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.

Fourier transforms yield a fully rigorous proof. 
First recall that, as explained here, for every integer valued random variable $Z$,
$$
P(Z=0)=\int_{-1/2}^{1/2}E(\mathrm{e}^{2\mathrm{i}\pi tZ})\mathrm{d}t.
$$
Hence, if $X_\lambda$ and $Y_\lambda$ are independent Poisson random variables with parameter $\lambda$,
$$
f(\lambda)=P(X_\lambda=Y_\lambda)=\int_{-1/2}^{1/2}E(\mathrm{e}^{2\mathrm{i}\pi tX_\lambda})E(\mathrm{e}^{-2\mathrm{i}\pi tY_\lambda})\mathrm{d}t.
$$
For Poisson distributions, one knows that $E(s^{X_\lambda})=\mathrm{e}^{-\lambda(1-s)}$ for every complex number $s$. This yields
$$
f(\lambda)=\int_{-1/2}^{1/2}\mathrm{e}^{-2\lambda(1-\cos(2\pi t))}\mathrm{d}t=\int_{-1/2}^{1/2}\mathrm{e}^{-4\lambda\sin^2(\pi t)}\mathrm{d}t.
$$
Consider the change of variable $u=2\pi\sqrt{2\lambda}t$. One gets
$$
f(\lambda)=\frac1{\sqrt{4\pi\lambda}}\int_\mathbb{R} g_\lambda(u)\mathrm{d}u,
$$
with
$$
g_\lambda(u)=\frac1{\sqrt{2\pi}}\mathrm{e}^{-4\lambda\sin^2(u/\sqrt{8\lambda})}\,[|u|\le\pi\sqrt{2\lambda}].
$$
When $\lambda\to+\infty$, $g_\lambda(u)\to g(u)$ where $g$ is the standard Gaussian density, defined by
$$
g(u)=\frac1{\sqrt{2\pi}}\mathrm{e}^{-u^2/2}.
$$
Furthermore, the inequality 
$$4\lambda\sin^2(u/\sqrt{8\lambda})\ge2u^2/\pi^2,
$$ 
valid for every $|u|\le\pi\sqrt{2\lambda}$, shows that the functions $g_\lambda$ are uniformly dominated by an integrable function. Lebesgue dominated convergence theorem and the fact that $g$ is a probability density yield finally
$$
\int_\mathbb{R} g_\lambda(u)\mathrm{d}u\to1,\qquad\text{hence}\ \sqrt{4\pi\lambda}f(\lambda)\to1.
$$
