Time pdf of number of poisson proscces events

Question

I know that the time between two events distributes exponentially with parmeter lambda, but, how will the time distributes between n events for example? I would guess there should be a simple connection like pdf of exp(lambda/n) but I cant get something mathematicly (my intuistion may be wrong and there is not such connection)

Answer 1

Let $\{N(t):t\geqslant0\}$ be a Poisson process with jump times $\{T_n:n=1,2,\ldots\}$. Then $T_n$ has an Erlang distribution with parameters $n$ and $\lambda$; i.e. $T_n$ has density $$ f_n(x) = \lambda\frac{(\lambda x)^{n-1}}{(n-1)!}e^{-\lambda x}\mathsf 1_{(0,\infty)}(x).\tag1 $$ We show this by induction. For $n=1$, the distribution of $T_1$ is exponential with rate $\lambda$, and hence $$ f_1(x) = \lambda e^{-\lambda x}\mathsf 1_{(0,\infty)}(x) = \lambda(\lambda x)^{1-1}e^{-\lambda x}\mathsf 1_{(0,\infty)}. $$ Assume now that $f_n(x) = \lambda(\lambda x)^{n-1}e^{-\lambda x}\mathsf 1_{(0,\infty)}(x)$ for some positive integer $n$. Then we compute the density of $f_{n+1}$ by convolution: \begin{align} f_{n+1}(x) &= f_n\star f_1(x)\\ &= \int_{\mathbb R} f_n(y)f_1(x-y)\ \mathsf dy\\ &= \int_0^x \lambda\frac{(\lambda y)^{n-1}}{(n-1)!}e^{-\lambda y}\lambda e^{-\lambda (x-y)}\ \mathsf dy\\ &= \frac{\lambda^2 e^{-\lambda x}}{(n-1)!} \int_0^x (\lambda y)^{n-1}\ \mathsf dy\\ &=\frac{\lambda^2 e^{-\lambda x}}{(n-1)!}\left(\frac{(\lambda x)^n}{n\lambda}\right)\\ &= \lambda\frac{(\lambda x)^n}{n!}e^{-\lambda x}. \end{align} Since $n$ was arbitrary, $(1)$ holds for all positive integers $n$.