The question comes from this post Poisson Process with Randomly Distributed Time.
Let $X\sim\text{Poisson}(Z)$ where $Z$ is exponentially distributed. Suppose that $X$ and $Z$ share the same parameter $\lambda$. Then, following the post above, we can see that $$\mathbb{P}(X(Z)=n)=\dfrac{1}{2^{n+1}},$$ which is geometric distribution with parameter $\frac{1}{2}$ with $X(Z)$ measuring the total number of failures in Bernoullis trails before the first success coming out
I would like to see if I can conclude the same thing from characteristic function. It is not hard to derive the characteristic function. Recall that $\varphi_{X}(t)=e^{\lambda (e^{it}-1)}$ and $\varphi_{Z}(t)=\frac{\lambda}{\lambda-it}$. Then, we have $$\mathbb{E}(e^{itX})=\mathbb{E}(\mathbb{E}(e^{itX}|Z))=\mathbb{E}(e^{Z(e^{it}-1)})=\varphi_{Z}\Big(\dfrac{e^{it}-1}{i}\Big)=\dfrac{\lambda}{\lambda-e^{it}+1}.$$ However, I do not know how to reverse this. I checked the Fourier transform of this and did not find anything special..
Edit 1: I just found out that if we put $\lambda=1$, then $$\mathbb{E}(e^{itX})=\dfrac{1}{2-e^{it}}=\sum_{n=0}^{\infty}e^{itn}\frac{1}{2^{n+1}},$$ so we can go back to geometric distribution. However, before I was not assuming any specific value of $\lambda$. What is happening??
The context seems to be: $N=(N_t)_{t \geq 0}$ is a Poisson process with parameter $\lambda$ and $\tau\sim \textrm{Exp}(\lambda)$ independent of $N$. We want to compute $\xi \mapsto E[e^{i\xi N_\tau}]$. Then $$\lambda \int_{[0,\infty)} e^{\color{red}{\boxed{\lambda}} z(e^{i\xi}-1)}e^{-\lambda z}dz=\lambda \int_{[0,\infty)}e^{-z\lambda (2-e^{i\xi})}dz=\frac{1}{2-e^{i\xi}}=\frac{1/2}{1-e^{i\xi}/2}$$ So $N_\tau\sim \textrm{Geom}_{\mathbb{N}_0}(1/2)$.