I am trying to find the distribution of the 13th event of a Poisson process with $\lambda = 4$.
My first approach is to look at the distribution of the first event given that no event occurred before.
$$P\{T1 < s|N(t) = 1\} = P \{T1 < s,N(t) = 1\}/P\{N(t) = 1\}$$ $$ = P\{1\text{ event in }[0, s), 0\text{ events in }[s, t ]\}/P\{N(t) = 1\} $$ $$ = P\{1\text{ event in } [0, s)\}P\{0\text{ events in }[s, t ]\}/P\{N(t) = 1\}$$ $$ = \lambda s e^{−\lambda s} e^{−\lambda(t−s)}/\lambda t e^{−\lambda t} $$ $$= s/t \sim U(0,t) $$
To generalise I take joint probability distribution that there are $n$ events given $N(t)=n$
$P(T_1 < s_1, T_2 < s_2,..., T_n < s_n \mid N(t) = n)$ and then apply Poisson distribution as above.
I know that the result be $n!/t^n$, but I do not understand how to get rid of the different intervals that remain.
It makes intuitive sense to be that as each event is distributed it has a $1/t$ chance to occur within the interval $(0,t]$ and that there should exists $n!$ ways of attributing an event within that time, but the proof is not clear to me.