From Poisson Points to Transformed Points
A Poisson random measure describes random counts in measurable regions: counts in disjoint regions are independent, and each count has a Poisson distribution. Two useful operations preserve this structure. One sends every point through a measurable map; the other attaches a random mark to each point, with the mark distribution allowed to depend on the point's location. In both cases, the new intensity is obtained by tracking how the original intensity is redistributed.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. Let \((E,\mathcal{E})\) be a measurable space, and let \(\nu\) be a sigma-finite measure on it. A Poisson random measure \(N\) on \(E\) with intensity \(\nu\) is a random counting measure such that, for each measurable \(A\) with \(\nu(A)<\infty\), \(N(A)\) has a Poisson distribution with mean \(\nu(A)\), and the counts of pairwise disjoint such sets are independent. A counting measure counts multiplicities: if several original points map to the same location, they contribute several counts there.
There is no time index in this definition: the underlying space \(E\) may represent locations, times, or another measurable domain. If \(E\) is a time domain, the counts in time sets are the corresponding process counts. We first consider a measurable map \(f:E\to F\), where \((F,\mathcal{F})\) is another measurable space. The image counting measure is defined by \(N_f(B)=N(f^{-1}(B))\) for \(B\in\mathcal{F}\).
Proof. Let \(B_1,\ldots,B_m\) be pairwise disjoint measurable subsets of \(F\) with \(\nu_f(B_r)<\infty\). Their preimages \(f^{-1}(B_r)\) are pairwise disjoint measurable subsets of \(E\), and \(\nu(f^{-1}(B_r))=\nu_f(B_r)\). Therefore the variables \(N_f(B_r)=N(f^{-1}(B_r))\) are independent Poisson variables with means \(\nu_f(B_r)\). These are exactly the defining count properties of a Poisson random measure with intensity \(\nu_f\). Sigma-finiteness is assumed so this intensity is an admissible Poisson random-measure intensity. \(\square\)
The formula \(\nu_f(B)=\nu(f^{-1}(B))\) is often the safest way to calculate: a target region receives the intensity of all original points that map into it. The map need not be one-to-one. If it is many-to-one, the output is still a Poisson random measure as a counting measure; whether it is a simple point process, with no repeated locations, is a separate question.
Worked Example: Scaling Locations
Let \(N\) be a Poisson random measure on \(E=[0,2]\) with intensity \(\nu(dx)=3\,dx\), and map each location \(x\) to \(f(x)=2x\). Find the image intensity and the expected image count in \([0,2)\) and \([2,4]\).
For a measurable target set \(B\subseteq[0,4]\), the intensity is \(\nu_f(B)=3\int_0^2\mathbf{1}_{B}(2x)\,dx\). Under the substitution \(y=2x\), \(dx=dy/2\), so \(\nu_f(B)=\int_B\tfrac32\,dy\). Thus the image has constant intensity density \(\tfrac32\) on \([0,4]\). The total intensity is \(\tfrac32\cdot4=6\), matching the original total intensity \(3\cdot2=6\).
For \([0,2)\), the preimage is \([0,1)\), so its image count has mean \(3\cdot1=3\). Equivalently, the image intensity is \(\tfrac32\cdot2=3\). For \([2,4]\), the preimage is \([1,2]\), with intensity \(3\cdot1=3\), again equal to \(\tfrac32\cdot2=3\). The two image counts are independent because the target intervals are disjoint. The endpoint \(2\) has zero intensity, so its inclusion in the second interval does not alter these values.
Check: The two intervals partition \([0,4]\) up to their shared zero-intensity boundary, and their means sum to \(3+3=6\), the total image mean.
Independent Marks and the Marked Intensity
Now let \((M,\mathcal{M})\) be a measurable mark space. A probability kernel \(K(x,dy)\) specifies the mark distribution at location \(x\): for each \(x\), \(K(x,\cdot)\) is a probability measure on \(M\), and for each measurable \(C\subseteq M\), \(K(x,C)\) is measurable as a function of \(x\). Conditional on the original points, attach marks independently, giving a point at \(x\) a mark distributed as \(K(x,dy)\). The resulting marked points lie in \(E\times M\).
