Martingale Representation Theorem¶

On a Brownian filtration the only randomness available is the path of $W$. So if $M_t$ is a martingale -- a process whose updates are pure surprise -- those surprises must themselves come from increments of $W$. Reassembling the surprises gives a predictable integrand $\phi$ such that $M_t = M_0 + \int_0^t \phi_s\,dW_s$. This is the Martingale Representation Theorem (MRT): every square-integrable Brownian martingale is a stochastic integral against $W$. In the unifying framework it is full control -- the algebraic engine of hedging and the bridge between expectations and PDEs.

Rather than repeating basic definitions already covered in earlier chapters, this section focuses on:

The precise statement of the theorem
Its structural meaning
Why it is fundamental for hedging and PDE connections
How it fits into change-of-measure theory

Prerequisites

This section assumes familiarity with:

Statement of the Theorem¶

Let $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{0 \leq t \leq T}, \mathbb{P})$ be a filtered probability space supporting a standard Brownian motion $W_t$, and assume $\{\mathcal{F}_t\}$ is the augmented Brownian filtration—that is, $\mathcal{F}_t = \sigma(W_s : s \leq t) \vee \mathcal{N}$, where $\mathcal{N}$ contains all $\mathbb{P}$-null sets.

Definition: Square-Integrable Martingale¶

A martingale $M = (M_t)_{0 \leq t \leq T}$ is square-integrable on $[0,T]$ if it belongs to $L^2$ at every time:

\[ \forall\, t \leq T, \quad \mathbb{E}[M_t^2] < \infty \]

For martingales on a finite horizon, this is equivalent to the single terminal condition

\[ \mathbb{E}[M_T^2] < \infty \]

which is the form most commonly used in textbooks. The equivalence follows from the martingale property: $M_t = \mathbb{E}[M_T \mid \mathcal{F}_t]$, so by Jensen's inequality $\mathbb{E}[M_t^2] \leq \mathbb{E}[M_T^2]$ for all $t \leq T$. In fact, Doob's $L^2$ inequality gives the stronger uniform bound $ \mathbb{E}[\sup_{t \leq T} M_t^2] \leq 4\,\mathbb{E}[M_T^2]$.

Recall (see § Itô Integral Properties): if $\phi$ is predictable with $\mathbb{E}[\int_0^T \phi_s^2\,ds] < \infty$, then $I_t = \int_0^t \phi_s\,dW_s$ is a square-integrable martingale. The Martingale Representation Theorem is the converse: every such martingale arises this way.

Martingale Representation Theorem¶

Martingale Representation Theorem

Let $M_t$ be a square-integrable $\mathcal{F}_t$-martingale on $[0,T]$. Then there exists a unique predictable process $\phi_t$ such that:

\[ \boxed{M_t = M_0 + \int_0^t \phi_s \, dW_s \quad \text{for all } t \in [0,T]} \]

with

\[ \mathbb{E}\left[ \int_0^T \phi_s^2 \, ds \right] < \infty \]

Uniqueness is understood up to indistinguishability (i.e., $\phi$ is unique in $L^2(\Omega \times [0,T])$).

Why This Is a Strong Result¶

This theorem asserts that:

Brownian motion is the only source of randomness in its own filtration
Any martingale adapted to this filtration must be built entirely from Brownian increments
There are no "hidden" or orthogonal martingale components

This property is often called the predictable representation property (PRP).

PRP and market completeness

In the standard Brownian setting, the predictable representation property corresponds to market completeness: every contingent claim measurable with respect to the filtration can be replicated by trading in the underlying assets.

Brownian filtration is essential

The MRT relies on the filtration being generated by Brownian motion. If other sources of randomness are present (jump processes, additional noise), orthogonal martingale components exist and PRP fails — the market becomes incomplete, and not every claim can be hedged.

Intuition

If you observe only Brownian motion, then every "fair game" (martingale) you can construct must come from betting on Brownian increments. There's no other source of uncertainty to bet on.

Proof Sketch¶

The proof relies on Hilbert space ideas and the Itô isometry.

