1
$\begingroup$

Let $(\mathcal{X},\mathcal{H},\mu)$ be a $\sigma$-finite measure space.

Suppose that $0<\mu(\mathcal{X})<\infty$. Let $(N;X_j:j=1,2,\ldots)$ be independent on a probability space $(\Omega,\mathcal{F},P)$ with each $X_j$ having distribution $\mu/\mu(\mathcal{X})$ on $(\mathcal{X},\mathcal{H})$ and $N$ having a Poisson distribution with mean $\mu(\mathcal{X})$.

Set $$V(\omega)= (X_j(\omega):j\leq N(\omega))$$ a multi set of members of $\mathcal{X}$.

How can I show that this construction gives a Possion point process $V$ in $(\mathcal{X},\mathcal{H})$ with intensity $\mu$.

Here I note that $V(\omega)=\emptyset$ if $N(\omega)=0$ and I understand that in general, a point process is a random variable $N$ from some probability space $(\Omega,\mathcal{F},P)$ to a space of counting measures on ${\bf R}$, say $(M,\mathcal{M})$. So each $N(\omega)$ is a measure which gives mass to points $$ \ldots < X_{-2}(\omega) < X_{-1}(\omega) < X_0(\omega) < X_1(\omega) < X_2(\omega) < \ldots $$ of ${\bf R}$ (here the convention is that $X_0 \leq 0$. The $X_i$ are random variables themselves, called the points of $N$.

The intensity of a point process is defined to be $$ \lambda_N = {\bf E}[N(0,1]]. $$

It´s really hard for me this problem, could someone help me pls.

Thanks for your time and help.

$\endgroup$
  • $\begingroup$ To start, show that if $B\in\mathcal H$ with $\mu(B)<\infty$, then the random variable $M(B):=\#\{j\le N: X_j\in B\}$ has the Poisson distribution with parameter $\mu(B)$. Do this by first conditioning on the value of $N$. $\endgroup$ – John Dawkins Nov 20 '17 at 23:57
1
$\begingroup$

Technically, you should put $M(\omega):=\sum_{v\in V(\omega)}\delta_v=\sum_{j=1}^{N(\omega)}\delta_{X_j(\omega)}$, and this random variable will be a Poisson point process of intensity $\mu$. To prove this, we need to show that if $A_1,\ldots,A_n\in\mathcal H$ are disjoint, then $$\mathbf P(M(A_i)=k_i,1\le i\le n)=\prod_{i=1}^ne^{-\mu(A_i)}\frac{\mu(A_i)^{k_i}}{k_i!}.$$ Break up the problem into steps. First, set $k:=\sum_{i=1}^nk_i$ and $A:=\bigcup_{i=1}^nA_i$. For $m\ge k$, conditioned on $N=m$, the vector $$\Big(M(A_1),\ldots,M(A_n),M(\mathcal X\setminus A)\Big)$$ is multinomial with $m$ trials and respective event probabilities $\frac{\mu(A_1)}{\mu(\mathcal X)},\ldots,\frac{\mu(A_n)}{\mu(\mathcal X)},\frac{\mu(\mathcal X\setminus A)}{\mu(\mathcal X)}$. Thus, $$\mathbf P(M(A_i)=k_i,1\le i\le n|N=m)=\frac{m!}{k_1!k_2!\ldots k_n!(m-k)!}\left(\frac{\mu(A_1)}{\mu(\mathcal X)}\right)^{k_1}\ldots\left(\frac{\mu(A_n)}{\mu(\mathcal X)}\right)^{k_n}\left(\frac{\mu(\mathcal X\setminus A)}{\mu(\mathcal X)}\right)^{m-k}\\ =m!(\mu(\mathcal X))^{-m}\prod_{i=1}^n\frac{\mu(A_i)^{k_i}}{k_i!}\cdot\frac{\mu(\mathcal X\setminus A)^{m-k}}{(m-k)!}.$$ Since $\mathbf P(N=m)=e^{-\mu(\mathcal X)}\frac{\mu(\mathcal X)^m}{m!}=\frac{\mu(\mathcal X)^m}{m!}\prod_{i=1}^ne^{-\mu(A_i)}\cdot e^{-\mu(\mathcal X\setminus A)}$ \begin{align*} \mathbf P(M(A_i)=k_i,1\le i\le n, N=m) &=\mathbf P(M(A_i)=k_i,1\le i\le n|N=m)\mathbf P(N=m)\\ &=\prod_{i=1}^ne^{-\mu(A_i)}\frac{\mu(A_i)^{k_i}}{k_i!}\cdot e^{-\mu(\mathcal X\setminus A)}\frac{\mu(\mathcal X\setminus A)^{m-k}}{(m-k)!}. \end{align*} Now simply sum over all $m\ge k$ and use the fact that $\sum_{m=k}^\infty \frac{x^{m-k}}{(m-k)!}=e^x$.

$\endgroup$
  • $\begingroup$ Thanks, I even can't understand this "Poisson point process " concept but I will try to solve in more detail your answer. In the first part you are proving that $A_1,...,A_n$ are disjoint in $\mathcal{H}$? $\endgroup$ – Rachel Nov 22 '17 at 22:04
  • $\begingroup$ sorry but what does it mean that the vector is multinomial? Is it a generalization of the binomial distribution? $\endgroup$ – Rachel Nov 23 '17 at 2:04

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.