Martingale Representation Theorem and Complete Markets

What the Theorem Guarantees

The martingale representation theorem says that every square-integrable martingale in a Brownian filtration can be written as a stochastic integral against the Brownian motion. In a suitably non-degenerate Brownian market, this representation can be translated into a self-financing trading strategy.

The qualifications matter. The theorem does not say that every claim in every market can be hedged. The claim must be measurable with respect to the model's Brownian information, satisfy the chosen integrability condition, and the traded assets must span the Brownian shocks. Market frictions are absent.

This is the last rung in a useful pricing ladder. A Radon-Nikodym density defines a new measure, Girsanov identifies Brownian motion under it, Feynman-Kac can identify the claim's Markov value function, and martingale representation supplies the hedge.

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Brownian Martingale Representation

Let $W$ be a one-dimensional Brownian motion and let $(\mathcal F_t^W)_{0\leq t\leq T}$ be its completed, right-continuous natural filtration. If $M$ is a square-integrable $\mathcal F^W$-martingale, then there is a predictable process $H$, unique up to equality $dt\otimes d\mathbb P$-almost everywhere, such that

$$\mathbb E\!\left[\int_0^T H_s^2\,ds\right]<\infty$$

and

$$M_t=M_0+\int_0^tH_s\,dW_s, \qquad 0\leq t\leq T.$$

Equivalently, every $X\in L^2(\mathcal F_T^W)$ has the terminal representation

$$X=\mathbb E[X]+\int_0^TH_s\,dW_s.$$

The process $H$ is predictable, not merely adapted. For the standard Itô integral, predictability is the right measurability condition.

For a $d$-dimensional Brownian motion,

$$M_t=M_0+\sum_{j=1}^d\int_0^tH_s^j\,dW_s^j.$$

The Imperial College mathematical-finance notes state this version and connect it directly to Black-Scholes completeness.

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Why the Filtration Matters

A filtration records the information available through time. Brownian representation holds for the filtration generated by the Brownian motion, after the usual augmentation. If the filtration also reveals an independent coin toss at $T$, the payoff of that coin cannot be an integral against $W$.

There are extensions. Brownian filtration has the predictable representation property, and some larger filtrations have a representation with several driving martingales. But adding information can destroy representation with respect to $W$ alone.

An equivalent change of probability measure does not change the underlying sigma-algebras, and the Brownian predictable representation property is preserved under suitable equivalent measure changes, with the transformed Brownian motion and the appropriate integrability formulation. Saying only that "the filtration is the same" skips that theorem.

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A Clean Proof Outline

The terminal form is most economical. For deterministic $h\in L^2[0,T]$, define the exponential random variable

$$\mathcal E_T(h) =\exp\!\left( \int_0^T h_s\,dW_s-\frac12\int_0^T h_s^2\,ds \right).$$

Its conditional-expectation process satisfies

$$d\mathcal E_t(h)=\mathcal E_t(h)h_t\,dW_t,$$

so each exponential has the required representation. The linear span of these stochastic exponentials is dense in $L^2(\mathcal F_T^W)$.

The Itô isometry,

$$\mathbb E\!\left[ \left(\int_0^T H_s\,dW_s\right)^2 \right] =\mathbb E\!\left[\int_0^T H_s^2\,ds\right],$$

shows that the space of terminal stochastic integrals is closed in centred $L^2$. Density and closedness then give the representation for every centred square-integrable terminal variable. Taking conditional expectations produces the process form.

Uniqueness follows from the same isometry. If two integrands represent the same martingale, the expected integral of their squared difference is zero.

This argument is more direct than claiming every Wiener chaos term can simply be "reduced" to one integral. Iterated integrals can indeed be grouped by their outermost time variable, but the measurability and closure steps still need proof.

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A Representation You Can Verify

Take the centred terminal variable

$$X=W_T^2-T.$$

Itô's formula gives

$$d(W_t^2-t)=2W_t\,dW_t.$$

After integration,

$$W_T^2-T=\int_0^T2W_t\,dW_t.$$

So the representing integrand is $H_t=2W_t$, and the martingale process is $M_t=W_t^2-t$. The square-integrability check is explicit:

$$\mathbb E\!\left[\int_0^T(2W_t)^2dt\right] =4\int_0^Tt\,dt =2T^2<\infty.$$

The example is simple, but it separates representation from expectation. Since $\mathbb E[W_T^2]=T$, the centred variable has no constant term; all of it is generated by trading exposure to Brownian increments.

For sufficiently Malliavin-differentiable $X$, the Clark-Ocone formula can identify the integrand:

$$H_t=\mathbb E[D_tX\mid\mathcal F_t].$$

Here $D_tX=2W_T\mathbf 1_{\{t\leq T\}}$, whose conditional expectation is $2W_t$. Clark-Ocone is a useful computational refinement, not an assumption of the basic martingale representation theorem.

