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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.

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    $\begingroup$ From first glance it looks like the solution would be some sort of Bessel function if you group the $2k$ exponents. $\endgroup$
    – Dale C
    Commented Nov 5, 2020 at 14:35
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    $\begingroup$ *Modified Bessel function. Whether or not you consider this a "better analytical solution" is debatable though $\endgroup$
    – Dale C
    Commented Nov 5, 2020 at 14:40
  • $\begingroup$ @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). $\endgroup$ Commented Nov 5, 2020 at 14:59

1 Answer 1

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$X$ 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 $-1$).

Image example

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

comparison

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

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