Step 1: The Space of Stochastic Integrals¶

Define the space of stochastic integrals:

\[ \mathcal{I} = \left\{ \int_0^T \phi_s \, dW_s : \phi \text{ predictable}, \, \mathbb{E}\left[\int_0^T \phi_s^2\,ds\right] < \infty \right\} \]

By the Itô isometry, this is isometrically isomorphic to $L^2(\Omega \times [0,T], \mathcal{P}, \mathbb{P} \otimes dt)$:

\[ \mathbb{E}\left[\left(\int_0^T \phi_s \, dW_s\right)^2\right] = \mathbb{E}\left[\int_0^T \phi_s^2 \, ds\right] \]

Step 2: Density of Simple Processes¶

Simple predictable processes (piecewise constant, adapted) are dense in $L^2(\Omega \times [0,T])$. Therefore, stochastic integrals of simple processes are dense in $\mathcal{I}$.

Step 3: I Is Closed¶

Since the Itô isometry preserves the $L^2$ norm, $\mathcal{I}$ is a closed subspace of $L^2(\Omega, \mathcal{F}_T, \mathbb{P})$.

Step 4: Identification of the Space¶

Using density arguments (e.g., via simple predictable processes) and the fact that the Brownian filtration is generated by $W$, one shows that the space of stochastic integrals is dense in the space of mean-zero square-integrable random variables. Since this space is also closed (by the Itô isometry), it must coincide with the entire space. Therefore, every square-integrable random variable admits a representation as a stochastic integral.

Step 5: Extend to Martingales¶

For a square-integrable martingale $M_t$ with terminal value $M_T$, we have $M_T - M_0 \in \mathcal{I}$, so:

\[ M_T - M_0 = \int_0^T \phi_s\,dW_s \]

By the martingale property, $M_t = \mathbb{E}[M_T \mid \mathcal{F}_t] = M_0 + \int_0^t \phi_s\,dW_s$. $\square$

Reference: For the complete proof, see Karatzas–Shreve, Brownian Motion and Stochastic Calculus, Theorem 3.4.15, or Revuz–Yor, Continuous Martingales and Brownian Motion, Theorem V.3.4.

Interpretation: Geometry of Martingales¶

The theorem reveals a beautiful geometric structure:

The space of square-integrable martingales (with $M_0 = 0$) is a Hilbert space
This space is isometrically isomorphic to $L^2(\Omega \times [0,T])$
Each martingale corresponds to a unique "direction" $\phi_t$ along which Brownian noise is accumulated
The Itô isometry is the inner product preserving map

This is why stochastic calculus has such strong parallels with Fourier analysis and functional analysis.

Connection to Hedging¶

The MRT supplies the integrand $\psi_t$ in $\tilde{V}_t = \tilde{V}_0 + \int_0^t \psi_s\,dW_s^{\mathbb{Q}}$ for the discounted claim value; matching against $d\tilde{S}_t = \sigma \tilde{S}_t\,dW_t^{\mathbb{Q}}$ identifies the share holding $\Delta_t = \psi_t/(\sigma \tilde{S}_t)$. The MRT thus delivers existence and uniqueness of the hedge; the full pricing/hedging construction is developed in § Risk-Neutral Construction and § Risk-Neutral Valuation Principle.

Relation to Change of Measure¶

The Martingale Representation Theorem and Girsanov's Theorem play complementary roles:

Theorem	What It Does
Girsanov	Changes the measure; explains how drift of $W$ changes under $\mathbb{Q}$
MRT	Works within a fixed measure; represents all martingales as stochastic integrals

Together, they provide:

Existence of an equivalent martingale measure (Girsanov)
Uniqueness of the replicating/hedging strategy (MRT)
Completeness of the market (combined)

Multi-Dimensional Extension¶

For a $d$-dimensional Brownian motion $W_t = (W_t^1, \ldots, W_t^d)$ with independent components:

Multi-Dimensional MRT

Every square-integrable $\mathcal{F}_t^W$-martingale $M_t$ can be written as:

\[ M_t = M_0 + \sum_{i=1}^d \int_0^t \phi_s^i \, dW_s^i = M_0 + \int_0^t \boldsymbol{\phi}_s \cdot d\mathbf{W}_s \]

where $\boldsymbol{\phi}_t = (\phi_t^1, \ldots, \phi_t^d)$ is a predictable $\mathbb{R}^d$-valued process.

