I'm reading about exponential martingale from these notes.
4.1 Exponential martingale
Let $M$ be a continuous square-integrable martingale and let $Y$ be the process defined as $$ Y_t=\exp \left(M_t-\frac{\langle M\rangle_t}{2}\right), \quad t \in \mathbb{R}_{+}. $$
Notice that $Y$ is not necessarily a martingale, a priori.
Fact. (to be proven later) If there exists a constant $K>0$ such that $$ \langle M\rangle_t \leq K t \quad \text { a.s., } \quad \forall t \in \mathbb{R}_{+}, \quad \quad (3) $$ then $Y$ is a continuous square-integrable martingale. $Y$ is said to be the exponential martingale associated to $M$.
Example. If $M_t=B_t$, then $\langle B\rangle_t=t$ and $Y_t=\exp \left(B_t-\frac{t}{2}\right)$ is indeed a martingale.
Remarks.
- There exists a more general condition than (3) which ensures that the process $Y$ is a martingale up to a finite time horizon $T>0$. This more general condition, called Novikov's condition, reads: $$ \mathbb{E}\left(\exp \left(\frac{\langle M\rangle_T}{2}\right)\right)<\infty . $$
- Under condition (3), one can apply Ito-Doeblin's formula to conclude that $$ Y_t=1+\int_0^t Y_s d M_s \quad \text { a.s., } \quad \forall t \in \mathbb{R}_{+} . $$ i.e., $Y$ is solution of the SDE: $$ d Y_t=Y_t d M_t, \quad Y_0=1. $$
Could you explain how to get $$ Y_t=1+\int_0^t Y_s d M_s \quad \text { a.s., } \quad \forall t \in \mathbb{R}_{+} . $$ under (3)?
Define $f(x,y) := e^{x-\frac 12 y}$ and observe that $Y_t = f(M_t, \langle M\rangle_t).$ Now, Ito's formula gives \begin{align*} Y_t &= f(M_t, \langle M\rangle_t) \\ &= 1+ \int_0^t \partial_x f(M_s,\langle M\rangle_s) dM_s + \int_0^t \partial_y f(M_s,\langle M\rangle_s)d\langle M\rangle_s + \frac 12 \int_0^t \partial_{xx} f(M_s,\langle M\rangle_s) d\langle M\rangle_s \\ &= 1 + \int_0^t f(M_s,\langle M\rangle_s) dM_s -\frac 12 \int_0^t f(M_s,\langle M\rangle_s)d\langle M\rangle_s + \frac 12 \int_0^t f(M_s,\langle M\rangle_s) d\langle M\rangle_s \\ &= 1 + \int_0^t Y_s dM_s.\end{align*}
There are no second order cross terms because $\langle M\rangle_t$ has finite variation.