Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 33 / 3

On a measurable space $(E,\mathcal{E})$ with a sigma-finite measure $\mu$, a Poisson random measure $N$ is a random counting measure such that $N(A)$ has the Poisson distribution with parameter $\mu (A)$ for every measurable $A$ of finite intensity, and the counts on finitely many pairwise disjoint measurable sets are independent. The sample measures are countably additive; repeated atoms are allowed if the intensity has atoms.

When $\mu (A)=\infty$, exhaustion by finite-intensity subsets gives $N(A)=\infty$ almost surely. Indeed their counts increase, and the Poisson distribution probability of a count at most any fixed integer tends to zero as the parameter tends to infinity. Independent Poisson counts on disjoint sets, with means prescribed by $\mu$, characterize the Poisson random measure.

For a measurable $A$ with $(\mu +\nu )(A)<\infty$, independence of the two Poisson random measures givesThus its count has the Poisson distribution with the summed parameter. If $A_1,\dots ,A_m$ are disjoint, the variables within each family of counts are independent, and independence of $\Phi$ and $\Psi$ makes the two families independent of each other. Hence the pairs $(\Phi (A_j),\Psi (A_j))$, and therefore their sums, are independent over $j$.

Adding sample counting measures preserves countable additivity. The definition in (a) therefore provesThis proves the required Superposition theorem for Poisson point processes directly from count distributions.

Interpret the second process as an independent copy of the unit-rate Poisson process. First, the sums are finite almost surely. On $[0,1]$ there are finitely many arrivals, none at $0$, so their inverse-square contributions are finite. On $[1,\infty )$,For completeness, this expectation identity follows first for nonnegative simple functions from $\mathbb{E}N(A)=|A|$, and then for all nonnegative measurable functions by monotone convergence. Thus no unproved Poisson-integral formula is needed, and the tail sum is finite almost surely.

By (b), merging the two processes produces a rate-$2$ Poisson process with arrival times $V_i$. Its rescaled arrivals $2V_i$ form a unit-rate Poisson process: for a measurable set $A$, their count is the merged count on $A/2$, with parameter $2|A/2|=|A|$, and disjoint-set independence is preserved. ConsequentlyTherefore $\boxed{c=4}$. This is the scaling of an inverse-power sum over Poisson arrivals with exponent $2$.

The PDF does not explicitly repeat the second process's rate. If that rate were $\rho$ rather than $1$, the same count argument would give $X+Y\stackrel{d}{=}(1+\rho {)}^{2}X$. The numerical value $4$ uses the intended independent-unit-rate-copy interpretation.

Use the independent-mark interpretation: the Brownian motions attached to the atoms are independent of one another and of the initial Poisson random measure. Fix $t$ and defineConditionally on the initial atoms, each atom is retained as a dangerous atom if its own path meets the target by time $t$. These tests are independent, with position-dependent retention probability $p_t(x)$. The permitted Poisson thinning property makes the dangerous atoms a Poisson random measure of intensity $p_t(x)dx$.

Its total intensity is computed without invoking a marking or displacement theorem. The event in the definition of $p_t(x)$ says that $x$ belongs toBy the Tonelli theorem,Symmetry of Brownian motion gives $-\xi \stackrel{d}{=}\xi$ as processes. It therefore identifies the last expectation with $\mathbb{E}\mathrm{vol}(W(t))$, where the sign of the deterministic path $f$ is unchanged. This swept set is a Wiener sausage with an added deterministic path.

The total intensity is finite on each bounded time interval. To see this explicitly, write $K_t=1+{\sup }_{s\le t}|f(s)|$ and $R_t={\sup }_{s\le t}|\xi (s)|$. The sausage is contained in a ball of radius $K_t+R_t$. On a finite time grid, splitting according to the first crossing of a positive level $u$ by a one-dimensional Brownian motion shows that the chance of ending above $u$, conditional on the crossing, is at least $1/2$: the remaining independent increment is symmetric. Thus the grid maximum exceeds $u$ with probability at most twice the final Gaussian upper tail. Dense grids and continuity give the same bound for the path maximum. Applying this to each coordinate and its negative givesThe last inequality follows from $\mathbb{E}{e}^{\theta N(0,t)}=e^{{\theta }^{2}t/2}$ and minimizing the exponential Markov bound. Integration of this tail gives finite moments of $R_t$ of every positive order, so $\mathbb{E}\mathrm{vol}(W(t))<\infty$.

There are consequently only finitely many dangerous atoms on each compact time interval, almost surely. For closed balls, continuity of the paths makes each finite hitting-time infimum attained. The event $T>t$ is therefore the absence of dangerous atoms up to time $t$.

The open-ball convention has the same survival probability at each fixed $t$. Indeed for any compact path range $K$, the difference between its closed and open radius-one neighborhoods is the shell $\{x:\mathrm{dist}(x,K)=1\}$, which has zero Lebesgue measure. To see this, choose a nearest $y\in K$ for a shell point $x$. For every sufficiently small $\epsilon >0$, the ball of radius $\epsilon /2$ centered at $x-\epsilon (x-y)$ lies in the open neighborhood and inside the ball of radius $3\epsilon /2$ about $x$. Thus the shell has a fixed proportional hole at every scale; the Lebesgue density theorem forces its measure to be zero. By the Tonelli theorem, closed-touch and open-entry retention probabilities have equal integrals. Independent thinning therefore gives no atom that touches only the boundary before this fixed horizon, almost surely. This also excludes a difference at an open-entry infimum equal to $t$.

The zero-count probability of a Poisson distribution is $e^{-\lambda }$, and henceThis survival among independently moving Poisson traps formula has been derived using only independent thinning and the defining Poisson zero-count probability, with the expectation/volume step supplied by Tonelli's theorem.