Probability that two independent Poisson random variables with same paramter are equal

Question

What is the probability that two independent Poisson random variables with parameter $\lambda<\infty$ have the same value?

My solution:

\begin{align} P = \sum_{i=0}^{\infty} \left(\frac{e^{-\lambda T} (\lambda T)^i}{i !} \right)^2 = e^{-2\lambda T} \sum_{i=0}^{\infty} \frac{ (\lambda T)^{2i}}{(i !)^2} \end{align}

I don't know how to continue from here. How can I evaluate the above summation? Can you help?

Answer 1

Let $X$ be Poisson-distributed with mean $\lambda_1$, and let $Y$, independent of $X$, be Poisson-distributed with mean $\lambda_2$. Then $X-Y$ has the Skellam distribution with parameters $\lambda_1$ and $\lambda_2$. In particular if $\lambda_1 = \lambda_2 = \lambda$, then

$$ P(X=Y) = P(X-Y = 0) = e^{-2\lambda} I_0(2\lambda)$$

where $I_0$ is the modified Bessel function of the first kind.