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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?

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    $\begingroup$ The distribution has density $$c_np(x_1)p(x_2)\cdots p(x_{n-1})q(T-x_1-\cdots-x_{n-1})\mathbf 1_{x_1+\cdots+x_{n-1}<T}\prod_{k=1}^{n-1}\mathbf 1_{x_k>0},$$ with respect to the Lebesgue measure on $\mathbf R^{n-1}$, with $$q(x)=\int_x^\infty p(y)dy,$$ and $$\frac1{c_n}=\iint\cdots\int p(x_1)p(x_2)\cdots p(x_{n-1})q(T-x_1-\cdots-x_{n-1})\mathbf 1_{x_1+\cdots+x_{n-1}<T}\prod_{k=1}^{n-1}\mathbf 1_{x_k>0}dx_k.$$ $\endgroup$ – Did Jan 19 '16 at 7:34
  • $\begingroup$ Thanks for that. Can you give me more details why that is true? $\endgroup$ – Susan_Math123 Jan 19 '16 at 17:30
  • $\begingroup$ That is a very general answer and I need more details. Can you provide that please? $\endgroup$ – Susan_Math123 Apr 6 '16 at 3:50
  • $\begingroup$ "That is a very general answer" Quite the contrary, the answer in my comment is exactly suited to the frame of the question. "and I need more details" Please explain. $\endgroup$ – Did Apr 6 '16 at 4:54
  • $\begingroup$ Observing $n$ points on $[0,T]$ is the same as requiring that $x_n \leq T < x_{n+1}$. So the obvious conclusion that $x_1 + \ldots + x_n = T$ is not so obvious for me. $\endgroup$ – D. Thomine May 2 '16 at 23:15

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