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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)?

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    $\begingroup$ 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. $\endgroup$ Commented Feb 24, 2023 at 14:18
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    $\begingroup$ @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... $\endgroup$
    – Akira
    Commented Feb 24, 2023 at 14:24
  • $\begingroup$ @user6247850 I guess the author forgot to include $M_0=0$. $\endgroup$
    – Akira
    Commented Feb 24, 2023 at 14:25

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Akira
    Commented Feb 25, 2023 at 14:44
  • $\begingroup$ Karatzas and Shreve's Brownian motion and Stochastic Calculus, Revuz and Yor's Continuous Martingales and Brownian Motion, or just using the integration by parts formula in the notes you provided. $\endgroup$ Commented Feb 26, 2023 at 13:16
  • $\begingroup$ Thank you so much for your help! $\endgroup$
    – Akira
    Commented Feb 26, 2023 at 14:36

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