density of an ordered Poisson-Point-Process

Question

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...

Answer 1

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*}

Answer 2

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?

Answer 3

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!