The martingale representation problem in its simplest form is the following. Given a filtration generated by a martingale $$M$$ and given another martingale $$N$$ adapted to the filtration, can we express $$N$$ as a stochastic integral with $$M$$ as the integrator? The martingale $$N$$ is generally closed, i.e. it can be expressed as the conditional expectation of a terminal variable $$N_{T}$$. In this case, the integrand $$H_{t}$$ of the stochastic integral representation is heuristically the sensitivity of $$N_{T}$$ to the shock $$dM_{t}$$.The Brownian filtration is the most important example where a Martingale Representation Theorem holds.

The theory of martingale representation is concerned with the following problem.

Consider a filtered probability space $$(\Omega,{\cal F},P)$$ with index space $$\mathbb{T}=[0,T]$$ where $$T$$ is finite. Such a space supports a set of martingales $${\cal M}$$ against which we can compute stochastic integrals for predictable integrands.

We are given an $${\cal F}_{T}$$-measurable random variable $$X_{T}$$. It induces a martingale $$(E_{t}[X_{T}])_{t \in \mathbb{T}}$$. This process represents, within the model, the anticipation of $$X_{T}$$ at any point $$t$$. The changes in $$E_{t}[X_{T}]$$ as a function of $$t$$ reflect the real time acquisition of information on $$X_{T}$$. New information comes as surprises as modeled in martingale differences (see this post). Heuristically, martingale representation asks the following question: can we represent the surprises in $$(E_{t}[X_{T}])_{t \in \mathbb{T}}$$ for any $$X_{T}$$ as a linear function of the (contemporaneous) surprises embedded in our set $${\cal M}$$ of martingales. More precisely, can we represent the martingale $$(E_{t}[X_{T}])_{t \in \mathbb{T}}$$ as a sum of stochastic integrals against some martingales in $${\cal M}$$.

A striking incarnation of this issue is found when the filtered probability space is generated by a Brownian motion1.

Theorem (Martingale Representation for the Brownian Filtration):Let $${\cal F}$$ be the smallest right continuous and complete filtration generated by a univariate Brownian motion $$(B_{t})_{t \in \mathbb{T}}$$. Let $$X_{T}$$ be an $${\cal F}_{T}$$-measurable random variable with finite second moment $$E_{0}[X_{T}^{2}]<\infty$$. Then there is a predictable process $$(H_{t})_{t \in \mathbb{T}}$$ with $$\int_{0}^{T}H_{s}^{2}ds<\infty$$ such that:$X_{T}=E[X_{T}]+\int_{0}^{T}H_{s}dB_{s}.$

$${\blacksquare}$$

In the same context as above, we have a simple yet important corollary:

Corollary: For any square integrable2right continuous martingale $$(M_{t})_{t \in \mathbb{T}}$$ with $$M_{0}=0$$, there exists a predictable process $$(H_{t})_{t \in \mathbb{T}}$$ with $$\int_{0}^{T}H_{s}^{2}ds<\infty$$ such that:$M_{t}=\int_{0}^{t}H_{s}dB_{s}.$

$${\blacksquare}$$

In other words, all square integrable right continuous martingales with initial value zero are Brownian stochastic integrals. Actually, in our context, all square integrable martingales have a version which is still a martingale and is right continuous with left limits. They can therefore be represented as Brownian integrals. Since Brownian integrals have continuous trajectories, all square integrable martingales in this setup have a continuous version. Finally, one can extend the above result to show that all local martingales can be represented as a Brownian stochastic integral.

It is quite easy to generate setups where the filtration is the minimal filtration generated by a given martingale $$(M_{t})_{t \in \mathbb{T}}$$, and yet, the filtration supports other martingales which cannot be written as sotchastic integrals of $$(M_{t})_{t \in \mathbb{T}}$$. In this post, an example is given where $$\mathbb{T}$$ is discrete and $$(M_{t})_{t \in \mathbb{T}}$$ has standardized gaussian increments. If, on the other hand, $$(M_{t})_{t \in \mathbb{T}}$$ has binomial increments, the martingale representation holds with the set $${\cal M}$$ consisting of $$(M_{t})_{t \in \mathbb{T}}$$. A solution to recover a martingale representation result when it does not hold for $${\cal M}=\{ (M_{t})_{t \in \mathbb{T}} \}$$ is to add other martingales in $${\cal M}$$, based on higher order moments of $$(M_{t})_{t \in \mathbb{T}}$$ for instance. Indeed, the problems generally come from the difficulty of generating non linear functions of $$(M_{t})_{t \in \mathbb{T}}$$ through the stochastic integral which, in the end, is just a linear reweighting of the increments of $$(M_{t})_{t \in \mathbb{T}}$$.

