I am trying to compute the characteristic function of the following:
Let $X$ and $Y$ be random variables such that $Y\mid X = x\sim N(0, x)$ with $X\sim\mathrm{Po}(\lambda)$. Find the characteristic function of $Y$.
I know the characteristic function is $\varphi_X(t)=E[e^{itX}]=\int e^{itx}f_X(x) \, dx$. How do I take into account the fact that $Y$ is conditional on $X$ to compute the characteristic function?
In my case I need to find $f_Y(y)$ in order to solve $\varphi_Y(t)=E[e^{itY}]=\int e^{ity}f_Y(y) \, dy$.
To find the probability density of y, I need to solve $P(Y=y)=\sum_{n=0}^{\infty}P(Y|X=x)*P(X=x)$, but I don't know how to reduce this.
\begin{align} \operatorname{E}(e^{itY}\mid X=x) & = \operatorname{E}(e^{it\sqrt{x} \, Z})\quad\text{where }Z\sim N(0,1), \\[10pt] & = \varphi_Z(t\sqrt x) = \exp \left( \frac{-1}2 t^2 x \right). \end{align}
\begin{align} \operatorname{E}(e^{itY}) & = \operatorname{E} \left( \operatorname{E}(e^{itY}\mid X) \right) = \operatorname{E}\left( \exp\left( \frac{-1}2 t^2 X \right) \right) \\[10pt] & = \sum_{x=0}^\infty e^{-t^2 x/2} \frac{\lambda^x e^{-\lambda}}{x!} = e^{-\lambda} \sum_{x=0}^\infty \frac{\left( e^{-t^2/2} \lambda \right)^x}{x!} \\[10pt] & = e^{-\lambda} e^{e^{-t^2 \lambda/2}}, \end{align} or a bit more legibly, $$ \exp( -\lambda)\exp\left(\lambda \exp \left( \frac{-t^2} 2 \right)\right). $$