Let $X \sim Pois(\lambda)$ and $Y \sim Pois(\mu)$ be independent random variables that follow two different Poisson distributions, with $\lambda < \mu$, then we have $P(X-Y \geq 0) \leq e^{-(\sqrt\mu-\sqrt\lambda)^2}$, as demonstrated on the Wikipedia with the following link https://en.wikipedia.org/wiki/Poisson_distribution#Poisson_races.
My question is what is the lower-bound and the upper bound of the probability $P(X-Y > 0)$.
The bounds provided in the Wikipedia article
$$\frac{e^{-(\sqrt{\mu} - \sqrt{\lambda})^2}}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2 \sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4 \lambda \mu} \le \Pr[X - Y \ge 0] \le e^{-(\sqrt{\mu} - \sqrt{\lambda})^2} \tag{1}$$
presumably can be modified as desired by noting $$\Pr[X - Y \ge 0] = \Pr[X - Y = 0] + \Pr[X - Y > 0], \tag{2}$$ then computing $$\Pr[X = Y] = \sum_{k=0}^\infty \Pr[X = k]\Pr[Y = k] = e^{-(\lambda + \mu)} \sum_{k=0}^\infty \frac{(\lambda \mu)^k}{(k!)^2}. \tag{3}$$ This sum can be written in terms of a modified Bessel function of the first kind, or a hypergeometric function:
$$\Pr[X = Y] = e^{-(\lambda + \mu)} I_0 (2 \sqrt{\lambda \mu}) = e^{-(\lambda + \mu)} {}_0 F_1 (1; \lambda \mu). \tag{4}$$ Subtracting $(4)$ from $(1)$ results in a bound for $\Pr[X - Y > 0]$. If $(4)$ is undesirable because of the use of non-elementary functions, a looser lower bound can be obtained by truncating the series $(3)$ to a finite number of terms.