In particular, for a location region \(A\subseteq E\) and mark region \(D\subseteq M\), the expected number of marked points in \(A\times D\) is \(\int_A K(x,D)\,\nu(dx)\). If the mark law does not depend on \(x\), this reduces to \(\nu(A)K(D)\). Location-dependent marks generally make the marked intensity a measure that does not factor into a location measure times a single mark distribution.
Proof. Take pairwise disjoint measurable sets \(C_1,\ldots,C_m\) in \(E\times M\) with finite \(\widetilde{\nu}(C_r)\). For each \(x\), define \(q_r(x)=\int_M\mathbf{1}_{C_r}(x,y)K(x,dy)\). The sets \(C_r\) are disjoint, so \(\sum_r q_r(x)\leq1\). For \(0\leq z_r\leq1\), conditional on the original locations, the probability generating function of the marked counts \(Z_r=\widetilde{N}(C_r)\) is the product over original points \(x\) of \(1+\sum_r q_r(x)(z_r-1)\): each point either contributes to one of the disjoint sets or contributes to none.
For completeness, the Poisson random-measure product formula used here is \(\mathbb{E}\prod_{x\in N}h(x)=\exp\{-\int_E(1-h(x))\,\nu(dx)\}\) for measurable \(0\leq h\leq1\). For a simple \(h\) on finitely many disjoint finite-intensity sets, this follows by multiplying the Poisson generating functions \(\mathbb{E}[a^{N(A)}]=\exp\{\nu(A)(a-1)\}\). Approximation by simple functions and exhaustion by finite-intensity sets gives the formula for measurable \(h\); the products and integrals converge by bounded convergence on each finite restriction, followed by the exhaustion limit.
Apply this formula to \(h(x)=1+\sum_r q_r(x)(z_r-1)\), which lies in \([0,1]\). The unconditional joint generating function is $$ \mathbb{E}\prod_{r=1}^m z_r^{Z_r} =\exp\left\{-\int_E\sum_{r=1}^m q_r(x)(1-z_r)\,\nu(dx)\right\} =\prod_{r=1}^m\exp\{(z_r-1)\widetilde{\nu}(C_r)\}. $$ The last expression is the joint generating function of independent Poisson variables with respective means \(\widetilde{\nu}(C_r)\). Since this holds for every such disjoint collection, the marked counts have the defining properties of a Poisson random measure with intensity \(\widetilde{\nu}\). \(\square\)
Worked Example: Location-Dependent Categorical Marks
Let \(N\) have intensity \(6\,dx\) on \([0,1]\). Give each point a mark in \(M=\{A,B\}\), with \(K(x,\{A\})=x\) and \(K(x,\{B\})=1-x\). Find the intensity of each mark type and the distribution of the total counts of types \(A\) and \(B\).
The kernel is valid at every \(x\in[0,1]\): both probabilities are nonnegative and \(x+(1-x)=1\). The intensity measure for type \(A\), viewed on the location interval, is \(6x\,dx\); for type \(B\), it is \(6(1-x)\,dx\). Integrating gives $$ \widetilde{\nu}([0,1]\times\{A\})=6\int_0^1x\,dx=6\left(\frac12\right)=3,\qquad \widetilde{\nu}([0,1]\times\{B\})=6\int_0^1(1-x)\,dx=6\left(\frac12\right)=3. $$
The two marked regions are disjoint, so their counts are independent Poisson variables, each with mean \(3\). The total number of marked points therefore has mean \(3+3=6\). More directly, every original point receives exactly one of the two marks, so this total is the original count, which is Poisson with mean \(6\).
Check: The marked intensities add to the original intensity density: \(6x+6(1-x)=6\) for every \(x\). Integrating this sum over \([0,1]\) gives \(6\), as required.
