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I was able to show the following:

$X$ a semimartingale, $X_0=0$ then the SDE

$$ dZ_t=Z_tdX_t$$

with $ Z_0=1$ has the unique solution $Z_t:=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}$.

I was able to show that this is indeed the a solution. However I got stuck at uniqueness. What I did so far: Suppose there are two solution, $Z,Z^'$. Apply Itô to $f(x,y):=\frac{x}{y}$ and with the $\mathbb{R}^2$ semimartingale $(Z,Z^')$. After some calculations, I'm at the point:

$$df(Z_t,Z^'_t)=\frac{Z^'_t}{Z^3_t}d\langle Z\rangle_t -\frac{1}{Z^2_t}d\langle Z,Z^'\rangle_t$$

The last two term should be equal, then uniqueness follows. Now why is this, i.e. why is it true that

$$d\langle Z\rangle_t = (Z_t)^2d\langle X\rangle_t$$


$$ d\langle Z,Z^' \rangle_t = Z^'Zd\langle X\rangle_t$$

Thank you for your help.

share|cite|improve this question
I found this equation in lecture notes, without a comment why they are true.(also in the example of stochastic exponential). – user20869 Jun 5 '12 at 17:42
Speaking only for continuous martingales, if $dZ_t = \sigma_t dX_t$ then the quadratic variation/compensator of $Z_t$ is $\int \sigma_t^2 d[X_t,X_t]$. If you are working from the def $Z^2 - [Z_t,Z_t]$ should be a martingale it is almost Ito's isometry, and if you are working directly from def of quadratic variation, start with step functions, etc. – mike Jun 5 '12 at 19:22

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