Representation theorem for local martingales

Question

I want to prove the following local martingale representation theorem. For the statement of the theorems to come we fix a filtered probability space $(\Omega,\mathcal{A},\mathcal{F},\mathbb{P})$ where $\mathcal{F}$ is the augmented Brownian filtration generated by a (for simplicity 1D) Brownian motion $W$.

The theorem I want to prove is the following: Let $M$ be a local martingale on $[0,T]$. Then there is a progressively measurable process $\phi$ satisfying $\int_0^T\phi(s)^2 ds<\infty$ a.s. such that $$M(t)=M(0)+\int_0^T\phi(s)dW(s)$$

For the proof I am basically only allowed to use the following martingale representation theorem for $L^2$-martingales: Let $M$ be an $L^2$-martingale on $[0,T]$. Then there is a progressively measurable process $\phi\in L^2(\lambda\otimes\mathbb{P})$ such that $$M(t)=M(0)+\int_0^T\phi(s)dW(s).$$

In the case that I a priori that my local martingale above is continuous I was able to find a reducing sequence of stopping times which makes the emerging stopped processes bounded: Use $\tau_k:=\inf\left\{t\in[0,T]: |M(t)|>k\right\}\wedge T$. Hence I am in the setting of the result I am allowed to use. I am also pretty sure that this approach also works for left-continuous processes. But what about the general case? Is there a way to find a reducing sequence which yields $L^2$-martingales as resulting stopped processes? Thanks you!

Answer 1

Since the augmented filtration is right-continuous, we may assume that $(M_t)_{t \geq 0}$ has càdlàg sample paths. Since $(M_t)_{t \geq 0}$ is a local martingale, there is a sequence of stopping times $(\tau_k)$ such that $\tau_k \uparrow \infty$ and $(M_{t \wedge \tau_k})_{t \geq 0}$ is a martingale. Set

$$Y := M_{T \wedge \tau_k}$$

for fixed $k \in \mathbb{N}$ and $T>0$. Since $Y \in L^1$ and $L^2$ is dense in $L^1$, we can find a sequence $(Y_n)_{n \in \mathbb{N}} \subseteq L^2(\mathcal{F}_T)$ such that $Y_n \to Y$ in $L^1$. Obviously,

$$M_t^n := \mathbb{E}(Y_n \mid \mathcal{F}_t), \qquad t \leq T,$$

is an $L^2$-bounded $\mathcal{F}_t$-martingale. By the martingale representation theorem for $L^2$-martingales, there exists $\phi_n \in L^2(\lambda_T \otimes \mathbb{P})$ such that

$$M_t^n = \int_0^t \phi_n(s) \, dW_s, \qquad t \leq T.$$

In particular, $(M_t^n)_{t \leq T}$ has continuous sample paths. By the maximal inequality,

$$\mathbb{P} \left( \sup_{t \leq T} |M_{t \wedge \tau_k}-M_t^n| > \epsilon \right) \leq \epsilon^{-1} \mathbb{E}|Y-Y_n| \to 0,$$

i.e. $\sup_{t \leq T} |M_{t \wedge \tau_k}-M_t^n|$ converges in probability to $0$. Extracting a convergent subsequence, we conclude that $(M_{t \wedge \tau_k})_{t \leq T}$ has continuous sample paths. Since both $k$ and $T$ are arbitrary, we find that $(M_t)_{t \geq 0}$ has a.s. continuous sample paths. Now the claim follows using the argumentation described in the question.

For martingales with not necessarily continuous sample paths (which are not adapted to a filtration generated by a Brownian motion), we need more general representation results; the following result is due to Ikeda-Watanabe.

Let $(M_t)_{t \geq 0} \in \mathcal{M}_2$ a martingale with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$ generated by a Lévy process. Then there exists predictable processes $f,g$ as well as a Brownian motion $(W_t)_{t \geq 0}$ and a Poisson random measure $N$ such that $$M_t - M_0 = \int_0^t f(s) \, dW_s + \int_0^t g(s) \, d\tilde{N}_s$$ where $\tilde{N}$ denotes the compensated Poisson random measure.

See also this question.