Equality of outcomes in two Poisson events I have a Poisson process with a fixed (large) $\lambda$.  If I run the process twice, what is the probability that the two runs have the same outcome?
That is, how can I approximate
$$f(\lambda)=e^{-2\lambda}\sum_{k=0}^\infty\frac{\lambda^{2k}}{k!^2}$$
for $\lambda\gg1$?  If there's a simple expression about $+\infty$ that would be best, but I'm open to whatever can be suggested.

So I've come across this problem before, and ran into the Bessel-function exact solution; but I figured that there must be some heuristic explanation. Here's the best I was able to come up with.
Let $X$ and $Y$ be independent Poisson with mean $\lambda$ (and therefore variance $\lambda$). You want $P(X=Y)$.
Now, $X$ and $Y$ are both approximately normal with mean $\lambda$ and variance $\lambda$, if $\lambda$ is large.
Let $U = X + \lambda$ and $V = 3 \lambda - Y$. Then $U$ and $V$  are both approximately normal with mean $2 \lambda$ and variance $\lambda$. But these are the mean and variance of a binomial distribution with parameters $4\lambda$ and $1/2$.   So $U$ and $V$ are both approximately binomial with parameters $4\lambda$ and $1/2$; thus $U + V$ is approximately binomial with parameters $8\lambda$ and $1/2$. But by construction $U + V = 4 \lambda + X - Y$. So $X = Y$ if and only if $U + V = 4 \lambda$.
The probability that $U + V$ is $4 \lambda$ is approximately ${8 \lambda \choose 4 \lambda} 2^{-8\lambda}$. And by Stirling's approximation this is of order $(4 \pi \lambda)^{-1/2}$, which is what we wanted.
(The fact that the probability should decay like $\lambda^{-1/2}$ is fairly easy to see -- since $X$ and $Y$ have standard deviation $\lambda^{1/2}$, the number of values they take ``regularly'' is of order $\lambda^{1/2}$, so the chance of collision is of order $\lambda^{-1/2}$. The constant $(4\pi)^{-1/2}$ is in my opinion much harder to guess.)
