Jump-Process Itô Formula for a Poisson-Driven Exponential

# Ito's formula for Jump process


# Ito's formula for Jump process












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 $dX_t$?

## Answer by user16651 (score 7, accepted)

https://quant.stackexchange.com/a/18592

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}

## Answer by quasi (score 7)

https://quant.stackexchange.com/a/18599

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.