Let $X_i$ $i=1,\dots,n$ be independent poisson random variables with $X_i \sim \text{Poisson}(\lambda_i)$
then we define $X = \max_i X_i$
how does $X$ distribute?
Is easy to see that
$$\mathbb{P}(X \leq k) = \prod_{i=1}^n \sum_{j=1}^k \frac{e^{-\lambda_i}}{j!}\lambda_i^j$$
But i don't know how to find a close formula.
Any help will be appreciated.
Best Answer
In the case of a common parameter, see, e.g.,
http://arxiv.org/pdf/0903.4373.pdf
One can show that
$\Pr(X \leq k) = (\Pr(X_1 \leq x))^n = (Q(k+1, \lambda))^n$,
where $Q(k+1, \lambda) \equiv \frac{\Gamma(k+1, \lambda)}{\Gamma(k+1)}$,
and $\Gamma(x, \lambda)$ is the (upper) incomplete gamma function.
It's also possible to derive a bound on the maximum of the expectation which does not depend on the independence. See problem 2.18 in "Concentration Inequalities: A Nonasymptotic Theory of Independence" by Boucheron, Lugosi, and Massart.