Is there any known polynomial bound of the Erlang distribution? I'd like to say that, given $k$ and $\lambda$ with probability p the r.v. is going to be less than some value x.
bound of Erlang distribution
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0Let $X_k \sim \mathrm{Erlang}(k,\lambda)$. Then $\mathbb P(X_k \leq t) = \mathbb P(N(t,\lambda) \geq k)$ where $N(t,\lambda) \sim \mathrm{Pois}(\lambda t)$. Now you can use your favorite tail bound on the Poisson distribution. – 2011-11-10
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That is simply the cumulative distribution function, given in WP by $\gamma(k,k\lambda)/(k-1)! = 1-\sum_{n=0}^{k-1}\mathrm e^{-\lambda x}(\lambda x)^{n}/n! $, where $\gamma$ is the incomplete gamma function.
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0Why that constraint? If $k$ is large, you can use [Stirling approximation](http://en.wikipedia.org/wiki/Stirling_approximation). There are probably approximations for the incomplete gamma function too. – 2011-04-21