TITLE: Estimating the probability that one Poisson RV is larger than another
QUESTION [7 upvotes]: Let $X$ and $Y$ be Poisson random variables with means $\lambda$ and $1$, respectively.  The difference of $X$ and $Y$ is a Skellam random variable, with probability density function
$$\mathbb P(X - Y = k) = \mathrm e^{-\lambda - 1}  \lambda^{k/2}  I_k(2\sqrt{\lambda}) =: S(\lambda, k),$$
where $I_k$ denotes the modified Bessel function of the first kind.  Let $F(\lambda)$ denote the probability that $X$ is larger than $Y$: $$F(\lambda) := \mathbb P(X > Y) = \sum_{k=1}^{\infty} S(\lambda, k) = \mathrm e^{-\lambda - 1} \sum_{k=1}^\infty \lambda^{k/2} I_k(2\sqrt{\lambda}).$$  According to Mathematica, the graph of the function $F$ looks like
(source: nyu.edu)
My question:Is there a closed-form expression for the function $F$?If not, what are $\lim_{\lambda \to 0} F'(\lambda)$ and $F'(1)$?  What is the asymptotic behavior as $\lambda \to \infty$?

REPLY [2 votes]: By simple computations : The definition of the modified Bessel function of the first kind yields
$$
I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k}
$$ 
so that we get (the sums transpositions are clearly allowed)
$$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum
$$ F'(\lambda) =  e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big)
= e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n
$$
we obtain the closed form
$$
F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}).
$$
One finally get  $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$
and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$
$$
I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big),
$$
that
$$
F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big)
$$
when $\lambda\rightarrow+\infty$.