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"Points $\{A_j\}_{j\in\Phi(\lambda)}$ are assumed to be distributed according to a homogeneous PPP with intensity $\lambda$, denoted $\Phi(\lambda)=\{X_j\}$, where $X_j$ is the location of the $j$th point." I was reading this in some paper.

My question is what is PPP? Of course I look for it in google and I found that means Poisson Point Process. I read the first definition of it in Wikipedia page but I still do not get it.

What is the intensity $\lambda$? Why we cannot simply say that these points are randomly located at $X_j$ with Poisson distribution?

What I understand is this: These points are in fact generated randomly with poisson distribution but if one picks a random region, then the density (or the number) of the point will be approximatively equals $\lambda$ in average. Is my understanding right?

Any help please? Thank you very much for your time.

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  • $\begingroup$ I think you mean Poisson point process. $\endgroup$
    – user122283
    Apr 9 '14 at 20:18
  • $\begingroup$ search for "Poisson process" $\endgroup$
    – Memming
    Apr 9 '14 at 20:18
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    $\begingroup$ Please do not use the tag (poisson-geometry) for questions related to Poisson processes or Poisson random variables. These are unrelated. $\endgroup$
    – Did
    Mar 7 '17 at 7:44
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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 $$ \lambda_N = {\bf E}[N(0,1]]. $$

There are many different possible point processes, but the Poisson point process with intensity $\lambda$ is the one for which the number of points in an interval $(0,t]$ has a Poisson distribution with parameter $\lambda t$: $$ P[N(0,t] = k] = \frac{(\lambda t)^k e^{-\lambda t}}{k!} $$ and which is stationary. Stationarity is a little more involved to go into here, but in this context you can think of it as meaning that the measure of two different intervals of equal length is the same, thus $$ P[N(s,s+t] = k] = \frac{(\lambda t)^k e^{-\lambda t}}{k!}, \ \ \forall s. $$

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    $\begingroup$ The intensity is a measure on $\mathbb{R}$ (one can also take other spaces): \begin{equation} \lambda_N([a,b]) := \mathbf{E}[N([a,b])]\, , \quad a<b \in \mathbb{R} \, . \end{equation} $\endgroup$ Apr 11 '14 at 6:26
  • $\begingroup$ Agreed, but the intensity of a point process usually refers to the measure of $(0,1]$ under the measure you've specified. $\endgroup$
    – user139388
    Apr 11 '14 at 6:28
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    $\begingroup$ If you are considering only the measure on $(0,1]$, then you are considering stationary Poisson point processes on $\mathbb{R}$ only. Your second last formula is fine (in the more general context), if you replaced $\lambda t $ by $\lambda ( (0,t])$. $\endgroup$ Apr 11 '14 at 6:34
  • $\begingroup$ Oh I see, you are interpreting the "intensity $\lambda$" in the question to mean "intensity measure $\lambda$". I didn't think that was what the OP was asking about. I'll update the question shortly. Thanks! $\endgroup$
    – user139388
    Apr 11 '14 at 6:36
  • $\begingroup$ One should note that point processes apply as well in $R^n$ or more general spaces. Also, it does not make much sense to talk about intensity (and not intensity measure) if one does not assume first stationarity. $\endgroup$ May 10 '14 at 19:29
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From Wikipedia:

For any disjoint bounded subsets $B_1,\ldots,B_n$ and non-negative integers $k_1,\ldots,k_n$ we have that $$\Pr[\xi(B_i) = k_i, 1 \leq i \leq n] = \prod_i e^{-\lambda \|B_i\|}\frac{(\lambda \|B_i\|)^{k_i}}{k_i!}.$$ The constant $\lambda$ is called the intensity of the Poisson point process. Note that the Poisson point process is characterised by the single parameter $\lambda$.

So, let the expectation measure be $Eξ$. Then, $Eξ(\cdot)=\lambda\|\cdot\|$, where $\|\cdot\|$ is the Lesbegue measure and $\lambda$ is the intensity.

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