Probability of one Poisson variable being greater than another

Given two Poisson distributions with different λ values, if each were to produce a single random variable, is there closed-form expression for calculating the probability of one random variable being greater than the other?

Solution 1:

Take two poisson random variables $A$ and $B$ with means $\lambda_A$ and $\lambda_B$ respectively. We see that

$$P(A > B) = \sum_{k = 0}^{\infty} P(A > B | B = k)P(B = k)$$ $$=\sum_{k=0}^{\infty} P(A \geq k + 1)P(B = k) = \sum_{k= 0}^{\infty} \left(\sum_{l=k+1}^{\infty} \frac{\lambda_A^{l} e^{-\lambda_A}}{l!} \right)\frac{\lambda_B^k e^{-\lambda_B}}{k!}.$$

Generally, this is difficult to calculate for a general result, but the key idea behind this is to condition on the value of one of the variables.