# Poisson distribution process

Meteorites hit the surface of the moon, treat as an infinite plane.

The meteorites that have hit the moon over the last 1 million yrs can be modeled as a Poisson process with constant intensity lambda.

Suppose that each meteorite leaves a circular crater of random radius, and the craters' radii are i.i.d. from a distribution having density f. Assume that f(r) = 0 for r suciently large. For a bounded (Borel) set A, let V (A) be the area of A that is not covered by a crater from a meteorite that hit in the last million years.

Since this is a Poisson process, how can we get the

E(Area of A not covered by a crater from a meteorite that hit in the last million years)

and the Var(Area of A not covered by a crater from a meteorite that hit in the last million years)

I do note that E(X) = 1/p

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I am not happy with your notation, but one approach is to look at a ring of inner radius $r$ and outer radius $r+\delta r$ (and so of area about $2 \pi r \delta r$ in the limit), calculate the probability either that the ring is not hit or that it is hit and all the meteorites hitting the ring (reducing to 1 in the limit) leave a crater of radius less than $r$. This is the probability that the centre of the ring is not covered by a crater caused by a hit on the ring.

You now want to multiply all the different ring probabilities together. One way to do this is by adding up the logarithms of the probabilities and then taking the anti-logarithm, so in the limit integrate the logarithms of the probabilities over $r$ and take the anti-logarithm.

That will give you the probability that a particular point is not covered by a crater, and so the expected fraction of $A$ which is not covered by a crater. The variance is harder.

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There is a long road to compute the expectation of the non covered area, and a shorter road. Both roads start with the random set $C$ of the points of the plane which are not covered by a meteorite and with the expression $$\mathrm E(|A\cap C|)=\mathrm E\int_A[x\in C]\,\mathrm dx=\int_A\mathrm P(x\in C)\mathrm dx=p\,|A|,$$ with $p=\mathrm P(0\in C)$ since the invariance by translation of the meteorite process shows that $\mathrm P(x\in C)$ does not depend on $x$. Hence the task is to compute $p$.

The long road: Let $\mathcal M$ denote the random set of the centers of meteorites. Let $\mathcal R=(R(x))_{x\in\mathcal M}$ where, for every $x$ in $\mathcal M$, $R(x)$ denotes the radius of the meteorite centered at $x$. Then, $$\mathrm P(0\in C\mid \mathcal M,\mathcal R)=\prod_{x\in \mathcal M}[R(x)\lt\|x\|],$$ hence, since conditionally on $\mathcal M$, $\mathcal R$ is a collection of i.i.d. random variables, $$\mathrm P(0\in C\mid \mathcal M)=\prod_{x\in \mathcal M}\mathrm P(R\lt\|x\|),$$ where $R$ denotes any random variable distributed like the radii $R(x)$.

Now comes into play the hypothesis that $R$ is almost surely bounded, say by $r$. Then the product over $x$ in $\mathcal M$ can be restricted to a product over $\mathcal M\cap B$ where $B=\{x\mid\|x\|\leqslant r\}$. This new Poisson process has total intensity $\mu=\lambda\pi r^2$ and, conditionally on its size, its points are uniformly distributed in $B$. Thus, $$\mathrm P(0\in C)=\sum_{n\geqslant0}\mathrm e^{-\mu}\frac{\mu^n}{n!}\int_{B^n}\prod_{k=1}^n\mathrm P(R\lt\|x_k\|)\cdot\prod_{k=1}^n\frac{\mathrm dx_k}{\pi r^2}.$$ The integral over $B^n$ is a product, hence $p=\mathrm e^{-\mu+\mu J/(\pi r^2)}$, with $$J=\int_{B}\mathrm P(R\lt\|x\|)\mathrm dx=\mathrm E\int_B[R\lt\|x\|]\mathrm dx=\mathrm E(\pi r^2-\pi R^2).$$ Finally, $$p=\mathrm e^{-\lambda\pi\mathrm E(R^2)},\quad\text{hence}\quad\color{red}{\mathrm E(|A\cap C|)=\mathrm e^{-\lambda\pi\mathrm E(R^2)}\,|A|}.$$ The shorter road: Consider the process of the centers of the meteorites covering the point $0$. This is a nonhomogenous Poisson process, obtained by a thinning of the original Poisson process, where one keeps a point $x$ with probability $\mathrm P(R\gt\|x\|)$. Thus the intensity at $x$ of the thinned Poisson process is $\lambda(x)=\lambda\mathrm P(R\gt\|x\|)$ and its total intensity is $$\nu=\int_{\mathbb R^n}\lambda(x)\mathrm dx=\lambda\int_{\mathbb R^n}\mathrm P(R\gt\|x\|)\mathrm dx=\lambda\mathrm E\int[R\gt\|x\|]\mathrm dx=\lambda\mathrm E(\pi R^2),$$ To conclude, observe that $0$ is not covered if and only if the thinned Poisson process is empty, and that this happens with probability $\mathrm e^{-\nu}$.

Note: The argument above extends to any distribution of $R$. It shows that $\mathrm E(|A\cap C|)$ is always given by the formula above. Thus, if $R^2$ is integrable, $|A\cap C|$ has positive measure with positive probability, for every $A$ of positive measure, but, if $R^2$ is not integrable, $|C|=0$ almost surely.

The same applies to any dimension and to any random objects instead of disks. The condition to check in this extended setting is whether the random volume $V$ of these objects is integrable or not and the formula for any point $x$ of the space to be not covered becomes $$\color{purple}{\mathrm P(x\in C)=\mathrm e^{-\lambda\mathrm E(V)}}.$$

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