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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?

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  • $\begingroup$ @G.Sassatelli Thanks. But they are independent. $\endgroup$
    – Susan_Math123
    Mar 13, 2017 at 19:59
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    $\begingroup$ Your sum should start at $0$. And are you looking for a closed-form? I doubt there is one. $\endgroup$
    – Clement C.
    Mar 13, 2017 at 20:02
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    $\begingroup$ Hint: The modified Bessel function $$I_0(t) = \sum_{i=0}^\infty \frac{4^{-k} t^{2k}}{(k!)^2} $$ $\endgroup$
    – Robert Israel
    Mar 13, 2017 at 20:07
  • $\begingroup$ 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 $\lambda$? $\endgroup$
    – Michael Lugo
    Mar 13, 2017 at 20:10
  • $\begingroup$ @MichaelLugo Yes, I need an exact answer. $\lambda$ may be very large, but it is finite. $\endgroup$
    – Susan_Math123
    Mar 13, 2017 at 20:26

1 Answer 1

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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.

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  • $\begingroup$ Thanks. Can you add the final value? $\endgroup$
    – Susan_Math123
    Mar 13, 2017 at 20:32
  • $\begingroup$ Well, that will depend on what $\lambda$ is. $\endgroup$
    – Michael Lugo
    Mar 13, 2017 at 20:33
  • $\begingroup$ The answer is $I_0(2 \lambda T)$. $\endgroup$
    – Susan_Math123
    Mar 13, 2017 at 20:36

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