Given the above remarks, the Brownian martingale representation theorem looks like a nice accident. I now sketch the proof. An $${\cal F}_{T}$$-measurable random variable is, roughly speaking, a function of the increments of the Brownian motion. A simple example would be a function $$f(B_{t_{1}}-B_{t_{0}},\cdots,B_{t_{n}}-B_{t_{n-1}})$$ where the time intervals $$[t_{i},t_{i-1}]$$ do not overlap. Such functions can however be recovered through Fourier transform from products of complex exponentials3: $\exp(iu_{1}(B_{t_{1}}-B_{t_{0}}))\cdots \exp(iu_{n}(B_{t_{n}}-B_{t_{n-1}})).$ It is conceivable that if a martingale representation were to hold for such a function, the representation could be extended by limiting arguments to all $${\cal F}_{T}$$-measurable random variables. However, Ito calculus implies that: $\exp(iu_{k}(B_{t}-B_{t_{k-1}})+\frac{1}{2}u_{k}^{2}(t-t_{k-1}))=1+$ $\int_{t_{k-1}}^{t_{k}}iu_{k}\exp(iu_{k}(B_{s}-B_{t_{k-1}})+\frac{1}{2}u_{k}^{2}(s-t_{k-1}))dB_{s},$ i.e. $d\left(\exp(iu_{k}(B_{t}-B_{t_{k-1}})+\frac{1}{2}u_{k}^{2}(t-t_{k-1}))\right)=\exp(iu_{k}(B_{t}-B_{t_{k-1}})+\frac{1}{2}u_{k}^{2}(t-t_{k-1}))dB_{t}.$ This complex exponential is a geometric martingale with initial value $$1$$ at $$t=t_{k-1}$$.

From this, we get (taking $$t=t_{k}$$ and rearranging terms): $Z_{k-1}=\exp(iu_{k}(B_{t_{k}}-B_{t_{k-1}}))=\exp(-\frac{1}{2}u_{k}^{2}(t_{k}-t_{k-1}))+$ $\int_{t_{k-1}}^{t_{k}}iu_{k}\exp(iu_{k}(B_{s}-B_{t_{k-1}})+\frac{1}{2}u_{k}^{2}(s-t_{k}))dB_{s}$ $=F_{k-1}+\int_{t_{k-1}}^{t_{k}}H_{k-1}(s)dB_{s},$ where $$Z_{k-1}$$ is the random variable of interest, $$F_{k-1}$$ is a function of non random parameters only and $$H_{k-1}$$ is the integrand within the stochastic integral. We thus have the right representation for a single exponential of a Brownian increment.

When multiplying two such terms attached to non overlapping intervals, say $$[t_{k-1},t_{k}]$$ and $$[t_{k},t_{k+1}]$$, the product rule entails no covariation terms because the stochastic integrals refer to non overlapping time intervals: $[\int_{t_{k-1}}^{t_{k}}H_{k-1}(s)dB_{s},\int_{t_{k}}^{t_{k+1}}H_{k}(s)dB_{s}]=0.$ We thus have the following representation for the product: $Z_{k-1}Z_{k}=F_{k-1}F_{k}+\int_{t_{k-1}}^{t_{k}}F_{k}H_{k-1}(s)dB_{s}+\int_{t_{k}}^{t_{k+1}}Z_{k-1}H_{k}(s)dB_{s},$ which still has the right structure. It is now clear that any product involving a finite number of such exponentials involving non overlapping intervals has a martingale representation. The rest of the proof is a matter of spelling out the limiting arguments that allow to extend4 the representation to any function $$f(B_{t_{1}}-B_{t_{0}},\cdots,B_{t_{n}}-B_{t_{n-1}})$$ and then to any $${\cal F}_{T}$$-measurable random variable (through a density argument).

In the Brownian context thus, Brownian integrals allow to generate all the local martingales supported by the filtration5. Amongst them are all the martingales generated by moments $$B^{\alpha}_{t}$$, for instance $$X_{t}=B^{2}_{t}-t=2\int_{0}^{t}B_{s}dB_{s}$$.

A striking illustration of this involves Hermite polynomial functions. If $$H_{n}(x,y)=\left(\frac{y}{2}\right)^{\frac{n}{2}}h_{n}(\frac{x}{\sqrt{2y}})$$ ($$n \geq 0$$) where $$h_{n}(\cdot)$$ are Hermite polynomials6, then $$H_{n}(B_{t},t)$$ are martingales and we have the following integral representation: $H_{n}(B_{t},t)=\int_{0}^{t}nH_{n-1}(B_{u},u)dB_{u }=n!\int_{0}^{t}\int_{0}^{t_{n-1}}\cdots\int_{0}^{t_{1}}dB_{s}dB_{t_{1}} \cdots dB_{t_{n-1}}.$ This result can be found for instance in Chung, chapter 6.

Reference: Chung K.L and R.J. Williams, 1990 : An introduction to Stochastic Integration, Birkhauser.

Bass R.F., 2011, Stochastic Processes, Cambridge University Press

1. The following results can be found in Bass, p. 80.

2. $$E_{0}[M_{T}^{2}]<\infty$$.

3. In our context, the Fourier transform amounts to mixing functions indexed by $$(u_{1},\ldots,u_{n})$$ using a weighting scheme $${\hat f}(u_{1},\ldots,u_{n})$$.

4. Through the Fourier transform, which amounts to integrating the integral representations attached to different parameters $$(u_{1},\ldots,u_{n})$$, using a weighting scheme $${\hat f}(u_{1},\ldots,u_{n})$$.

5. It is important that the filtration be the minimal filtration generated by the Brownian motion, i.e. the smallest right continuous and complete filtration generated by the Brownian motion.

6. $$H_{0}(x,y)=1,H_{1}(x,y)=x,H_{2}(x,y)=x^{2}-y,\ldots$$