Is there a nice formulation for the convolution of a Bernoulli and Poisson random variable? Are there any special properties about this convolution that permit an easy formulation for the resulting R.V? I feel like there should be, given that:


*

*The Poisson process can be thought of as the continuous analog of the Bernoulli process.

*There are nice formulas for the convolution of each of these processes, e.g. the sum of Bernoulli R.V.s with same probability of success is the binomial distribution, and the sum of Poisson R.V.s is another Poission RV.


I note that from this Q/A that such a formulation does not exist if we consider a binomial R.V. instead of a Bernoulli R.V.
As noted in the comments, it is easy to write down the convolution of these R.V.s:
Let $P(X = k)$ be Poisson distributed with mean $\lambda$. Let $Q(X = k)$ denote the shifted R.V of $P(X = k)$ such that $Q(X = k) = P(X = k - 1)$.
Then, the R.V. $R(X = k)$ we are after has mass function given by:
$R(X = k) = pQ(X = k-1) + (1-p)P(x = k)$.
where $p$ is the probability of success for the Bernoulli R.V.
 A: If $X \sim \text{Bernoulli}(p)$ and $Y \sim \text{Poisson}(\lambda)$
then $Z = X+Y$ takes on values $0, 1, 2, \ldots$.
It follows straightforwardly that
\begin{align}P\{Z = k\} &= \begin{cases}P\{X=0, Y=0\},& k = 0,\\
 \\
P\{X=0, Y=k\} + P\{X=1, Y=k-1\}, & k \geq 1,\end{cases}
\end{align}
No further simplification is possible unless the joint
probabilities have specific properties. For example, if
$X$ and $Y$ are independent random variables, then the joint
probabilities factor into the product of the marginal
probabilities and we have that
\begin{align}P\{Z = k\} &= \begin{cases}P\{X=0\}P\{Y=0\},& k = 0,\\
 \\
P\{X=0\}P\{Y=k\} + P\{X=1\}P\{Y=k-1\}, & k \geq 1,\end{cases}\\
 \\
&= \begin{cases}
(1-p)e^{-\lambda}, & k = 0,\\
 \\
(1-p)e^{-\lambda}\frac{\lambda^{k}}{k!} + p\cdot e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!}, & k \geq 1.\end{cases}
\end{align}
I suppose that we could write that last line as 
$$(1-p)e^{-\lambda}\frac{\lambda^{k}}{k!} + p\cdot e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!} 
= \left.\left.e^{-\lambda}\frac{\lambda^{k-1}}{(k-1)!}
\right[p + (1-p)\frac{\lambda}{k} \right]$$
and count it as a simplification. Or, perhaps even better,
use the version in OP Alex's comment to combine both
lines and write
$$P\{Z = k\} = \left.\left.e^{-\lambda}\frac{\lambda^{k}}{k!} \right[\frac{pk}{\lambda} + 1-p\right], k = 0, 1, 2, \ldots $$
but I don't think this sort of stuff is referred to as a
deformation of a Poisson random variable.
