# Why can any positive martingale be written as the exponential of an Ito integral w.r.t. Brownian motion?

Why is any positive martingale the exponential of an Ito integral w.r.t. Brownian motion?

Here is a little proof.

For any positive P-martingale M, $dM_t = Mt ·1/M_tdM_t$. By Martingale Representation Theorem, $dMt = \Gamma_tdW(t)$ for some adapted process $\Gamma_t$. So $dM_t = M_t(\Gamma_t/M_t)dW$, i.e. any positive martingale must be the exponential of an integral w.r.t. Brownian motion.

Where in this proof are we using the fact that $M$ is positive?

You are using the fact that $M$ is positive because you divide by $M$.

Dividing by zero is the root of evil.

• So, $M_t$ can be a negative Martingale?
– user16651
Sep 5, 2016 at 15:58
• Replace $M$ by $-M$ Sep 6, 2016 at 6:00
• Dividing by zero , is not sufficient
– user16651
Sep 6, 2016 at 16:08
• $dM_t=M_tdB_t$ and $M_0=-1$. You can solve it. Sep 7, 2016 at 3:43
• Everything lies in the in the sign of the initial condition indeed. Sep 7, 2016 at 5:08