This extends to markets with multiple sources of risk (multiple stocks, stochastic volatility, etc.).

Beyond the Basic Setting¶

Clark–Ocone Formula¶

In Malliavin calculus, the integrand $\phi_t$ can be written explicitly using conditional expectations of Malliavin derivatives.

Clark–Ocone Formula

If $F \in \mathbb{D}^{1,2}$ (the Malliavin–Sobolev space), then:

\[ F = \mathbb{E}[F] + \int_0^T \mathbb{E}[D_t F \mid \mathcal{F}_t] \, dW_t \]

where $D_t F$ is the Malliavin derivative of $F$ at time $t$.

Application: For $F = \Phi(S_T)$, the Clark–Ocone formula can sometimes give explicit expressions for the hedging strategy $\phi_t = \mathbb{E}[D_t \Phi(S_T) \mid \mathcal{F}_t]$.

Kunita–Watanabe Decomposition (Incomplete Markets)¶

When the filtration is larger than the Brownian filtration (e.g., jump processes, stochastic volatility with unhedgeable risk), the MRT fails. Instead, we have:

Kunita–Watanabe Decomposition

Let $M_t$ be a square-integrable martingale and $N_t$ a given martingale (e.g., a traded asset). Then:

\[ M_t = M_0 + \int_0^t \phi_s \, dN_s + L_t \]

where:

$\int_0^t \phi_s \, dN_s$ is the hedgeable part
$L_t$ is a martingale orthogonal to $N$: $\langle L, N \rangle_t = 0$

The orthogonal component $L_t$ represents unhedgeable risk. This decomposition is fundamental for:

Variance-optimal hedging
Mean-variance portfolio theory
Incomplete market pricing

Summary¶

\[ \boxed{M_t = M_0 + \int_0^t \phi_s \, dW_s} \]

Aspect	Description
What it says	Every square-integrable martingale is a stochastic integral
Key requirement	Filtration must be generated by Brownian motion (augmented)
Uniqueness	The integrand $\phi_t$ is unique (in $L^2$)
Financial meaning	Hedging strategies exist and are unique (market completeness)
Proof method	Hilbert space theory + Itô isometry
Extensions	Clark–Ocone (explicit $\phi$), Kunita–Watanabe (incomplete markets)

Key Takeaway

The Martingale Representation Theorem formalizes the idea that Brownian motion generates all randomness in its filtration. In finance, this translates to market completeness: every derivative can be perfectly hedged, and the hedging strategy is unique. When the theorem fails (incomplete markets), unhedgeable risk remains, and pricing/hedging become more complex.

Exercises¶

Exercise 1. Let $M_t = W_t^2 - t$, where $W_t$ is a standard Brownian motion. Verify that $M_t$ is a martingale. Then find the predictable process $\phi_t$ such that $M_t = \int_0^t \phi_s\,dW_s$. (Hint: apply Itô's formula to $W_t^2$.)

Solution to Exercise 1

First, verify $M_t = W_t^2 - t$ is a martingale. By Itô's formula applied to $f(x) = x^2$ and $X_t = W_t$:

\[ d(W_t^2) = 2W_t\,dW_t + dt \]

Therefore $dM_t = d(W_t^2) - dt = 2W_t\,dW_t$. Since $M_t$ has no $dt$ term and is a stochastic integral, it is a local martingale. Moreover, $\mathbb{E}[M_t^2] = \mathbb{E}[(W_t^2 - t)^2] < \infty$, so it is a true (square-integrable) martingale.

Integrating:

\[ M_t = M_0 + \int_0^t 2W_s\,dW_s = 0 + \int_0^t 2W_s\,dW_s \]

Hence the predictable process is $\phi_t = 2W_t$.

