Cumulants of Poisson random variable conditioned on a Bernoulli random variable Consider a Bernoulli distributed random variable $Y$, which is 1 with probability $p$ and 0 with probability $1-p$. Further there is a random variable $X$ where the conditional probability distribution of $X$ on $Y$ is given by a Poisson random variable with parameter $\lambda$ if $Y=1$ and $X=0$ if $Y=0$.
I want to get analytic expressions for the cumulants of $X$.
The mean and variance are easy to calculate using expectations.
$K_1 = p\lambda$
$K_2 = p\lambda +p(1-p)\lambda^2$
I'm struggling with the law of total cumulance to give me reasonable simple expressions.
Does a cumulant-generating function exist for $X$? If it exists, what is it?
Can the higher order cumulants be expressed as analytic direct or recursive formulas?
 A: Use the characteristic function (cf), because the cf of a mixture is a mixture of the cfs. 
By definition, the cf of a distribution $F$ is
$$\phi_F(t) = \mathbb{E}_{X\sim F}\left(\exp(itX)\right).$$
Because this is an expectation, it is a linear functional of $F.$  This is what makes the next steps possible (and simple).
Consider a bivariate random variable $(U,X)$ whose conditional cfs are known.  That is, associated with each possible value $u$ of $U$ there is a conditional distribution $F_u$ governing the conditional response $X_u$ with cf
$$\phi(t,u) = \mathbb{E}_{X_u\sim F_u}\left(\exp(itX_u)\right).$$
Integrating out $u$ (governed by its marginal distribution $G$) must yield the marginal cf for $X$:
$$\phi(t) = \mathbb{E}\left(\exp(itX)\right) =  \mathbb{E}_{U\sim G}\left(\mathbb{E}_{X_\sim F_u}\left(\exp(itX_u)\right)\right)=\int \phi(t,u) dG(u).\tag{*}$$
In the question, $U$ has a Bernoulli$(p)$ distribution, which makes the right hand side of $(*)$ a simple sum of two cfs. $X_0$ is an atom at zero (with characteristic function $\phi_0(t)=1$), and $X_1$ has a Poisson$(\lambda)$ distribution with cf 
$$\phi_\lambda(t) = \exp\left(\lambda\left(e^{it}-1\right)\right).\tag{**}$$
Therefore
$$\phi(t) = (1-p) + p\exp\left(\lambda\left(e^{it}-1\right)\right).$$
By definition, the cumulant generating function (cgf) is the logarithm of $\phi,$
$$\psi_F(t) = \log \phi_F(t).$$
Letting $\psi_\lambda$ be the cgf of the Poisson distribution given at $(**),$ we may write

$$\psi(t) = \log\left(1-p + p \exp(\psi_\lambda(t)\right).$$

The terms of its Maclaurin series $\psi(t) = \kappa_1 t + \frac{\kappa_2 t}{2!} + \cdots + \frac{\kappa_n t^n}{n!} + \cdots$ are the cumulants
$$(\kappa_1,\kappa_2,\kappa_3, \ldots) = \left(p\lambda,\ p\lambda(1+\lambda(p-1)),\ p\lambda (1+\lambda  (p-1) (\lambda  (2 p-1)-3)),\ \ldots\right).$$
You can obtain direct formulas by computing the logarithms of truncated MacLaurin series of the exponential.  (I haven't done the analysis to see whether there is some nice formula, but clearly $\kappa_{2n}$ and $\kappa_{2n+1}$ are polynomials of degree $n$ in $\lambda^2$ with few or no zero terms.)
