# Closed form of double-sum series

Does the following series have a closed-form (or at least simpler) expression:

$$\sum_{a=1}^{\infty} \left( \frac{x^a}{a!} \sum_{b=0}^{a-1} \frac{y^b}{b!} \right)$$

Since the inner summation would be empty for $a=0$, I suppose the outer summation could start at zero without changing the meaning, i.e.:

$$\sum_{a=0}^{\infty} \left( \frac{x^a}{a!} \sum_{b=0}^{a-1} \frac{y^b}{b!} \right)$$

This problem came up in computing the probability of a win in a football match with each team's goal scoring modeled as a Poisson process. The outer summation relates to the goals scored by team A (the winning team) and the inner summation relates to the (lesser number of) goals scored by team B (the losing team).

Thanks,

John

You can simplify a little since the inner summation $$\sum_{b=0}^{a-1} \frac{y^b}{b!}=\frac{\Gamma (a,y)}{\Gamma (a)}\,e^y$$ where appears the incomplete gamma function.
This makes that you are left with $$\sum_{a=1}^{\infty} \left( \frac{x^a}{a!} \sum_{b=0}^{a-1} \frac{y^b}{b!} \right)=e^y\,\sum _{a=1}^{\infty } \frac{ \Gamma (a,y)}{a! (a-1)!} x^a$$ which, at least to me, will not simplify further.
• Let $J_0(x) = \sum_{a \ge 0} \frac{x^{2a}}{a!^2}$ a Bessel function. Then $$e^{2x} = \sum_{a\ge 0, b \ge 0} \frac{x^ax^b }{a!b!} = \sum_{a= b} + \sum_{b < a}+\sum_{a > b} \frac{x^ax^b }{a!b!}= J_0(x)+2 \sum_{a \ge 0, 0 \le b < a} \frac{x^ax^b }{a!b!}$$ – reuns Oct 14 '17 at 5:30