Sum of Binomial and Poisson random variables

Question

If we have two independent random variables $X_1 \sim \mathrm{Binom}(n,p)$ and $X_2 \sim \mathrm{Pois}(\lambda)$, what is the probability mass function of $X_1 + X_2$?

NB This is not homework for me.

Answer 1

You will end up with two different formulas for $p_{X_1+X_2}(k)$, one for $0 \leq k < n$, and one for $k \geq n$. The easiest way of doing this problem is to compute the product of $\sum_{i=0}^n p_{X_1}(i)z^k$ and $\sum_{j=0}^{\infty}p_{X_2}(j)z^j$. Then, $p_{X_1+X_2}(k)$ is the coefficient of $z^k$ in the product. No simplification of the sums is possible.

Answer 2

Giving the closed formula in terms of generalized hypergeometric functions (GHF) hinted at in other answers (the GHF in this case is really only a finite polynomial, so is a shorthand for the finite sum.) I used maple to sum the convolution, with this result: $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(X_1+X_2=k)= \sum_{x_1=0}^{\min(n,k)} \binom{n}{x_1} p^{x_1}(1-p)^{n-x_1} e^{-\lambda} \frac{\lambda^{k-x_1}}{(k-x_1)!}= {\frac { \left( 1-p \right) ^{n}{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \Gamma \left( k+1 \right) } {\mbox{$_2$F$_0$}(-k,-n;\,\ ;\,-{\frac {p}{ \left( p-1 \right) \lambda}})} } $$

Answer 3

Dilip Sarwate stated 7 years ago that no simplification is possible, although this has been challenged in comments. However, I think it is useful to note that even without any simplification the computation is quite straightforward in any spreadsheet or programming language.

Here is an implementation in R:

# example parameters
n <- 10
p <- .3
lambda <- 5

# probability for just a single value
x <- 10  # example value
sum(dbinom(0:x, n, p) * dpois(x:0, lambda))

# probability function for all values
x0  <- 0:30   # 0 to the maximum value of interest
x   <- outer(x0, x0, "+")
db  <- dbinom(x0, n, p)
dp  <- dpois(x0, lambda)
dbp <- outer(db, dp)
aggregate(as.vector(dbp), by=list(as.vector(x)), sum)[1:(max(x0)+1),]