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I have $$\frac{dM_t}{M_t}=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t, \tag{1}$$ where $B_t$ and $W_t$ are two independent Brownian Motions, which was further presented as $$ M_t=\exp \left( -\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s -\frac{1}{2} \int_0^t \gamma_s^2 ds \right) \tag{2}$$

can anybody explain how $(1)$ was transitioned into $(2)$?

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  • $\begingroup$ Made some stylistic changes and typo corrections. $\endgroup$ – Gordon May 13 '16 at 17:35
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Apply Ito's lemma to $\ln M_t$, we obtain that \begin{align*} d\ln M_t &= \frac{1}{M_t} dM_t -\frac{1}{2} \frac{1}{M_t^2} d\langle M, M\rangle_t\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \frac{1}{M_t^2}\left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right) dt.\tag{Eq. 1} \end{align*} Here, since $dM_t=M_t\big[-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t\big]$, \begin{align*} d\langle M, M\rangle_t = \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt. \end{align*} From $(\textrm{Eq.} 1)$, \begin{align*} \ln M_t - \ln M_0 &= \int_0^t\left[-\frac{\mu}{\sigma} dW_s + \gamma_s dB_s -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_s^2\right) ds\right]\\ &=-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds. \end{align*} That is, \begin{align*} M_t = M_0 \exp\left(-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds\right). \end{align*}

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  • $\begingroup$ where the $M_t^2$ in the second line comes from? $\endgroup$ – Michal May 13 '16 at 19:03
  • $\begingroup$ @Michal: See my updates. $\endgroup$ – Gordon May 13 '16 at 19:20
  • $\begingroup$ I can see that integrating both sides and solving for $M_t$ $$ \begin{align*} d\ln M_t &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right) dt. \end{align*}$$ gives the equation (1), but I don't quite follow the condition for $M_0=1$ $\endgroup$ – Michal May 13 '16 at 19:33
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    $\begingroup$ @Michal: $M_t$ can not be zero based on the given dynamics. I do not know what is the safe rate. For any quantity, the initial value is assumed to be known. You do not have to assume that $M_0=1$, but, then you need to have a multiplier for your equation $(2)$. $\endgroup$ – Gordon May 13 '16 at 20:08
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    $\begingroup$ @Michal: Do you mean that $M_t$ is a constant? Note that, you may be provide with the initial $M_0$, or you can assume that $M_0=1$. I do not know why you want to say that $M_t=1$. If that is the case, you do not have to model it using an SDE as it is a constant. My answer above is based on the SDE you provided. You can force it to be a constant by assuming that $\mu=0$ and $\gamma_t=0$, for all $t$. $\endgroup$ – Gordon May 13 '16 at 20:28

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