for the set of terminal Ito integrals $\int_0^T H_s\,dB_s$, where $H\in\mathcal{L}^2$. The Ito isometry gives $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. Consequently, $\mathcal{I}$ is a closed subspace of the mean-zero part of $L^2(\mathcal{F}_T)$: if terminal integrals converge in $L^2$, their integrands are Cauchy in $\mathcal{L}^2$, and the isometry sends their limit to the limiting terminal variable. The same isometry proves uniqueness of an integrand.

It remains to show that every mean-zero $X\in L^2(\mathcal{F}_T)$ belongs to $\mathcal{I}$. Suppose $Y\in L^2(\mathcal{F}_T)$ is orthogonal both to constants and to $\mathcal{I}$. For any deterministic step function $u$ on $[0,T]$, define $Z_t=\exp(i\int_0^t u(s)\,dB_s+\tfrac12\int_0^t u(s)^2\,ds)$. This complex-valued process satisfies $Z_0=1$ and, by Ito’s formula, $dZ_t=i u(t)Z_t\,dB_t$. Its integrand is square-integrable: $|Z_t|=\exp(\tfrac12\int_0^t u(s)^2\,ds)$ is bounded by a deterministic constant on $[0,T]$, and $u$ is bounded. Applying the Ito isometry to real and imaginary parts shows that $Z_T-1$ lies in the complexification of $\mathcal{I}$. Thus $\mathbb{E}[Y\overline{Z_T}]=0$.

As $u$ ranges over step functions, the random variables $\int_0^T u(s)\,dB_s$ include every real linear combination of increments over a finite time partition. Therefore $Y$ is orthogonal to the complex exponentials of all finite vectors of Brownian increments. To see why this implies $Y=0$, fix such a vector $V$ and define the finite signed measure $\nu(A)=\mathbb{E}[Y\mathbf{1}_{\{V\in A\}}]$. Its Fourier transform is zero at every vector of frequencies. By the standard uniqueness theorem for Fourier transforms of finite measures, $\nu=0$; the proof of that general theorem is beyond this tutorial. Hence $\mathbb{E}[Y\mid V]=0$. Brownian continuity means that the Brownian path up to $T$ is generated by its values at rational times together with $B_T$. The standard density of cylinder functions, or equivalently the Hilbert-space convergence of conditional expectations onto these increasing finite-coordinate sigma-fields, now gives $Y=0$ in $L^2$.

We have proved that the orthogonal complement of the constants plus $\mathcal{I}$ is $\{0\}$. Since that sum is closed, it is all of $L^2(\mathcal{F}_T)$. Thus $X=\mathbb{E}[X]+\int_0^T H_s\,dB_s$ for some $H\in\mathcal{L}^2$. Taking conditional expectations at time $t$, the Ito integral after $t$ has conditional expectation zero: it is a square-integrable martingale, as follows from the Ito integral construction and its conditional-increment property. This gives the asserted formula. $\square$

The theorem’s proof also explains the importance of the square-integrability condition. It puts the claim and the stochastic integral in a Hilbert space, where the Ito isometry makes the set of terminal integrals closed. For a general integrable claim that is not square-integrable, a representation may require additional localization and integrability arguments; that case is not asserted here.

Worked Representations

Worked Example: Representing the Brownian Terminal Value

Let $X=B_T$. Since $\mathbb{E}[B_T]=0$ and $\mathbb{E}[B_T^2]=T<\infty$, the claim is square-integrable. For every $t\le T$, Brownian independent increments give $\mathbb{E}[B_T\mid\mathcal{F}_t]=B_t+\mathbb{E}[B_T-B_t\mid\mathcal{F}_t]=B_t$. Because $B_t=\int_0^t 1\,dB_s$, the representation integrand is $H_s=1$.

Check: at $t=0$ the conditional value is $0$, and at $t=T$ it is $B_T$. Also $\mathbb{E}\int_0^T 1^2\,ds=T$, so the integrand belongs to $\mathcal{L}^2$.

