Martingale Representation Theorem¶

In the unifying framework of this section, the martingale representation theorem represents full control — every source of randomness can be expressed as a stochastic integral, which means every contingent claim can be hedged.

The Martingale Representation Theorem (MRT) states that, under a Brownian filtration, every square-integrable martingale can be expressed as a stochastic integral with respect to the driving Brownian motion.

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 \(M_t\) is square-integrable on \([0,T]\) if:

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

Equivalently (by the martingale property and Doob's \(L^2\) inequality): \(\sup_{t \leq T} \mathbb{E}[M_t^2] < \infty\).

Recall: Itô Integrals Are Martingales¶

Before stating the main theorem, recall the converse direction (proved in Itô Integral Construction):

Itô Integrals Are Martingales

If \(\phi_t\) is a predictable process with \(\mathbb{E}\left[\int_0^T \phi_s^2 \, ds\right] < \infty\), then the Itô integral:

\[ I_t = \int_0^t \phi_s \, dW_s \]

is a square-integrable martingale with \(I_0 = 0\).

This tells us that stochastic integrals produce martingales. The Martingale Representation Theorem is the converse: every 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

PRP holds if and only if the market is complete. Every contingent claim admits a replicating strategy precisely when the driving filtration supports no orthogonal martingale components.

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: I Equals All Mean-Zero Square-Integrable Martingales¶

The key step is showing that \(\mathcal{I}\) contains all mean-zero square-integrable \(\mathcal{F}_T\)-measurable random variables.

Idea: If there existed a square-integrable random variable \(X\) orthogonal to \(\mathcal{I}\), then \(\mathbb{E}[X \cdot \int_0^T \phi_s\,dW_s] = 0\) for all \(\phi\). Taking \(\phi_s = \mathbb{E}[X \mid \mathcal{F}_s]\), one can show \(X\) must be constant. Since \(\mathcal{F}_T\) is generated by \(W\), no non-trivial orthogonal complement exists.

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 and Black–Scholes¶

In a complete market, the MRT provides the mathematical foundation for hedging.

Setup¶

Under the risk-neutral measure \(\mathbb{Q}\), the discounted stock price \(\tilde{S}_t = e^{-rt}S_t\) is a martingale. For a European claim with payoff \(\Phi(S_T)\), the discounted claim value:

\[ \tilde{V}_t = e^{-rt}V_t = \mathbb{E}^{\mathbb{Q}}[e^{-rT}\Phi(S_T) \mid \mathcal{F}_t] \]

is also a martingale.

Applying MRT¶

By the Martingale Representation Theorem:

\[ \tilde{V}_t = \tilde{V}_0 + \int_0^t \psi_s \, dW_s^{\mathbb{Q}} \]

for some predictable process \(\psi_t\).

Identifying the Hedge¶

Since \(d\tilde{S}_t = \sigma \tilde{S}_t \, dW_t^{\mathbb{Q}}\) (under \(\mathbb{Q}\)), we can write:

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

Substituting:

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

The hedging strategy (number of shares) is:

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

Connection to Delta¶

For the Black–Scholes model, one can show:

\[ \Delta_t = \frac{\partial V}{\partial S}(t, S_t) \]

The MRT integrand \(\psi_t\) encodes the sensitivity of the option value to Brownian shocks, which translates to delta hedging.

This is the mathematical statement of market completeness: every contingent claim can be replicated by dynamic trading in the underlying asset.

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 Ito'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 Ito 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 \(V_t = S_t\Phi(d_1) - Ke^{-r(T-t)}\Phi(d_2)\), 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:

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

Dimensional check: \(\psi_t\) has the dimension of "dollars" (same as \(\tilde{V}_t\), since \(dW_t\) is dimensionless). Indeed, \(\Delta_t = \Phi(d_1)\) 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.