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Suppose we have a point process in $\mathbb{R}$ with intensity $\lambda(x)$. Then, for a given compact set ${ S}$ we have

$$\Lambda({ S})=\int_{\rm S} \lambda(x) \, dx,$$

where $\Lambda({ S})$ is the number of points in the region ${ S}$. I have the estimate of $\lambda(x)$ (via kernel smoothing), but is there a way that I can convert $\lambda(x)$ to the points' probability density function $f(x)$, say in ${ S}$? Is it just $\lambda(x)/\Lambda({ S})$?

Here's what I think not sure if its correct. Suppose that events were observed to occur at the locations $x_1,\dots,x_n$ on a linear network $L$. The objective is to estimate the probability density $f(x)$ from which the locations are assumed to be drawn. $f(x)\ge0$ for all $x\in L$, $\int_Lf(x)dx=1$. The probability that a random point $x$ with density $f$ falls in a subset $B\subset L$ is $P(x\in B)=\int_Bf(x)dx$.

Equivalently $\boldsymbol{x}=\{x_1,\dots,x_n\}$ is a realization of a point process $\boldsymbol{X}$ on the network, and the objective is to estimate the intensity function $\lambda(x)$ of $\boldsymbol{X}$. The intensity function is defined so that the number $n(\boldsymbol{X}\cap B)$ of points of $\boldsymbol{X}$ falling in $B\subset L$ has expectation $E(n(\boldsymbol{X}\cap B))=\int_B\lambda(u)du$. $\lambda(u)$ is the expected number of random points per unit length of network, in the vicinity of location $u$. Estimation of $f$ and of $\lambda$ is essentially equivalent because, if $n$ is fixed, $\lambda(x)=nf(x)$ for all locations $x\in L$.

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  • $\begingroup$ An intensity is an expected number of points per unit length. Could you explain the sense in which you understand that as being a "probability density function"? As an example of why this seems mysterious to me, consider (say) a homogeneous Poisson process on $\mathbb R$ with $\lambda=1.$ That's an expectation of one point per unit length. The total number of points therefore must be unbounded and clearly there is no way this density of points can be normalized into any probability density on the reals. In this example, then, what is your "probability density" intended to be? $\endgroup$ – whuber Nov 16 '18 at 22:59
  • $\begingroup$ intensity is number of point in a region divided by size of the region (length), in the limit as size of the region goes to 0. What I would like is given a line segment, what's the density of the points? For example, if it's a homogeneous poisson process, given a fixed number of points, the positions of the points follow a uniform distribution. $\endgroup$ – dynamic89 Nov 16 '18 at 23:54
  • $\begingroup$ That's not a probability density, that's a point density. The two are quite different! With this understanding, your question seems tautologically to ask "how do I convert the point density $\lambda$ into a point density?" $\endgroup$ – whuber Nov 17 '18 at 16:30
  • $\begingroup$ I think the model above is for the inhomogeneous Poisson point process. @whuber, If the integral is over a sub-region of the space (not infinite space), would that change the interpretation. I have also seen this expression described as a probability density (...for an inhomogeneous Poisson point process). $\endgroup$ – coreydevinanderson Nov 20 '18 at 5:17
  • $\begingroup$ @dynamic89. When you say "what's the density of points?" (over some interval)...do you really mean, what's the expected number of points (in some interval given some intensity)? Perhaps this is the source of confusion? $\endgroup$ – coreydevinanderson Nov 20 '18 at 5:24

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