Exercise 2. In the Black–Scholes model, the discounted option price $\tilde{V}_t = e^{-rt}V_t$ is a $\mathbb{Q}$-martingale. By the MRT, $\tilde{V}_t = \tilde{V}_0 + \int_0^t \psi_s\,dW_s^{\mathbb{Q}}$. Explain how the hedging strategy $\Delta_t = \psi_t / (\sigma S_t e^{-rt})$ is derived from the MRT integrand, and why the uniqueness of $\psi_t$ implies the uniqueness of the replicating portfolio.

Solution to Exercise 2

By the MRT, the discounted option price satisfies $\tilde{V}_t = \tilde{V}_0 + \int_0^t \psi_s\,dW_s^{\mathbb{Q}}$. Under $\mathbb{Q}$, the discounted stock satisfies $d\tilde{S}_t = \sigma \tilde{S}_t\,dW_t^{\mathbb{Q}}$, so:

\[ dW_t^{\mathbb{Q}} = \frac{d\tilde{S}_t}{\sigma \tilde{S}_t} \]

Substituting into the MRT representation:

\[ d\tilde{V}_t = \psi_t\,dW_t^{\mathbb{Q}} = \frac{\psi_t}{\sigma \tilde{S}_t}\,d\tilde{S}_t \]

For a self-financing portfolio holding $\Delta_t$ shares, $d\tilde{V}_t = \Delta_t\,d\tilde{S}_t$. Comparing coefficients:

\[ \Delta_t = \frac{\psi_t}{\sigma \tilde{S}_t} = \frac{\psi_t}{\sigma S_t e^{-rt}} \]

The uniqueness of $\psi_t$ in the MRT (unique in $L^2$) directly implies the uniqueness of $\Delta_t$. Since the hedging strategy is uniquely determined by the MRT integrand, the replicating portfolio is unique. This is the mathematical content of market completeness: every claim has exactly one hedging strategy.

Exercise 3. Consider a market with a stock $S_t$ and an independent Poisson process $N_t$. Explain why the Martingale Representation Theorem fails in this setting. What does the Kunita–Watanabe decomposition tell us about the hedging problem for a claim that depends on $N_T$?

Solution to Exercise 3

The MRT requires the filtration to be generated by Brownian motion (augmented). When we add an independent Poisson process $N_t$, the filtration $\mathcal{F}_t = \sigma(W_s, N_s : s \leq t)$ is strictly larger than the Brownian filtration. There exist $\mathcal{F}_t$-martingales that cannot be written as stochastic integrals with respect to $W$ alone — for example, the compensated Poisson process $\tilde{N}_t = N_t - \lambda t$ is a martingale orthogonal to all Brownian stochastic integrals.

The Kunita–Watanabe decomposition gives:

\[ M_t = M_0 + \int_0^t \phi_s\,dS_s + L_t \]

where $\int_0^t \phi_s\,dS_s$ is the hedgeable part (correlated with the stock) and $L_t$ is a martingale orthogonal to $S$, with $\langle L, S \rangle = 0$. For a claim depending on $N_T$, the component $L_t$ captures the unhedgeable risk — the part of the claim's randomness that comes from the Poisson process and cannot be replicated by trading the stock. Perfect hedging is impossible, and variance-optimal or other hedging criteria must be used.

Exercise 4. Prove that the Itô isometry map $\phi \mapsto \int_0^T \phi_s\,dW_s$ is an isometry from $L^2(\Omega \times [0,T])$ to $L^2(\Omega)$. Explain why this means $\mathcal{I}$ (the space of stochastic integrals) is a closed subspace of $L^2(\Omega, \mathcal{F}_T, \mathbb{P})$.

Solution to Exercise 4

The Itô isometry states:

\[ \mathbb{E}\left[\left(\int_0^T \phi_s\,dW_s\right)^2\right] = \mathbb{E}\left[\int_0^T \phi_s^2\,ds\right] = \|\phi\|^2_{L^2(\Omega \times [0,T])} \]

Define the map $\Psi: L^2(\Omega \times [0,T]) \to L^2(\Omega, \mathcal{F}_T, \mathbb{P})$ by $\Psi(\phi) = \int_0^T \phi_s\,dW_s$. The Itô isometry says $\|\Psi(\phi)\|_{L^2(\Omega)} = \|\phi\|_{L^2(\Omega \times [0,T])}$, so $\Psi$ preserves norms — it is an isometry.

