Estimating the probability that one Poisson RV is larger than another

Question

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 $\lambda \to \infty$?

Answer 1

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 $$F'(1)=e^{-2} \sum_{k=0}^\infty \frac{1}{(k!)^2}$$

Answer 2

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?

Answer 3

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.

Answer 4

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 $\lambda\rightarrow+\infty$.