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?


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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