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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.

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,

The joint process is a two-dimensional Compound Poisson process whose Lévy measure isTwo coordinates of a Lévy process are independent exactly when its Lévy measure charges only the coordinate axes and its Gaussian covariance has no cross term. Here there is no Gaussian part, so independence is equivalent toSince $g_1{g}_2$ is continuous, this is equivalent to pointwise vanishing. Conversely, when the product vanishes, the mark sets where $g_1$ and $g_2$ are nonzero are disjoint; independent thinning of the Poisson random measure gives independent coordinate processes. Hence

Let $B$ be Brownian motion and let $N(ds,dx)$ be an independent Poisson random measure with intensity $dsK(dx)$. Writing $\tilde{N}=N-dsK(dx)$ for its compensated version, the Lévy–Itô decomposition constructsThe four terms are independent drift, Gaussian, compensated small-jump and compound-Poisson large-jump components.

For a measurable map $f$ and a Poisson random measure of intensity $\nu$, image counts in disjoint sets equal original counts in their disjoint preimages. They are therefore independent Poisson variables with the pushforward intensity. The result is a Poisson random measure even if images collide; diffuse pushforward intensity gives a simple point process.

For a Poisson random measure on $E$ with finite intensity $\nu (E)$, its integral of a finite-valued measurable mark function is the finite sum of the marks at its random atoms. Its characteristic function is $\mathbb{E}\exp (iu\int fdN)=\exp (\int (e^{iuf}-1)d\nu )$. On a time-mark space with intensity $ds\nu (dz)$ and finite mark intensity, restriction to each bounded time interval has finite intensity and defines a Compound Poisson process. Zero-valued marks do not affect the integral and need not appear in its Lévy measure.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.

Start traps at a unit-intensity Poisson random measure in Euclidean space and attach mutually independent Brownian motions, independent of the initial measure. For a deterministic target path $f$, independently thin initial positions according to whether their marked path comes within radius one of the target by time $t$. The retention probability $p_t(x)$ integrates, by the Tonelli theorem, to the expected volume of the sausage along $f-B$. Symmetry of Brownian motion makes this the expected volume of the sausage along $f+B$. The Poisson zero-count probability proves the displayed survival formula. This derives the formula directly from independent thinning, without assuming a separate marking or displacement theorem.

The void probability functional assigns to a measurable set the probability that it contains no atom of a random point measure. For a Poisson random measure of intensity $\mu$, it is $e^{-\mu (A)}$. The exponential void probabilities characterize a Poisson random measure result uses simplicity; doubling every atom of a Poisson measure preserves void probabilities but changes count distributions.