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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Nov 24, 2022 at 19:21


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