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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.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

Assume individual incubation delays are independent of the Inhomogeneous Poisson process of infections and of one another. This marking assumption is needed in addition to specifying a marginal incubation density.

In the discrete cohort model let $N_i\sim \mathrm{Poisson}(h_i)$ be the independent infection counts. Mark each infection by its eventual onset interval. For cohort $i$, the categories have probabilities $q_{ik}$ for observed intervals, with the remaining probability assigned to onsets outside the window. Poisson thinning makes the counts $N_{ik}$ in these categories mutually independent Poisson random variables with means $h_i{q}_{ik}$. One direct proof is their probability generating function:Different cohorts are independent. Summing their counts therefore gives independent Poisson onset counts in disjoint intervals:The same argument applies exactly in continuous time by the Independent marking theorem for Poisson point processes and mapping each marked infection to its onset time. The endpoint model approximates its means; its independent-Poisson conclusion is exact within that discrete model. A fixed cohort size would instead induce negatively correlated onset-bin counts, so the Poisson process infection assumption matters.

An Inhomogeneous Poisson process with intensity $\lambda$ is a counting process with $N_0=0$, independent increments, andThe continuity and positivity of $\lambda$ make $\Lambda$ strictly increasing.

For $M_t=N_{g(t)}$ and $0\le s<t$,has a Poisson distribution with meanDisjoint increments remain independent, so $M$ is a homogeneous rate-one Poisson process. Taking $g={\Lambda }^{-1}$ gives $N_t=M_{\Lambda (t)}$, and henceThis is the time change of an inhomogeneous Poisson process.

For the bicycle model, let an arrival occur at time $s\le t$ and have velocity $V$. At time $t$ its position is $V(t-s)$, so it lies in the first $x$ miles exactly whenConditional on its arrival time, independently marking the bicycle by whether this event occurs is Poisson thinning. Therefore the desired count is Poisson with meanThus

By Poisson thinning, allocating a Poisson$(365)$ population independently and uniformly among $365$ days gives independent Poisson$(1)$ counts. Hence

Let $X_t$ be the number of telephone lines in use. The memorylessness of the exponential distribution makes $(X_t{)}_{t\ge 0}$ a continuous-time Markov chain, specifically the M-M-infinity queue with generator rates

Suppose $X_0=n$ and put $p_t=e^{-\mu t}$. Each initial call is still active at time $t$ with the exponential survival probability $p_t$, independently of the others. Their surviving number is thereforeA call arriving at time $u\in [0,t]$ survives until $t$ with probability $e^{-\mu (t-u)}$. By Poisson thinning, the number of surviving new calls is independent of $Y_t$ and has lawHence $X_t=Y_t+Z_t$ in distribution. Multiplying the probability generating functions of the two independent random variables gives, for $-1\le s\le 1$,Equivalently, the transient distribution of an M-M-infinity queue iswith independent summands. As $t\to \infty$, the binomial term converges to zero and the Poisson mean tends to $\lambda /\mu$. Thereforewhich proves the stated large-time approximation.

A particle arriving at time $s\le t$ survives until $t$ with probability $e^{-\mu (t-s)}$. Independently retaining each Poisson arrival with this time-dependent probability is Poisson thinning. Hence the number alive at time $t$ is Poisson with meanThus

An arrival at $s\le t$ remains active with probability $1-(t-s)$ when $0\le t-s\le 1$, and zero later. By Poisson thinning, the detector count is Poisson with mean

Mark each member of an independent Poisson infection cohort by its symptom-onset interval, using independent delays. Poisson thinning gives independent onset-bin counts, and sums over independent cohorts remain independent Poisson counts. Fixed cohort totals instead induce multinomial dependence. The marking assumptions are essential.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.

If $X_0=n$ and $p_t=e^{-\mu t}$, thenwith independent summands. The binomial term counts surviving initial customers; Poisson thinning gives the surviving later arrivals.