# Probability of compound Poisson process

Let $$X$$ be a compound Poisson process with rate $$\lambda$$ and increments $$Y_i = \pm 1$$ with probability $$\frac{1}{2}$$. Find $$P(X(t) = 0)$$.

I tried conditioning on $$N(t)$$: $$P(X(t) = 0) = P(\sum\limits_{i=1}^{N(t)}Y_i=0) = \sum\limits_{k=0}^{\infty}P(\sum\limits_{i=1}^{k}Y_i = 0\mid N(t) = k)P(N(t) = k) = \sum\limits_{k=0}^{\infty}P(\sum\limits_{i=1}^{k}Y_i = 0)P(N(t) = k)$$ Here it is worth nothing that the sum of the $$Y_i$$ will never reach 0 when $$k$$ is odd, also if I interpret each $$Y_i$$ as a Bernoulli trial, getting the sum of those to $$0$$ is equivalent to half of my trials being a success, so finally considering the sum as binomial variable I got: $$P(X(t) = 0) = \sum\limits_{k=0}^{\infty}P(\sum\limits_{i=1}^{2k}Y_i=0)P(N(t) = 2k) = \sum\limits_{k=0}^{\infty}\binom{2k}{k}\frac{1}{2^{2k}}e^{-\lambda t}\frac{(\lambda t)^{2k}}{2k!}\\ = e^{-\lambda t}\sum\limits_{k=0}^{\infty}\frac{(\lambda t)^{2k}}{k!^2 2^{2k}}$$ And here I have no idea how to proceed to get a better analytic solution. Any help is greatly appreciated, thank you for reading.

• From first glance it looks like the solution would be some sort of Bessel function if you group the $2k$ exponents. Commented Nov 5, 2020 at 14:35
• *Modified Bessel function. Whether or not you consider this a "better analytical solution" is debatable though Commented Nov 5, 2020 at 14:40
• @DaleC I'm not familiar with those but the first form of the modified Bessel's does look like the series (from what I saw in wikipedia). Commented Nov 5, 2020 at 14:59

$$X$$ will be distributed as a Skellam distribution.

### Intuition

You can view it intuitively as following. For a Poisson process on some piece of length $$L$$, you randomly designate each event as $$Y_i = +1$$ or $$Y_i = -1$$ (in the image below this is shown as black/white circles on a line).

Effectively this is the same as generating two independent Poisson processes each on a piece of length $$L/2$$ and then mixing the points (you can verify that this correct by the following thought: The sum of two Poisson variables is another Poisson variable, and each point will have 1/2 probability of being $$+1$$ or $$-1$$).

So the number of $$Y_i = +1$$ and the number of $$Y_i = -1$$ are two Poisson variables with rate $$\lambda/2$$ and the difference of the two is a Skellam distribution.

### Computational

With the code below we can verify our intuition by comparing the Skellam distribution with a simulation

n_sim <- 10^4
lambda <- 20

### draw Poisson variables
P <- rpois(n_sim, lambda)
### for each Poisson variable compute sum of Binomial variables
X <- sapply(P, function(n) sum(rbinom(n,1,0.5)*2-1))
hist(X, breaks = c(-20:21)-0.5, freq = 0,
main = "comparing simulation (histogram) \n with Skellam distribution (line)")
x <- c(-20:20)
lines(x,skellam::dskellam(x, lambda1 = lambda/2, lambda2 = lambda/2))

• This is pretty dope because it also tells me to where that series converges. Thank you! Commented Nov 15, 2020 at 0:29