# Representation theorem for local martingales

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!

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.

• Thank you for your answer. So, I need to demand some kind of continuity of the given local martingale in advance. I was just wondering why this is not the case in the representation theorem for $L^2$-martingales that I quoted: There only square-integrability was required besides the fact that the martingale should be adapted w.r.t. the Brownian filtration. Does the latter already imply continuity implicitly? Sep 30, 2015 at 9:37
• I use Øksendal's book about SDEs. It's Theorem 4.3.4 there (the martingale representation in the $L^2$-setting). As far as I understand his arguments there is no assumption of continuity. The local martingale representation is a result from the lecture of mine which I tried to prove on my own. Sep 30, 2015 at 10:07