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

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?

• @G.Sassatelli Thanks. But they are independent. Mar 13, 2017 at 19:59
• Your sum should start at $0$. And are you looking for a closed-form? I doubt there is one. Mar 13, 2017 at 20:02
• Hint: The modified Bessel function $$I_0(t) = \sum_{i=0}^\infty \frac{4^{-k} t^{2k}}{(k!)^2}$$ Mar 13, 2017 at 20:07
• Do you want an exact answer (in which case you'll end up looking at Bessel functions) or an approximate one? If an approximate one, about how large is $\lambda$? Mar 13, 2017 at 20:10
• @MichaelLugo Yes, I need an exact answer. $\lambda$ may be very large, but it is finite. Mar 13, 2017 at 20:26

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.
• Well, that will depend on what $\lambda$ is. Mar 13, 2017 at 20:33
• The answer is $I_0(2 \lambda T)$. Mar 13, 2017 at 20:36