# Ito representation theorem

By the Ito representation theorem for a $$\mathcal{F}_T$$-measurable random variable $$X$$ there exists a $$L^2$$ process $$\eta$$ such that $$X = \mathbb{E}[X]+\int_0^T\eta_s\,dB_s$$ where $$B$$ is a $$\mathcal{F}_t$$ Brownian motion. From this I derive that, if $$M$$ is a $$\mathcal{F}_t$$ martingale then $$M_t = \mathbb{E}\left[M_T|\mathcal{F}_t\right] = \mathbb{E}\left[ \mathbb{E}[M_T]+\int_0^T\eta_s\,dB_s|\mathcal{F}_t\right] = \mathbb{E}[M_T]+\int_0^t\eta_s\,dB_s = \mathbb{E}[M_0]+\int_0^t\eta_s\,dB_s\quad(1)$$ where I have used the fact that the expected value of a martingale is constant and the fact that $$\int_0^t\eta_s\,dB_s$$ is a martingale (which follows from the fact that $$\eta$$ is $$\mathcal{L}^2$$).

I think there is something wrong in this reasoning since identity $$(1)$$ implies $$M_0=\mathbb{E}[M_0]$$, which is in general false for a martingale. Where am I wrong?

• The reason for $M_0=\mathbb E[M_0]$ is that the filtration in the martingale representation theorem cannot be any filtration for which $B$ is a BM. It must be the smaller filtration generated by $B$. This post is discussing the difference. Essentially, your ${\cal F}_0$ is, modulo augmentation with $\mathbb P$-null sets, the trivial $\sigma$-algebra $\{\emptyset,\Omega\}$ which leads so $M_0=\mathbb E[M_0|{\cal F}_0]=\mathbb E[M_0]\,,\quad\mathbb P$-a.s. Nov 24, 2022 at 19:21