For a point process with independent and identically distributed (i.i.d) inter-renewals, with distribution $p(x)$, we observed $N$ points on $[0,T]$. What is the joint probability distribution function of random variables $x_1,x_2,\cdots,x_n$? Obviously, $x_1+x_2+\cdots+x_n=T$
For a Poisson point process, it is simply the probability distribution function of the event that we put $N$ points independently and randomly with respect to uniform distribution on $[0,T]$.
For a renewal point process, where the inter-renewal times are i.i.d with distribution $p(x)$ the distribution function is:
\begin{align} &F_{X_1,\dots,X_n}\left(t_1,\cdots,t_2\right)\\&=\mathbb{P}\left(x_1\leq t_1,\cdots,x_{n-1}\leq t_{n-1}|\{\sum_{i=1}^{n-1}x_i<T\} \cap \{\sum_{i=1}^{n}x_i>T\} \right)\\ &=\frac{\mathbb{P}\left(x_1\leq t_1,\cdots,x_{n-1}\leq t_{n-1}\cap\{\sum_{i=1}^{n-1}x_i<T\} \cap \{\sum_{i=1}^{n}x_i>T\} \right)}{\mathbb{P}\left(\{\sum_{i=1}^{n-1}x_i<T\} \cap \{\sum_{i=1}^{n}x_i>T\} \right)}\\ &=\frac{1}{c}{\mathbb{P}\left(x_1\leq t_1,\cdots,x_{n-1}\leq t_{n-1}\cap\{\sum_{i=1}^{n-1}x_i<T\} \cap \{\sum_{i=1}^{n}x_i>T\} \right)}\\ &=\frac{1}{c}\int_{x_1=0}^{min(T,t_1)}\int_{x_2=0}^{min(T-x_1,t_2)}\cdots\int_{x_n=T-x_1-x_2\cdots-x_{n-1}}^{\infty} p(x_1)p(x_2) \cdots p(x_n) d_{x_1} \cdots d_{x_n} \end{align}
To obtain the pdf, I have to take derivatives with respect to $t_1,\cdots,t_n$. I don't think it is going to be a clean result. Any idea?