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At the instant $t = 0$ a certain radioactive focus starts emitting particles. The infinitesimal probability that the focus emits a particle in the differential interval is $\lambda dt$. Let $N_t$ also be the random variable 'number of particles emitted by the focus in the time interval $[0,t]$'. Hence, we have that the probability distribution that $N_t$ follows is a Poisson one:

$$P_n = e^{-\lambda t} \frac{(\lambda t)^n}{n!}$$

However, imagine we wanted to calculate the probability distribution of the continuous random variable $T_n$ 'moment $t$ at which the focus emits the nth particle'.

How would we calculate this probability distribution? How would it be related to the Poisson one above?

I know it has to be a gamma distribution, but I don't know how to get to that conclusion.

Many thanks.

Attempt:

$$P[T_n \leq t]=P[N_t\geq n]=1-P[N_t\leq n-1] = 1 - e^{-\lambda t} \sum_{i = 1}^{n-i = 0} \frac{(\lambda t)^{n-i}}{(n-i)!}$$

How could I get to the following expression?

$$\rho_n(t)= \frac{1}{(n-1)!}\lambda^nt^{n-1}e^{-\lambda t}$$

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    $\begingroup$ Hogg, Tannis & Zimmerman (I'm looking at 10th ed.) have a nice/accessible explanation for this connection. It is §3.2 (on pp 100-102). $\endgroup$
    – Gregg H
    Commented Feb 23, 2023 at 15:59
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    $\begingroup$ Welcome to CV! It suffices to show the waiting time between successive events is Exponential, because the sum of $n$ independent Exponentials has a Gamma$(n)$ distribution. I derive this fundamental result from first principles at stats.stackexchange.com/questions/214421. For the sum of Exponentials see stats.stackexchange.com/questions/577324 or stats.stackexchange.com/questions/72479 for example (Exponential variables have Gamma$(1)$ distributions.) $\endgroup$
    – whuber ♦
    Commented Feb 23, 2023 at 17:49
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    $\begingroup$ Thank you both @GreggH and @whuber! I've analysed both references and I understand the derivation. $\endgroup$
    – user9867
    Commented Feb 23, 2023 at 18:28
  • $\begingroup$ If you differentiate $P(T_n\le t)$ w.r.t. $t$, you get a telescoping sum (see en.wikipedia.org/wiki/Telescoping_series) which simplifies to $\rho_n(t)$. $\endgroup$
    – Jarle Tufto
    Commented Feb 23, 2023 at 19:50

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