To show $\mathcal{I} = \text{Range}(\Psi)$ is closed: let $(I_n)$ be a Cauchy sequence in $\mathcal{I}$ with $I_n = \Psi(\phi_n)$. Since $\Psi$ is an isometry:

\[ \|I_n - I_m\|_{L^2(\Omega)} = \|\phi_n - \phi_m\|_{L^2(\Omega \times [0,T])} \]

So $(\phi_n)$ is Cauchy in $L^2(\Omega \times [0,T])$, which is complete, so $\phi_n \to \phi$ for some $\phi \in L^2(\Omega \times [0,T])$. Then $I_n \to \Psi(\phi) \in \mathcal{I}$. Hence $\mathcal{I}$ is a closed subspace — the image of a complete space under an isometry is closed.

Exercise 5. The Clark–Ocone formula states $F = \mathbb{E}[F] + \int_0^T \mathbb{E}[D_t F | \mathcal{F}_t]\,dW_t$ for $F \in \mathbb{D}^{1,2}$. For $F = W_T^3$, use the Malliavin derivative $D_t(W_T^3) = 3W_T^2$ to write the explicit representation. Verify your answer by checking that $\mathbb{E}[(\int_0^T \phi_s\,dW_s)^2] = \mathbb{E}[F^2] - (\mathbb{E}[F])^2$.

Solution to Exercise 5

For $F = W_T^3$, the Malliavin derivative is $D_t(W_T^3) = 3W_T^2$ for $t \leq T$. The Clark–Ocone formula gives:

\[ W_T^3 = \mathbb{E}[W_T^3] + \int_0^T \mathbb{E}[3W_T^2 \mid \mathcal{F}_t]\,dW_t \]

Since $W_T^3$ is an odd function of the symmetric random variable $W_T$, $\mathbb{E}[W_T^3] = 0$.

Now compute $\mathbb{E}[W_T^2 \mid \mathcal{F}_t]$. Write $W_T = W_t + (W_T - W_t)$, where $W_T - W_t$ is independent of $\mathcal{F}_t$ with distribution $N(0, T-t)$:

\[ \mathbb{E}[W_T^2 \mid \mathcal{F}_t] = \mathbb{E}[(W_t + (W_T - W_t))^2 \mid \mathcal{F}_t] = W_t^2 + (T - t) \]

So $\phi_t = 3(W_t^2 + T - t)$ and:

\[ W_T^3 = \int_0^T 3(W_t^2 + T - t)\,dW_t \]

Verification: By the Itô isometry:

\[ \mathbb{E}\left[\left(\int_0^T \phi_t\,dW_t\right)^2\right] = \mathbb{E}\left[\int_0^T 9(W_t^2 + T - t)^2\,dt\right] \]

This should equal $\mathbb{E}[(W_T^3)^2] - (\mathbb{E}[W_T^3])^2 = \mathbb{E}[W_T^6] - 0 = 15T^3$ (using the sixth moment of a normal). One can verify by expanding $(W_t^2 + T - t)^2$ and integrating that this equals $15T^3$.

Exercise 6. In a two-dimensional Brownian filtration with $\mathbf{W}_t = (W_t^1, W_t^2)$, a square-integrable martingale $M_t$ has representation $M_t = M_0 + \int_0^t \phi_s^1\,dW_s^1 + \int_0^t \phi_s^2\,dW_s^2$. If only one stock is traded (driven by $W_t^1$), explain why the component $\int_0^t \phi_s^2\,dW_s^2$ represents unhedgeable risk and relate this to market incompleteness.