Worked Example: Representing the Squared Brownian Terminal Value

Let $X=B_T^2$. Brownian increments are independent of $\mathcal{F}_t$, have mean zero, and have variance $T-t$. Expanding $B_T=B_t+(B_T-B_t)$ therefore gives $\mathbb{E}[B_T^2\mid\mathcal{F}_t]=B_t^2+T-t$. Apply Ito’s formula to $f(x)=x^2$, whose derivatives are $f'(x)=2x$ and $f''(x)=2$: $$ d(B_t^2)=2B_t\,dB_t+\tfrac12(2)\,dt=2B_t\,dB_t+dt. $$ Subtracting the differential of $t$ yields $d(B_t^2-t)=2B_t\,dB_t$. Hence $\mathbb{E}[B_T^2\mid\mathcal{F}_t]=T+\int_0^t 2B_s\,dB_s$, so $H_s=2B_s$.

The integrability condition holds because $\mathbb{E}\int_0^T(2B_s)^2\,ds=4\int_0^T s\,ds=2T^2<\infty$. Check: at $t=T$, Ito’s identity gives $B_T^2=T+\int_0^T2B_s\,dB_s$, and the conditional value at $t=0$ is $T=\mathbb{E}[B_T^2]$.

Worked Example: Representing an Exponential Claim

Fix $\lambda\in\mathbb{R}$ and take $X=\exp(\lambda B_T-\tfrac12\lambda^2T)$. The normal moment-generating function gives $\mathbb{E}[X]=1$ and $\mathbb{E}[X^2]=e^{\lambda^2T}<\infty$. Independence of $B_T-B_t$ from $\mathcal{F}_t$ gives \[ M_t:=\mathbb{E}[X\mid\mathcal{F}_t] =\exp(\lambda B_t-\tfrac12\lambda^2t). \]

For $f(t,x)=\exp(\lambda x-\tfrac12\lambda^2t)$, the derivatives are $f_t=-\tfrac12\lambda^2f$, $f_x=\lambda f$, and $f_{xx}=\lambda^2f$. Ito’s formula, including its second-derivative correction, gives $$ dM_t=f_t\,dt+f_x\,dB_t+\tfrac12 f_{xx}\,dt =\lambda M_t\,dB_t. $$ Thus $H_t=\lambda M_t$. This integrand is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^T\lambda^2M_t^2\,dt=\int_0^T\lambda^2e^{\lambda^2t}\,dt=e^{\lambda^2T}-1$ when $\lambda\ne0$, and equals $0$ when $\lambda=0$. The same formula $e^{\lambda^2T}-1$ is $0$ at $\lambda=0$.

Check: integrating the differential gives $X=1+\int_0^T\lambda M_t\,dB_t$. The integral has mean zero by square-integrability, so the representation gives $\mathbb{E}[X]=1$, consistent with the direct normal-moment calculation. When $\lambda=0$, $X=1$ and the integrand vanishes.

How a Conditional Claim Determines Its Integrand

For many claims, the most efficient route is to calculate $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$ first, then apply Ito’s formula. If $M_t=v(t,B_t)$ for a sufficiently smooth function $v$, Ito’s formula gives diffusion coefficient $v_x(t,B_t)$. When $v$ also satisfies the backward heat equation $v_t+\tfrac12v_{xx}=0$, the drift vanishes, and the representation integrand is $H_t=v_x(t,B_t)$. The conditional-expectation identity must still be established; a formal PDE calculation alone does not identify the correct terminal claim.

