Time up to $n$th event in Poisson process distributed as $\frac{1}{2\lambda} \chi^2_{2n}$

Question

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 $$P(X \geq n) = \sum\limits^{+\infty}_{i = n} \frac{(\lambda t)^i \cdot e^{-\lambda t}}{i!} = P(\frac{1}{2\lambda} \cdot \chi^2_{2n} < t)$$

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 $2$.