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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?

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

Dividing by zero is the root of evil.

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    $\begingroup$ So, $M_t$ can be a negative Martingale? $\endgroup$
    – user16651
    Sep 5, 2016 at 15:58
  • $\begingroup$ Replace $M $ by $-M $ $\endgroup$
    – MJ73550
    Sep 6, 2016 at 6:00
  • $\begingroup$ Dividing by zero , is not sufficient $\endgroup$
    – user16651
    Sep 6, 2016 at 16:08
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    $\begingroup$ $dM_t=M_tdB_t $ and $M_0=-1$. You can solve it. $\endgroup$
    – MJ73550
    Sep 7, 2016 at 3:43
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    $\begingroup$ Everything lies in the in the sign of the initial condition indeed. $\endgroup$
    – Quantuple
    Sep 7, 2016 at 5:08

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