# Properties of Stochastic Exponential

Let $\{X_t\}_{t \ge 0},\{Y_t\}_{t \ge 0}$ be a continuous semi-martingale with $X_0 = Y_0 = 0$, let ${\cal E}(X)$ to be the unique solution of:
$dZ_t = Z_t dX_t$ with $Z_0=1$.

We can show that ${\cal E}(X)_t = exp(X_t - \frac{1}{2}[X]_t)$, but how to show that ${\cal E}(X){\cal E}(Y) = {\cal E}(X+Y+[X,Y])$ where $[X，Y]$ denotes the quadratic covariation between $X_t$ and $Y_t$. I really appreciate your help.

## 1 Answer

Let $dV_t = V_tdY_t$. We will consider $d(VZ)_t$. \begin{align} d(VZ)_t &= V_tdZ_t + Z_tdV_t + d[V,Z]_t \\ &= V_tZ_tdX_t + Z_tV_tdY_t + Z_tV_td[X,Y]_t \\ & = V_tZ_td(X + Y + [X,Y])_t \end{align} I used the product rule above and also the fact that stochastic integral is linear in the integrator.

We have $V_tZ_t = \mathcal{E}(X)\mathcal{E}(Y)$. On the other hand, $V_tZ_t = \mathcal{E}(X+Y+[X,Y])$ by the SDE above.