Ito lemma on product of Brownian motion and Poisson process If we are given that
$$
Z_{t}=\sigma N_{t}dW_{t}+\sigma W_{t}dN_{t}
$$
where $W_{t}$ is a Brownian motion, $N_{t}$ is a Poisson process with intensity $\lambda$, and both are independent.
How do apply Ito lemma on this? So far I've got:
$$
df(t,Z_t)=\left[\frac{\partial f}{\partial t}+\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}\right]dt+\left[\sigma N_{t}\frac{\partial f}{\partial z}\right]dW_{t}+\left[f\left(t,Z_{t^{-}}+\sigma W_{t}\right)-f\left(t,Z_{t^{-}}\right)\right]dN_{t}
$$
Is this correct? I am not sure about the coefficient on $\frac{\partial^{2}f}{\partial z^{2}}$. The result I am seeking to obtain is just $\frac{1}{2}\sigma^{2}\frac{\partial^{2}f}{\partial z^{2}}$, and not $\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}$.
Thanks.
 A: Most likely you meant $dZ_t=\sigma N_t\,dW_t+\sigma W_t\,dN_t\,.$
For non continuous processes Ito's lemma is better written in this form
\begin{align}
&f(T,Z_T)=f(0,Z_0)+\int_0^T\left[\frac{\partial f}{\partial t}+\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}\right](t,Z_t)\,dt\\&+\int_0^T\sigma N_{t}\frac{\partial f}{\partial z}(t,Z_{t-})\,dW_t\\
\tag{1}
&+\sum_{0\le t\le T}\Big[f(t,Z_t)-f(t,Z_{t-})\Big]-\sum_{0\le t\le T}\frac{\partial f}{\partial z}(t,Z_{t-})(Z_t-Z_{t-})\,.
\end{align}
Your coefficient of $\frac{\partial^{2}f}{\partial z^{2}}$ is correct.
Note however that $Z_t=\sigma N_t W_t\,.$ Writing $\Delta N_t=N_t-N_{t-}$
we get therefore
\begin{align}
Z_t-Z_{t-}&=\sigma W_t\Delta N_t\,,\\
f(t,Z_t)-f(t,Z_{t-})&=\Big[f(t,Z_t)-f(t,Z_{t-})\Big]\Delta N_t\\
&=\Big[f(t,Z_{t-}+\sigma W_t)-f(t,Z_{t-})\Big]\Delta N_t
\end{align}
because $\Delta N_t$ is either zero or one. Recall also that
$$
\sum_{0\le t\le T}X_{t-}\,\Delta N_t=\int_0^TX_{t-}\,dN_t\,.
$$
This allows to write the last two terms in (1) as
\begin{align}\tag{2}
\int_0^T\Big[f(t,Z_{t-}+\sigma W_t)-f(t,Z_{t-})\Big]\,dN_t-\int_0^T\frac{\partial f}{\partial z}(t,Z_{t-})\,\sigma W_t\,dN_t\,.
\end{align}
In your equation the last term of (2) was missing.
