Estimating the probability that one Poisson RV is larger than another 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 $\lambda \to \infty$?
 A: Regarding the asymptotic behavior when $\lambda \to \infty$:
To get an estimate one simply finds that the dominating element among
$$A_k= \mathbb{P}(Y=k+1)\mathbb{P}(X=k) = e^{-1-\lambda} \frac{\lambda^k}{k!(k+1)!}$$
is $A_{\sqrt{\lambda}}$ which gives roughly $e^{2\sqrt{\lambda}-\lambda}$. Probably there's also a Poly($\lambda$) factor here.
Oh, and $F'(\lambda)$ is simply $\mathbb{P}(X=Y)$ since the probability of adding 1 when increasing the intensity of a Poisson RV by $\epsilon$ is $\epsilon$. In the special case $\lambda=1$ we get
$$F'(1)=e^{-2} \sum_{k=0}^\infty \frac{1}{(k!)^2}$$
A: Numerically, $\lim_{\lambda \to 0} F^\prime(\lambda) = 1/e$. Heuristically this should be true because to have $X > Y$ when $\lambda$ is very small, the most likely case will be $X = 1, Y = 0$ by far; that occurs with probability $\lambda e^{-\lambda} e^{-1}$.
Maple gives an explicit formula for $F^\prime(1)$ involving sums; evaluating gives $F^\prime(1) = 0.3085083225$, and the Inverse Symbolic Calculator says this is somehow to $I_0(2) e^{-2}$. I'm not sure how to prove this at all but maybe knowing the answer helps?
A: An alternative way to see the $1/e$ is as follows.
Let $x=2\sqrt{\lambda}$. Recall that for small argument $0 < x \ll \sqrt{k+1}$ we have
$$I_k(x) \approx \frac{1}{\Gamma(k+1)}(x/2)^k$$
Using this, we see that
$$\sum_{k \ge 0} \lambda^{k/2}I_k(2\sqrt{\lambda}) \approx \sum_{k\ge 0}\frac{(x/2)^{2k}}{\Gamma(k+1)}.$$
This sum is nothing but $e^{x^2/4} = e^\lambda$. Multiplying with $e^{-\lambda-1}$ we obtain the said $1/e$ approximation.
Perhaps better approximations to $F$ can be obtained in a similar vein.
A: 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 $\lambda\rightarrow+\infty$.
