# How is exponential martingale the solution of $d Y_t=Y_t d M_t$?

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)?

• I don't think you need condition (3) to conclude that. Just apply Ito's formula to $Y_t$ to get $Y_t = Y_0 + \int_0^t Y_s dM_s$. To get $Y_0 = 1$, it looks like you need $M_0=0$, but that doesn't follow from condition (3) either. Commented Feb 24, 2023 at 14:18
• @user6247850 Assume $(Y_t, t \in [0, T])$ is a continuous square-integrable martingale and $f:\mathbb R \to \mathbb R$ twice continuously differentiable. By Ito's lemma, $f\left(Y_t\right)-f\left(Y_0\right)=\int_0^t f^{\prime}\left(Y_s\right) \mathrm{d} Y_s+\frac{1}{2} \int_0^t f^{\prime \prime}\left(Y_s\right) \mathrm{d}\langle Y\rangle_s$. I could not see how to pick $f$ to get the desired result... Commented Feb 24, 2023 at 14:24
• @user6247850 I guess the author forgot to include $M_0=0$. Commented Feb 24, 2023 at 14:25

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.
• The usual Itô's lemma I have seen so far is \begin{aligned} & f\left(t, M_t\right)-f\left(0, M_0\right) \\ = & \int_0^t f_t^{\prime}\left(s, M_s\right) \mathrm{d} s+\int_0^t f_x^{\prime}\left(s, M_s\right) \mathrm{d} M_s+\frac{1}{2} \int_0^t f_{x x}^{\prime \prime}\left(s, M_s\right) \mathrm{d} s. \end{aligned} Could you provide me with a reference containing the version you have used? Commented Feb 25, 2023 at 14:44