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:\[X_{T}=E[X_{T}]+\int_{0}^{T}H_{s}dB_{s}.\]

\({\scriptstyle \blacksquare}\)

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:\[M_{t}=\int_{0}^{t}H_{s}dB_{s}.\]

\({\scriptstyle \blacksquare}\)

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[1990], 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[2011], 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. \(H_{0}(x,y)=1,H_{1}(x,y)=x,H_{2}(x,y)=x^{2}-y,\ldots\)↩︎