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From a Claim to a Hedge

Consider the Black-Scholes market under its bank-account measure $\mathbb Q$:

$$dS_t=rS_t\,dt+\sigma S_t\,dW_t^{\mathbb Q}, \qquad B_t=e^{rt}, \qquad \sigma>0.$$

The discounted stock $\widetilde S_t=S_t/B_t$ satisfies

$$d\widetilde S_t=\sigma\widetilde S_t\,dW_t^{\mathbb Q}.$$

For a square-integrable payoff $X$, the undiscounted value is

$$V_t=B_t\mathbb E^{\mathbb Q}\!\left[ \frac{X}{B_T}\,\middle|\,\mathcal F_t \right].$$

It is $\widetilde V_t=V_t/B_t$, not $V_t$, that is a $\mathbb Q$-martingale:

$$\widetilde V_t =\mathbb E^{\mathbb Q}\!\left[ \frac{X}{B_T}\,\middle|\,\mathcal F_t \right].$$

Martingale representation supplies $H$ with

$$d\widetilde V_t=H_t\,dW_t^{\mathbb Q}.$$

Since $d\widetilde S_t=\sigma\widetilde S_t dW_t^{\mathbb Q}$, define the number of shares

$$\Delta_t=\frac{H_t}{\sigma\widetilde S_t}.$$

Then

$$d\widetilde V_t=\Delta_t\,d\widetilde S_t.$$

This is the discounted self-financing identity. The residual value is held in the bank account. At maturity $V_T=X$, so the strategy replicates the claim.

MRT produced $H$; the non-zero traded volatility converted $H\,dW$ into $\Delta\,d\widetilde S$. Without that spanning step, Brownian representation alone would not establish completeness.

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Checked Black-Scholes Delta

For a one-year at-the-money call with $S_0=K=100$, $r=5\%$, no dividends and $\sigma=20\%$,

$$d_1=0.35,\qquad d_2=0.15,$$

and the value is

$$C_0=100\Phi(0.35)-100e^{-0.05}\Phi(0.15) =10.4506.$$

The time-zero hedge is

$$\Delta_0=\Phi(d_1)=\Phi(0.35)=0.63683.$$

The martingale-representation integrand at time zero is therefore

$$H_0=\Delta_0\sigma\widetilde S_0 =0.63683\times0.20\times100 =12.7366.$$

If time is in years, $H_0$ has units currency per square-root year, because $H_0dW_t$ must have currency units. Delta is shares per option. These are related quantities, not the same object.

For a Markov European claim with a smooth value function, Itô's formula identifies $\Delta_t=\partial V/\partial S$. MRT is more general: it establishes an integrand even when a classical derivative representation is unavailable.

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Several Assets and Risk Factors

Suppose $n$ discounted risky assets are driven by a $d$-dimensional Brownian motion:

$$d\widetilde S_t =\operatorname{diag}(\widetilde S_t)\Sigma_t\,dW_t^{\mathbb Q}.$$

A claim martingale has representation $d\widetilde V_t=H_t^{\mathsf T}dW_t^{\mathbb Q}$. A trading strategy $\Delta_t$ replicates its Brownian exposure when

$$\Sigma_t^{\mathsf T}\operatorname{diag}(\widetilde S_t)\Delta_t=H_t.$$

Every Brownian claim can be spanned only if this linear system has a solution for every $H_t$, which requires rank $d$ almost everywhere. Having the same number of assets and Brownian motions is not enough if the volatility matrix is singular.

This rank test makes completeness concrete. It also shows why an untraded stochastic-volatility shock generally leaves residual risk even though the filtration itself may have a two-dimensional Brownian representation.

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Completeness and the Second FTAP

Harrison and Pliska's 1981 paper on martingales and continuous trading links attainable claims to representation with respect to discounted asset prices. Their 1983 follow-up proves, in its stated framework, that market completeness is equivalent to uniqueness of the martingale measure.

The slogan is:

$$\text{complete market} \quad\Longleftrightarrow\quad \text{unique equivalent martingale measure}.$$

It is a theorem under technical assumptions on the market model and admissible strategies, not a definition that can be applied without qualification. A unique expectation for a claim does not, by itself, construct a replicating strategy.

In Black-Scholes, the Brownian representation property plus non-zero stock volatility provides that construction. In an incomplete diffusion market, several market-price-of-risk vectors may produce martingale measures. No-arbitrage then gives a range of possible claim values rather than a unique one.

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Jumps and Other Limits

Jumps do not automatically make every market incomplete. A filtration generated by Brownian motion and a Poisson process has a representation using both a Brownian integral and an integral against the compensated jump measure. Completeness asks whether traded assets span all those integrands. A model with a continuum of possible jump sizes and only a few assets usually does not.

Transaction costs, trading constraints, discrete rebalancing and liquidity break the frictionless self-financing argument. Even in a mathematically complete model, practical hedging error remains.

For non-square-integrable claims, local-martingale and $L^1$ versions may be available, but admissibility becomes more delicate. The clean $L^2$ theorem should not be quoted as if all unbounded payoffs satisfy it.

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Frequently Asked Questions

What does the martingale representation theorem say?

Every square-integrable martingale in a Brownian natural filtration is an initial value plus an Itô integral against that Brownian motion. The integrand is unique in the predictable $L^2$ class.

Does MRT prove every contingent claim is hedgeable?

No. It represents Brownian-filtration claims. Hedgeability additionally requires traded discounted assets to span every Brownian integrand and requires admissible frictionless trading.

Why must the option value be discounted?

Under the bank-account measure, a tradeable value divided by the bank account is a martingale. The undiscounted value normally has financing drift $rV_tdt$.

Is the MRT integrand the option delta?

Not directly. In one-dimensional Black-Scholes, $H_t=\Delta_t\sigma\widetilde S_t$. Delta appears after converting Brownian exposure into units of the discounted stock.

Are stochastic-volatility markets incomplete?

A stock driven by one return shock plus an untraded volatility shock is generally incomplete because stock trading cannot span both Brownian risks. Adding sufficiently rich traded instruments can change that conclusion.

Does an equivalent measure change preserve MRT?

Under suitable conditions, the predictable representation property transfers to the equivalent measure with the transformed driving local martingale. Mere equality of sigma-algebras is not the whole argument.