Let $Y_1 \sim CP(\lambda_1, d F_Z ) $ and $Y_2\sim CP(\lambda_2, d F_W )$ be two Compound Poisson random variables and $X = \mathcal{N} + Y_1 + Y_2$ with $\mathcal{N}\sim N(0,1)$ such that $\mathcal{N}, Y_1$ and $Y_2$ independent , i.e.:
$$X = \mathcal{N} + \sum_{j=1}^{N_1} Z_j + \sum_{j=1}^{N_2} W_j , \quad N_i \sim Poisson (\lambda_i),\,\, i = 1,2$$
I want to calculate the characteristic function of $X$.
First, I will suppose that $\mathcal{N} =0$. My intuition says that $X = Y_1 + Y_2$ is Compound Poisson. In fact:
$$\varphi_{Y_1+Y_2}(s) = \varphi_{Y_1}(s)\varphi_{Y_2}(s)= e^{\lambda_1\left[ \varphi_Z(s)- 1\right]} e^{\lambda_2\left[ \varphi_W(s)- 1\right]}.$$
Since $$\lambda_1\left[ \varphi_Z(s)- 1\right] + \lambda_2\left[ \varphi_W(s)- 1\right]= \lambda \left[ \frac{1}{\lambda} \left( \lambda_1 \varphi_Z(s) + \lambda_ 2\varphi_W(s) \right) - 1 \right]$$ with $\lambda = \lambda_1+\lambda_2$, we have: $$\varphi_{Y_1+Y_2}(s) = e^{\lambda\left[ \varphi_\xi(s)- 1\right]} $$ where, $$\varphi_\xi (s) = \frac{1}{\lambda} \left( \lambda_1 \varphi_Z(s) + \lambda_ 2\varphi_W(s) \right) $$ i.e.: $$X = \sum_{j=1}^N \xi_j, \quad N \sim Poisson(\lambda), \,\, \lambda = \lambda_1 + \lambda_2$$ $$\xi \sim dF_\xi = \frac{1}{\lambda} \left( \lambda_1 dF_Z + \lambda_ 2 dF_W \right) $$
But, how about the case $\mathcal{N}\neq 0$? i.e., How to show that $$X = \sum_{j=1}^N \xi_j, \quad N \sim Poisson(\lambda), \,\, \lambda = 1+ \lambda_1 + \lambda_2$$ $$\xi \sim dF_\xi = \frac{1}{\lambda} \left( dF_\mathcal{N} + \lambda_1 dF_Z + \lambda_ 2 dF_W \right) $$