density of an ordered Poisson-Point-Process Consider the PPP $\Pi$ on $R$ with intensity $e^{-t} dt$. Since for any $x$ we have 
$$ \Pi (x,\infty) = Pois( \int_x^\infty e^{-t} dt) < \infty$$,
we can order the points of $\Pi$ ,e.g. $Y_1 > Y_2 > …$.
What is the joint density of $(Y_1,…,Y_k)$? I can set up the joint distribution function:
$$F(x_1,…,x_k) = P( Y_1 \le x_1,…,Y_k \le x_k) = P( \Pi(x_1,\infty)=0,…,\Pi(x_k,\infty) \le k-1),$$
but this will lead to a very complicated expression…
Any other ideas? There should be an easy expression for that...
 A: Given that $E_n := \{ |\Pi'(0, 1)| = n \}$ with $n \geq k$, points in $\Pi'(0, 1)$ have the same distribution as $n$ uniformly (and independently) chosen points on $(0, 1)$. This means that the first $n$ arrival times $X_1, \cdots, X_n$ have the order statistics of $n$ i.i.d. uniform variables on $(0, 1)$. Thus it follows that for $0 < x_1 < \cdots < x_k < 1$, we have
$$ f_{X_1, \cdots, X_k | E_n} (x_1, \cdots, x_k) = \frac{n!}{(n-k)!} (1 - x_k)^{n-k}. $$
Then given $E = \cup_{n \geq k} E_k = \{ |\Pi'(0, 1)| \geq k \}$, we get
\begin{align*}
f_{X_1, \cdots, X_k | E} (x_1, \cdots, x_k)
&= \sum_{n \geq k} f_{X_1, \cdots, X_k | E_n} (x_1, \cdots, x_k) \Bbb{P}(E_n \mid E) \\
&= \sum_{n\geq k} \frac{n!}{(n-k)!} (1 - x_k)^{n-k} \cdot \frac{1}{n!} e^{-1} \cdot \frac{1}{\Bbb{P}(E)} \\
&= \frac{e^{-x_k}}{\Bbb{P}(E)}.
\end{align*}
A: Okay I have another idea:
let $\Pi'$ be the standard homogenous PPP (intensity $\lambda$ (Lebgeuemeasure)) on $(0,\infty)$. Consider the map
$$T: (0,\infty) \to R,\quad x \mapsto - log(x)$$
Since T is a proper map the process $T(\Pi') = \sum_{x \in \Pi} \delta_{T(x)}$ is a PPP as well. Its intensity is $T(\lambda)$ and since
$$ T(\lambda)(a,b) = \lambda( (-log)^{-1}(a),(-log)^{-1}(b)) = \lambda(e^{-a},e^{-b}) = \int_a^b e^{-t} dt $$
it follows that $T(\Pi')$ is equal to $\Pi$ in distribution. If now we order $\Pi'$ e.g. $X_1 < X_2 < …$, it's left to determine the joint density of $(X_1,…X_k)$ (we have $Y_i = T(X_i)$). If $Z_1,…,Z_k$ are independent, exponential(1) distributed variables we have that 
$$(X_1,…,X_k) = (Z_1,Z_1+Z_2,…,Z_1+…+Z_k)$$
in distribution. So what is the joint density of the first summands of a sum of independent, exponential distributed variables? Will this lead to an easier expression? 
A: Okay so the answer is:
$$f_{Y_1,...,Y_k}(x_1,...,x_k) = e^{-(x_1+...+x_k)} e^{-e^{-x_k}} 1_{x_k < ... < x_1}$$
If someone is interested in details just leave a comment. Thanks!