Displacement as Marking Followed by Mapping
A displaced point is handled in two stages. First mark each original location with its displacement; then map the pair \((x,z)\) to the final location \(x+z\). The marking theorem gives the intensity on location-displacement pairs, and the mapping theorem pushes that intensity onto the destination space. The resulting formula is $$ \nu_{\mathrm{disp}}(B)=\int_E\int_{\mathbb{R}}\mathbf{1}_{B}(x+z)\,K(x,dz)\,\nu(dx). $$ $$
Worked Example: Two Possible Displacements
Let the parent points have intensity \(4\,dx\) on \([0,2]\). Independently give each point displacement \(Z=0\) or \(Z=1\), each with probability \(\tfrac12\). Find the expected numbers of displaced points in \([0,1)\), \([1,2)\), and \([2,3)\), and explain their joint distribution.
For any measurable \(B\subseteq\mathbb{R}\), the displaced intensity is $$ \nu_{\mathrm{disp}}(B)=2\int_0^2\mathbf{1}_{B}(x)\,dx+2\int_0^2\mathbf{1}_{B}(x+1)\,dx. $$ The first term has density \(2\) on \([0,2]\), and the second has density \(2\) on \([1,3]\). Thus the combined density is \(2\) on \([0,1)\), \(4\) on \([1,2)\), and \(2\) on \([2,3]\), with endpoint values irrelevant to the measure.
For \([0,1)\), only the zero-displacement term contributes, giving \(2\cdot1=2\). For \([1,2)\), the zero-displacement term contributes \(2\cdot1=2\), and the unit-displacement term contributes \(2\cdot1=2\), for a total of \(4\). For \([2,3)\), only the unit-displacement term contributes, giving \(2\cdot1=2\). These counts are independent Poisson variables because the three destination intervals are disjoint and the displaced measure is Poisson.
Check: The three means sum to \(2+4+2=8\), equal to the parent intensity \(4\cdot2=8\). Displacement changes where the intensity lies, but each parent point still contributes exactly one displaced point.
Conditions, Interpretation, and a Common Mistake
The mapping theorem requires a measurable map so that preimages of measurable sets are measurable. Its stated sigma-finiteness condition matters: without it, the proposed image measure may not be an admissible intensity measure on the target space. The marking theorem requires a measurable probability kernel and independent marking conditional on the original points. If marks are dependent across points, the proof's product of conditional contributions no longer applies in general.
Neither theorem says that the transformed process must be simple. For example, a map can collapse a positive-intensity region to a single target point; the image count there is then a Poisson count with multiplicity, not a single point. In continuous displacement examples, collisions may have probability zero, but the theorem itself does not require that fact.
Common mistake: using the forward scale factor for intensity. In the scaling example \(y=2x\), it is incorrect to multiply the source density \(3\) by \(2\) and claim image density \(6\). The preimage calculation gives the correction: for \(B=[0,2)\), \(f^{-1}(B)=[0,1)\), so the image mean is \(3\cdot1=3\). A proposed density \(6\) would instead give \(6\cdot2=12\) on that interval, contradicting the direct preimage count. The correct change of variables is \(dx=dy/2\), giving image density \(3/2\), and \((3/2)\cdot2=3\). Always derive transformed intensity from preimages or the full kernel integral before applying a Jacobian rule.
These results let one build spatial models from simpler Poisson inputs. A measurable transformation can represent rescaling, projection, or aggregation of locations; marks can represent types, labels, or random displacements. The practical sequence is to specify the original intensity, state the map or mark kernel precisely, and calculate the new measure on an arbitrary target set. The defining independence of disjoint Poisson counts is then preserved by the theorems, not inferred merely from matching expected counts.
Check Your Understanding
For each question, identify the relevant preimage or marked intensity and state any condition needed for the conclusion.
- A Poisson random measure has intensity \(5\,dx\) on \([0,1]\). Under \(f(x)=3x\), what is the image intensity density on \([0,3]\)? Verify its total mass.
- If marks have probabilities \(K(x,\{A\})=1-x\) and \(K(x,\{B\})=x\) on \([0,1]\), what are the two total mark intensities when the original intensity is \(4\,dx\)?
- Why does a many-to-one measurable map still give a Poisson random measure, and what feature of the output might fail?
- For parent intensity \(2\,dx\) on \([0,1]\) and displacement \(0\) or \(1\) with equal probabilities, calculate the displaced intensity of \([1,2)\).
- Which part of the marking theorem proof would fail if marks of different points were dependent conditional on their locations?