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In a book is a exercise to prove Yor's formula for stochastic exponential, i.e.

$$\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}=\mathcal{E}(X)\mathcal{E}(Y)$$

where $\mathcal{E}(X)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}$. Now if we would define the stochastic exponential as $\exp{(X_t-\frac{1}{2}\langle X\rangle)}$, then the above is simple algebra:

$$\mathcal{E}(X)_t\mathcal{E}(Y)_t=\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}\exp{(X_t-\frac{1}{2}\langle X\rangle_t)}=\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}$$

Using that $\langle X,Y\rangle=\frac{1}{2}(\langle X+Y\rangle-\langle X\rangle -\langle Y\rangle$ we would get:

$$\exp{(X_t+Y_t-\frac{1}{2}(\langle X\rangle_t-\langle Y\rangle_t))}=\exp{(X_t+Y_t+\langle X,Y\rangle -\frac{1}{2}\langle X+Y\rangle)}=\exp{(X_t+Y_t-\frac{1}{2}\langle X+Y\rangle)}\exp{(\langle X,Y\rangle)}=\mathcal{E}(X+Y)\exp{(\langle X,Y\rangle)}$$

If I would define the stochastic exponential as the (unique) solution $Z_t$ of the SDE $Z_tdX_t=dZ_t$. Then I guess I have to use Itô, to prove the statement, but how exactly? However, if we define the stochastic exponential as above, then my conclusion would be correct?

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What you've done looks fine. So it suffices to show that if $(Z_t)_{t\geq 0}$ is a solution to the given SDE, then $$Z_t = \exp\left(X_t-\frac{1}2{\langle X\rangle_t}\right),$$

since from there we can just follow your solution.

We may compute $d\log(Z_t)$ using the local version of Ito's formula (page 48 of these notes, for example), giving:

$$ \begin{align} d\log(Z_t) &= \frac{Z_tdX_t}{Z_t}-\frac{1}{2}\frac{d\langle Z\rangle_t}{Z_t^2}\\ &= dX_t - \frac{1}{2}\frac{Z_t^2d\langle X\rangle_t}{Z_t^2}\\ &= dX_t-\frac{1}{2}d\langle X\rangle_t. \end{align}$$

So $$\log(Z_t) = \log(Z_0)+X_t-\frac{1}{2}\langle X\rangle_t.$$

Exponentiating, we are done.

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