TITLE: Comparing Poisson variables
QUESTION [7 upvotes]: Let $X$ and $Y$ be two Poisson variables with different mean. 
Is there a better (as in more concise or numerically faster) way to compute $P(X\leq Y)$ than using
$$ P(X\leq Y) = \sum_{y=0}^\infty P(Y=y) P(X\leq y) $$

REPLY [2 votes]: Rasholnikov has shown you a method using modified Bessel functions.  You could also get that sort of result from a Skellam distribution, which is the difference between two Poisson distributions.  
If you want a quick and simple approximation for $X \sim \text{Poiss}(\lambda)$ and $Y \sim \text{Poiss}(\mu)$ then 
$$\text{Pr}(X \leq Y) \approx \Phi\left( \frac{1/2 - (\lambda - \mu )}{\sqrt{\lambda + \mu}} \right)$$ 
where $\Phi()$ is the cumulative distribution function of a standard normal distribution.  The $1/2$ is there as a continuity adjustment to deal with the possibility that $X=Y$. 
For example with $\lambda =5$ and  $\mu =10$, the true result is about 0.9256 while the approximation gives 0.9222.