2
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ To be a PPP means that for all Borelsets B we have $$\Pi(B) = Pois( \int_B e^{-t}dt)$$ and for all $A,B$ disjoint the random variables $$\Pi(A),\Pi(B)$$ are independent. $\endgroup$ – crankk Mar 21 '16 at 18:33
2
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ Thanks I guess this gives me a good hint $\endgroup$ – crankk Apr 6 '16 at 12:19
0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.