Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 32 / 6 / b

For $r\ge 0$, Lebesgue measure of the ball is $v_d{r}^d$, so the Poisson random measure gives $N_r\sim \mathrm{Poisson}(v_d{r}^d)$ andBy monotonicity of the balls, the set of radii with no points is an interval starting at zero. Thus $\{R\ge r\}=\{N_r=0\}$: if all smaller balls are empty then their union, the open ball of radius $r$, is empty; if that ball is empty then its radius belongs to the defining set. We obtainThe continuous tail gives $\mathbb{P}(R=0)=0$ and $\mathbb{P}(R=\infty )=0$. Differentiating the cumulative distribution function gives the probability density functionThe substitution $u=v_d{r}^d$ proves that this density integrates to one; equivalently $v_d{R}^d$ has the unit-rate exponential distribution. This is the first-neighbour case of the Kth-nearest-neighbour distance in a homogeneous Poisson point process.

First derive the Laplace functional of a Poisson random measure. For a simple nonnegative $g={\sum }_j{t}_j{1}_{A_j}$ on disjoint finite-intensity sets, independent Poisson random variables giveIncreasing simple approximations, monotone convergence for the integrals, and bounded convergence theorem for the random exponentials prove the same identity for every nonnegative measurable $g$.

Here the intensity is Lebesgue measure. Put $A=B(0,r)$ andThe integral is finite because $f$ has compact support. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$, $t\ge 0$:Differentiate from the right at $t=0$. The difference quotient on the random side is dominated by $N_r$, which is integrable with mean $v_d{r}^d$; the intensity integral is over a finite-volume ball. We obtain the count-weighted Laplace functional of a Poisson random measure:At $r=0$, the ball is empty and both sides are zero.

For $r>0$, write $m_r=v_d{r}^d$, $a_r={\int }_{B(0,r)}{e}^{-f(x)}dx$, and $L=\mathbb{E}{e}^{-M(f)}$. The count-weighted Laplace functional of a Poisson random measure and the Poisson distribution giveTo compute the other numerator, on $\{N_r=0\}$ the inside contribution to $M(f)$ vanishes. The restrictions of the Poisson random measure to the ball and its complement are independent; this follows first for their simple-function integrals from independent counts, then by approximation. HenceSubtract from $L$ and divide by $\mathbb{P}(N_r\ge 1)$ to findContinuity of $f$ at zero givesAlso $(1-e^{-z})/z\to 1$ as $z\downarrow 0$. Applying these facts to the two exact expressions proves the small-ball conditioning for a Poisson random measure:One can also see equality of the limits without the exact first formula: since $0\le e^{-M(f)}\le 1$, the difference of the two quantities lies between zero and $[m_r-(1-e^{-m_r})]/(1-e^{-m_r})\to 0$. The conditioning is used only at positive $r$, where its event has positive probability; it is not conditioning directly on a point at the origin.