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 $N$ 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 $N$ 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 $T$ '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.
I'll show the derivation for $n=1$, you can likely figure out how to extend this. Let $T_1$ be the time where the first particle is emitted and let $N_t$ be the number of particles emitted at time $t$. Note that $$\{T_1\leq t\}=\{N_t\geq 1\}$$ and hence $$\mathbb P[T_1\leq t]=\mathbb P[N_t\geq 1]=1-\mathbb P[N_t=0]=1-e^{-\lambda t}.$$ As you can see, $T_1$ follows an exponential distribution.
Now you should try to answer the questions:
What's the relationship between an exponential distribution and a Gamma distribution?
If $T_n$ denotes the time at which the $n$-th particle was emitted, what's the relationship between $T_n$ and $T_{n-1},\dots,T_1$?