Solution to Exercise 6

The martingale $M_t$ decomposes as:

\[ M_t = M_0 + \underbrace{\int_0^t \phi_s^1\,dW_s^1}_{\text{hedgeable}} + \underbrace{\int_0^t \phi_s^2\,dW_s^2}_{\text{unhedgeable}} \]

Since only the stock (driven by $W^1$) is traded, a self-financing portfolio can only generate gains of the form $\int_0^t \Delta_s\,dS_s$, which involves only $dW_s^1$. The component $\int_0^t \phi_s^2\,dW_s^2$ is driven by $W^2$, which is independent of $W^1$, so it is orthogonal to all trading gains: $\langle \int \phi^2\,dW^2, \int \Delta\,dS \rangle = 0$.

This means no trading strategy in the stock can replicate or reduce the risk from $W^2$. The variance of the hedging error is at least:

\[ \mathbb{E}\left[\left(\int_0^T \phi_s^2\,dW_s^2\right)^2\right] = \mathbb{E}\left[\int_0^T (\phi_s^2)^2\,ds\right] > 0 \]

whenever $\phi^2 \neq 0$. This is market incompleteness: one traded asset cannot span two independent sources of risk. Examples include stochastic volatility models where $W^2$ drives the volatility process.

Exercise 7. Consider a European call option in the Black–Scholes model. The delta hedge is $\Delta_t = \Phi(d_1(t, S_t))$. Identify the MRT integrand $\psi_t$ in terms of the Black–Scholes Greeks and model parameters. Verify dimensionally that $\Delta_t = \psi_t / (\sigma \tilde{S}_t)$ is consistent.

Solution to Exercise 7

The Black–Scholes call price is $V_t = S_t\mathcal{N}(d_1) - Ke^{-r(T-t)}\mathcal{N}(d_2)$, and the delta is $\Delta_t = \Phi(d_1(t, S_t))$.

The discounted option price satisfies:

\[ d\tilde{V}_t = \psi_t\,dW_t^{\mathbb{Q}} \]

Since $d\tilde{V}_t = \Delta_t\,d\tilde{S}_t = \Delta_t \sigma \tilde{S}_t\,dW_t^{\mathbb{Q}}$, we identify:

\[ \psi_t = \Delta_t \cdot \sigma \tilde{S}_t = \mathcal{N}(d_1) \cdot \sigma S_t e^{-rt} \]

Dimensional check: $\psi_t$ has the dimension of "dollars" (same as $\tilde{V}_t$, since $dW_t$ is dimensionless). Indeed, $\Delta_t = \mathcal{N}(d_1)$ is dimensionless, $\sigma$ is dimensionless (per $\sqrt{\text{time}}$), and $\tilde{S}_t = S_t e^{-rt}$ has dimension of dollars. So $\psi_t = \Delta_t \cdot \sigma \tilde{S}_t$ has dimension of dollars, consistent with the MRT representation.

Inversely, $\Delta_t = \psi_t / (\sigma \tilde{S}_t)$ is dimensionless (shares), consistent with it being the number of shares in the hedging portfolio.

Exercise 8. You are in a filtration generated by a Brownian motion $W_t$ and an independent Poisson process $N_t$. Only the Brownian-driven stock is tradable. A candidate says: "MRT still holds because Brownian motion is present." Explain why this is wrong and identify the financial consequence.

Solution to Exercise 8

MRT in its standard form requires the filtration to be generated by Brownian motion alone (augmented). Here the filtration $\mathcal{F}_t = \sigma(W_s, N_s : s \leq t)$ is strictly larger than the Brownian filtration, because the Poisson process contributes additional information.

The compensated Poisson process $\tilde{N}_t = N_t - \lambda t$ is a martingale in this filtration that is orthogonal to all Brownian stochastic integrals: $\langle \tilde{N}, \int \phi\,dW \rangle = 0$. This means there exist $\mathcal{F}_t$-martingales that cannot be represented as stochastic integrals with respect to $W$ alone.

The financial consequence is market incompleteness: claims depending on the Poisson component (e.g., jump risk) cannot be perfectly hedged by trading the Brownian-driven stock. The Kunita–Watanabe decomposition applies instead, splitting any claim into a hedgeable Brownian part and an orthogonal unhedgeable residual.