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?


2 Answers 2


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.


The Yor formula above is specialized to continuous processes but the actual Yor formula works for arbitrary semimartingales, $$\mathscr{E}(X)\mathscr{E}(Y) = \mathscr{E}(X+Y+[X,Y]).$$ To derive it, it is better NOT to use an explicit expression for the stochastic exponential in terms of natural exponential (even though such expression is available also for general semimartingales, see Doléans-Dade, C., Quelques applications de la formule de changement de variables pour les semi-martingales, Z. Wahrscheinlichkeitstheor. Verw. Geb. 16, 181-194 (1970). ZBL0194.49104.)

One simply writes out the Itô-Meyer formula for the product $$\mathrm{d}(\mathscr{E}(X)\mathscr{E}(Y)) = \mathscr{E}(Y)_-\mathrm{d}\mathscr{E}(X) + \mathscr{E}(X)_-\mathrm{d}\mathscr{E}(Y) + \mathrm{d}\mathscr{E}(X)\mathrm{d}\mathscr{E}(Y).$$ Now plug in $\mathrm{d}\mathscr{E}(X) = \mathscr{E}(X)_-\mathrm{d}X$ (and likewise for $Y$) to obtain $$\mathrm{d}(\mathscr{E}(X)\mathscr{E}(Y)) = \mathscr{E}(X)_-\mathscr{E}(Y)_-(\mathrm{d}X + \mathrm{d}Y + \mathrm{d}X\mathrm{d}Y).$$ This proof can be found, for example, in Proposition 1.18 of Eberlein and Kallsen (2019).


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