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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.

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Let $X_k \sim \mathrm{Erlang}(k,\lambda)$. 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. – cardinal Nov 10 '11 at 23:27
up vote 1 down vote accepted

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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I know, but I wanted to see if there is an approximation that does not involve factorials – ACAC Apr 21 '11 at 20:32
Why that constraint? If $k$ is large, you can use Stirling approximation. There are probably approximations for the incomplete gamma function too. – Emre Apr 21 '11 at 20:36

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