Worked Example: Representing a Sine Claim

Let $X=\sin(\lambda B_T)$, with $\lambda\in\mathbb{R}$. It is bounded, so it is square-integrable. The conditional normal distribution of $B_T-B_t$ gives $M_t=\mathbb{E}[\sin(\lambda B_T)\mid\mathcal{F}_t]=e^{-\lambda^2(T-t)/2}\sin(\lambda B_t)$. For $f(t,x)=e^{-\lambda^2(T-t)/2}\sin(\lambda x)$, its derivatives are $f_t=\tfrac12\lambda^2f$, $f_x=\lambda e^{-\lambda^2(T-t)/2}\cos(\lambda x)$, and $f_{xx}=-\lambda^2f$. Therefore Ito’s formula gives $$ dM_t=\left(f_t+\tfrac12f_{xx}\right)dt+f_x\,dB_t =\lambda e^{-\lambda^2(T-t)/2}\cos(\lambda B_t)\,dB_t. $$ The integrand is progressively measurable and bounded in absolute value by $|\lambda|$, so its squared time integral has expectation at most $\lambda^2T$. Thus it belongs to $\mathcal{L}^2$.

Check: at $t=T$, $M_T=\sin(\lambda B_T)=X$. At $t=0$, $M_0=0=\mathbb{E}[\sin(\lambda B_T)]$, since the centered normal distribution is symmetric. For $\lambda=0$, both the claim and integrand are identically zero.

Interpretation as a Hedge

The integrand is the amount of Brownian risk carried by the conditional claim value. To express this as a stock position, suppose a stock follows $dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}$ under a risk-neutral measure $\mathbb{Q}$, with constant $r$, nonzero constant $\sigma$, and Brownian motion $B_t^{\mathbb{Q}}$ under $\mathbb{Q}$. Write $\widetilde S_t=e^{-rt}S_t$. Then $d\widetilde S_t=\sigma\widetilde S_t\,dB_t^{\mathbb{Q}}$. If a discounted terminal payoff $X$ is square-integrable under $\mathbb{Q}$, and the filtration is the augmented natural filtration of $B^{\mathbb{Q}}$, its representation is $\mathbb{E}^{\mathbb{Q}}[X\mid\mathcal{F}_t]=\mathbb{E}^{\mathbb{Q}}[X]+\int_0^t H_s\,dB_s^{\mathbb{Q}}$. A position $\Delta_t=H_t/(\sigma\widetilde S_t)$ in the discounted stock has stochastic gain $\Delta_t\,d\widetilde S_t=H_t\,dB_t^{\mathbb{Q}}$. Thus the representation integrand determines the hedge in Brownian-risk units.

This interpretation distinguishes the probability measure from the pricing measure: a claim represented under $\mathbb{P}$ uses conditional expectations under $\mathbb{P}$, whereas the preceding pricing statement uses $\mathbb{Q}$. The risk-neutral conclusion also depends on the stated stock dynamics and filtration; it is not a claim that every market or every additional source of randomness can be hedged with one stock.

A Common Mistake

A frequent error is to differentiate $B_t^2$ by the ordinary chain rule and omit the quadratic-variation correction. That calculation would give $d(B_t^2)=2B_t\,dB_t$, and would suggest $B_T^2=\int_0^T2B_t\,dB_t$. This is wrong: the right-hand side has expectation zero, while $\mathbb{E}[B_T^2]=T$. Ito’s formula includes $\tfrac12 f''\,dt$, so the correct identity is $B_T^2=T+\int_0^T2B_t\,dB_t$. The missing deterministic term is exactly the difference between the Ito calculation and the ordinary chain rule.

Takeaway: For a square-integrable Brownian terminal claim, first form its conditional-expectation martingale. The theorem guarantees a unique Ito integrand, and an explicit conditional value often reveals that integrand as its spatial derivative.

Check Your Understanding

For each question, specify the filtration and integrability condition needed for the representation being used.

For $X=B_T^2$, compute $\mathbb{E}[X]$ and identify its representation integrand.
Why does the representation theorem require the augmented natural Brownian filtration in its stated form?
For $X=\exp(\lambda B_T-\tfrac12\lambda^2T)$, verify its second moment and state the representation integrand, including the case $\lambda=0$.
If $M_t=v(t,B_t)$, which differential equation removes the drift term in Ito’s formula, and what is the resulting integrand?
In the risk-neutral stock model above, express the stock position in terms of $H_t$, $\sigma$, and $\widetilde S_t$.