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Let $\{N_t\,|\,0\leq t\leq T\}$ be a Poisson process with intensity $\lambda>0$ defined on the probability space $(\Omega,\mathcal{F}_t,P)$ with respect to the filtration $\mathcal{F}_t$ and \begin{align} X_t=e^{(\lambda-\eta)\,t}\,\left(\frac{\eta}{\lambda}\right)^{N_t}, \end{align} where $\eta>0$.How can I obtain $dX_t$?

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Write $X_t = A_t B_t$ with $A_t = e^{(\lambda - \eta)t}$ and $B_t = \left(\frac{\eta}{\lambda} \right)^{N_t}$.

Then $dX_t = A_t dB_t + B_t dA_t$ by the product rule of calculus. There are no second order terms since both $A_t$ and $B_t$ are finite variation (i.e. $\langle A_t, B_t\rangle$= 0).

Next, $dA_t = (\lambda - \eta)A_t dt$, and $dB_t = B_t \cdot \left(\frac{\eta}{\lambda} - 1\right)dN_t$. The form of $dA_t$ follows from normal calculus, and the form of $dB_t$ follows from subtracting before and after values of the jump process.

Using the two paragraphs above, we get $$dX_t = A_tB_t \left( \left(\frac{\eta}{\lambda} - 1 \right)dN_t + (\lambda - \eta)dt\right) = X_t\left( \left(\frac{\eta}{\lambda} - 1 \right)dN_t + (\lambda - \eta)dt\right).$$

Edit for @Behrouz:

$dB_t = \left(\frac{\eta}{\lambda}\right)^{N_t} - \left(\frac{\eta}{\lambda}\right)^{N_{t-}}$.

When $N_t$ does not jump, this value is zero. When $N_t$ does jump, it is equal to

$\left(\frac{\eta}{\lambda}\right)^{N_t} - \left(\frac{\eta}{\lambda}\right)^{N_{t} - 1} = B_{t-}\left(\frac{\eta}{\lambda} - 1 \right)dN_t$

So actually in my original answer, I should have minuses for left hand limits to be technically correct.

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    $\begingroup$ Why $dB_t = B_t \cdot \left(\frac{\eta}{\lambda} - 1\right)dN_t$?Do you use Ito's lemma? $\endgroup$ – user16651 Jun 30 '15 at 21:36
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    $\begingroup$ No, it's just arithmetic...The poisson jumps by 1, so $B_t$ jumps correspondingly. $\endgroup$ – quasi Jun 30 '15 at 21:37
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    $\begingroup$ please prove it $\endgroup$ – user16651 Jun 30 '15 at 21:38
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    $\begingroup$ @AFK , I accepted,but , do these answers are different? $\endgroup$ – user16651 Jun 30 '15 at 21:53
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    $\begingroup$ @quasi Thank u,your solution was perfect by Hippocrates our discussion! $\endgroup$ – user16651 Jun 30 '15 at 21:58
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By Ito's lemma,

\begin{align} dX_t=\frac{\partial X_t}{\partial t}dt+\frac{\partial X_t}{\partial N(t)}dN_t+\frac{1}{2!}\frac{\partial^2 X_t}{\partial N^2_t}(dN_t)^2+\frac{\partial^2 X_t}{\partial N_t\partial t}{}dN_tdt+\frac{1}{3!}\frac{\partial^3 X_t}{\partial N^3_t}(dN_t)^3+... \end{align} Since $dN_t\,dt = 0, (dN_t)^2 = (dN_t)^3 = . . . = dN_t$, we have \begin{align} dX_t=\frac{\partial X_t}{\partial t}dt+\left(\frac{\partial X_t}{\partial N_t}+\frac{1}{2!}\frac{\partial^2 X_t}{\partial N^2_t}+\frac{1}{3!}\frac{\partial^3 X_t}{\partial N^3_t}+...\right)dN_t. \end{align} On the other hand \begin{align} &\frac{\partial X_t}{\partial t}=(\lambda-\eta)\,X_t\\ &\frac{\partial^n X_t}{\partial N_t^n}=\left[\ln \left(\frac{\eta}{\lambda}\right)\right]^nX_t,\\ \end{align} therefore

\begin{align} dX_t=(\lambda-\eta)\,X_tdt+\sum_{n=1}^{\infty}\frac{1}{n!}\left[\ln \left(\frac{\eta}{\lambda}\right)\right]^nX_t\,dN_t \end{align} We know $\sum_{n=1}^{\infty}\frac{1}{n!}\left[\ln \left(\frac{\eta}{\lambda}\right)\right]^n=exp\left(\ln \left(\frac{\eta}{\lambda}\right)\right)-1=\frac{\eta}{\lambda}-1=\frac{\eta-\lambda}{\lambda}$ thus we have \begin{align} dX_t=(\lambda-\eta)\,X_tdt+\frac{\eta-\lambda}{\lambda}X_t\,dN_t \end{align}

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    $\begingroup$ Why do the higher derivatives with respect to N matter? I just dont see it atm. $\endgroup$ – Phun Jun 30 '15 at 17:12
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    $\begingroup$ Ito's Lemma isn't really necessary, since Poisson processes have finite variation. $\endgroup$ – quasi Jun 30 '15 at 17:44
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    $\begingroup$ @Phun $\frac{\partial^n X_t}{\partial N_t^n}=\left[\ln \left(\frac{\eta}{\lambda}\right)\right]^nX_t $ for $n=1,2,3,...$ $\endgroup$ – user16651 Jun 30 '15 at 19:10
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    $\begingroup$ @ quasi Yes ,Poisson processes have finite variation, But if we don't use Ito lemma then we can not obtain $dX_t$ $\endgroup$ – user16651 Jun 30 '15 at 19:14
  • $\begingroup$ @Behrouz I think you can use the product rule of ordinary calculus. The jump term itself doesn't require calculus at all. $\endgroup$ – quasi Jun 30 '15 at 19:50

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