Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 31 (Poisson Processes)

Write $z=(x,y)$ and $\mu (dz)=\lambda (z)dz$. Treat the nonnegative plant weights as independent marks conditional on the locations, with a measurable mark law $K_z(dw)$ depending only on $z$. Its first and second moments areThis is the independent-marking interpretation appropriate for location-dependent weights.

The general results needed are the following. The Independent marking theorem for Poisson point processes says that such marks turn a Poisson point process of intensity $\mu$ into a Poisson point process on location-mark space with intensity $\nu (dz,dw)=\mu (dz)K_z(dw)$. The Campbell first-moment formula says that for any nonnegative measurable $g$, $\mathbb{E}{\sum }_{u\in \Pi }g(u)=\int gd\nu$, allowing infinite values. The second Poisson factorial moment measure identity says that for nonnegative measurable $h$,These statements apply to sigma-finite intensities and therefore do not require finitely many plants in the entire field. We use these general results without proving them.

Define the total as the nonnegative sum $W={\sum }_{(z,w)\in \Pi }w$, or equivalently as the increasing limit of its finite-intensity truncations. the Campbell theorem givesSince $M<\infty$, the nonnegative random variable $W$ cannot be infinite on an event of positive probability. Thus $W$ is finite almost surely, even if the number of plants is infinite.

Separate the square into its diagonal and ordered distinct-pair terms. Nonnegativity allows the use of Tonelli's theorem and increasing limits, givingThe assumed second integral is finite, so $W$ is square-integrable. Subtracting $(\mathbb{E}W{)}^2$ yieldsThe term $v$ accounts for random weights at fixed locations; the term $m^2$ accounts for the Poisson fluctuations in the locations and count. Nonnegativity is natural for weights. For signed marks, absolute first-moment integrability, rather than just a conditionally convergent integral of the signed mean, would be needed for the ordinary sum.

The intensity function of a point process is positive and continuous inside the interval, so its measure is finite on every compact subinterval. Consequently the Poisson point process is locally finite there. Its nonatomic intensity makes it simple and gives no point at $0$ almost surely. At the endpoints,Both half-intervals therefore have infinite intensity. More explicitly, exhaust either half by increasing compact intervals with means $t_j\to \infty$. For any fixed integer $k$,Taking the countable union over $k$ proves infinitely many points on each side almost surely.

Local finiteness excludes accumulation in the interior, including at $0$. Hence there is a largest negative point and a smallest positive point. For example, given any negative point, the compact interval between it and $0$ contains finitely many points and therefore a largest one. Call it $X_0$; call the smallest positive point $X_1$, and enumerate consecutively in both directions. Infinite counts give every integer index, and the lack of interior accumulation implies $X_n\uparrow 1$ as $n\to \infty$ and $X_n\downarrow -1$ as $n\to -\infty$. Thus the specified two-sided labeling exists almost surely.

Use the cumulative-intensity transformationIt has $f(0)=0$, positive derivative, and limits $-\infty ,+\infty$ at the two endpoints, so it is an increasing bijection onto $\mathbb{R}$. One explicit expression isFor any bounded interval $(a,b)$ in the target, its preimage has intensityDisjoint target sets have disjoint preimages, so their counts are independent. This proves directly, or by the Poisson mapping theorem, that the transformed points form a unit-rate Poisson process on the real line.

Put $Y_n=f(X_n)$. On the positive half-line, the successive spacings are independent exponential random variables of mean $1$, so $Y_n=E_1+\cdots +E_n$. The strong law of large numbers gives $Y_n/n\to 1$ almost surely. On the negative half-line, independently, the reflected points satisfy $-Y_{-k}=E_1^{\prime }+\cdots +E_{k+1}^{\prime }$, because $Y_0$ is the first point to the left of zero. Thus $-Y_{-k}/k\to 1$, or $Y_n/n\to 1$ also as $n\to -\infty$. The shift by one has no effect on these limits.

From the explicit formula, or by integrating the endpoint intensity asymptotics,Substitute $x=X_n$ and use the strong law for Poisson arrival times. At the right endpoint,soAt the left endpoint the corresponding result isEquivalently $8|n|(1+X_n)\to 1$. The different powers of $|n|$ reflect the different endpoint singularity orders, an instance of power-law boundary accumulation of a Poisson point process.

Identify a line with its unique parameter pair in $E=(0,\infty )\times [0,2\pi )$, usingA Poisson line process means a Poisson point process on the Borel parameter space $E$: for each measurable $B$ of finite mean measure, the number of its parameter points is Poisson with mean $\mu (B)$, and counts in finitely many disjoint sets are independent. Infinite-measure sets are handled by a countable finite-measure exhaustion. Here $\mu (dp,d\theta )=dpd\theta$ is sigma-finite and nonatomic, so there are countably many distinct lines almost surely. It is not an assertion that line counts are Poisson with respect to planar area.

Fix $r>0$ and let $D_r=\{z:|z|<r\}$. Exactly the lines with $0<p<r$ hit this disk, apart from the zero-probability tangent case. Thus the count of disk-hitting lines isEvery intersection in $D_r$ uses two of these lines. Writing $N_r=\#(\Phi \cap D_r)$, we haveThis also proves local finiteness of the intersection point process. Initially it gives a non-strict bound $\mathbb{P}(N_r\ge 1)\le 1-(1+2\pi r)e^{-2\pi r}$.

To establish strictness, choose the two parameter boxesEach has positive measure, and both lie among the disk-hitting lines. If one line from each box intersected at $z\in D_r$, thenBut their distance parameters differ by more than $r/4$, a contradiction. The event of exactly one line in each box and no other disk-hitting lines has probabilityOn this event $M_r=2$ but $N_r=0$. ThereforeThe following original diagram illustrates why two disk-hitting lines need not produce an intersection inside the disk.

To decide whether $\Phi$ is Poisson, compute its intensity measure and compare its void probability. Use the equivalent signed line coordinates $L(t,\theta )$ with $t\in \mathbb{R}$ and $0\le \theta <\pi$. If $t>0$ this is the original pair $(t,\theta )$; if $t<0$ it is $(-t,\theta +\pi )$. The measure is $dtd\theta$, with no extra factor of two. Translation by $a$ sends $t$ to $t+a\cdot n_{\theta }$, preserving this measure. Thus both the line and intersection processes are stationary.

Almost surely no two lines are parallel and no three are concurrent. In each finite-intensity parameter region, conditional line parameters have continuous joint densities. Parallel pairs and concurrent triples satisfy lower-dimensional equations and have probability zero; a countable exhaustion proves the assertion for all lines. Hence different unordered pairs have distinct intersections almost surely. For a bounded Borel set $A$, the second Poisson factorial moment measure identity givesFor nonparallel normals, their intersection $z$ satisfies $t=z\cdot n_{\theta }$, $s=z\cdot n_{\varphi }$. The absolute Jacobian determinant of this transformation is $|\sin (\varphi -\theta )|$. Consequentlysince the inner integral is $2$ for every $\theta$. In particular $\mathbb{E}{N}_r={\pi }^2{r}^2$.

If $\Phi$ were a Poisson point process, the count $N_r$ would therefore be Poisson with that mean, giving $\mathbb{P}(N_r=0)=e^{-{\pi }^2{r}^2}$. But the earlier necessary-two-lines argument givesAt $r=2/\pi$ this lower bound is $5e^{-4}$, whereas the supposed Poisson void probability is $e^{-4}$. This contradiction proves $\Phi$ is not a Poisson point process. Its points share parent lines, and the explicit void/mean comparison demonstrates the resulting dependence rather than merely asserting it.