Let's assume that a number X of some events over time $t$ is modeled by Poisson distribution with rate $\lambda$ (here, it's rate, not mean): $$ X \sim Poisson(\lambda \cdot t) ~~~~ (\lambda t ~\text{as whole denotes mean}). $$

Now, I'm interested in inverse sampling: for given number of events $n$, what's the time $t_n$ up to an occurrence of $n$th event. As I remember this, time $t_n$ is distributed as $$ t_n \sim \frac{1}{2\lambda} \cdot \chi^2_{2n}~~~~(\chi^2 ~\text{with 2n degrees of freedom})$$

If I remember correctly, this property was described by G.A. Barnard, but I cannot find it any more.

Could someone give me a hint how to prove that by myself or remind me the publication title?

Actually, I'm interested in proving that $$P(X \geq n) = \sum\limits^{+\infty}_{i = n} \frac{(\lambda t)^i \cdot e^{-\lambda t}}{i!} = P(\frac{1}{2\lambda} \cdot \chi^2_{2n} < t)$$


1 Answer 1


If I understand you well, you only need the fact that the waiting times between events are exponential so the total waiting time is the sum of exponentials which follows the gamma distribution. A proper standardization from here will give you your chi-squared distribution, i.e. the gamma distribution with shape parameter equal to $n$ and scale parameter equal to $2$.

  • $\begingroup$ Great answer. By the way, I would take an advantage and ask you how can I use this property $P(X \geq n) = P(\frac{1}{2\lambda} \chi^2_{2n} < t)$ with rather very high number of degrees of freedom, e.g. $2n = 10,000$. I'm aware that in the straight manner it won't work, but do you think that Wilson-Hilferty transformation would be sufficient to run this method? $\endgroup$ Feb 15, 2016 at 15:13
  • $\begingroup$ I'm afraid I cannot help you with that. Why not try a small simulation? $\endgroup$
    – JohnK
    Feb 15, 2016 at 15:18
  • 1
    $\begingroup$ The W-H transformation leads to excellent approximations of $\chi^2$ distributions, so I would expect it to be a good approach. $\endgroup$
    – whuber
    Feb 15, 2016 at 17:15

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