Nu-transform

Let ${\displaystyle \delta _{x}}$ denote the Dirac measure on the point ${\displaystyle x}$ and let ${\displaystyle \mu }$ be a simple point measure on ${\displaystyle S}$. This means that

${\displaystyle \mu =\sum _{k}\delta _{s_{k}}}$

for distinct ${\displaystyle s_{k}\in S}$ and ${\displaystyle \mu (B)<\infty }$ for every bounded set ${\displaystyle B}$ in ${\displaystyle S}$. Further, let ${\displaystyle \nu }$ be a Markov kernel from ${\displaystyle S}$ to ${\displaystyle T}$.

Let ${\displaystyle \tau _{k}}$ be independent random elements with distribution ${\displaystyle \nu _{s_{k}}=\nu (s_{k},\cdot )}$. Then the point process

${\displaystyle \zeta =\sum _{k}\delta _{\tau _{k}}}$

is called the ν-transform of the measure ${\displaystyle \mu }$ if it is locally finite, meaning that ${\displaystyle \zeta (B)<\infty }$ for every bounded set ${\displaystyle B}$^[1]

For a point process ${\displaystyle \xi }$, a second point process ${\displaystyle \zeta }$ is called a ${\displaystyle \nu }$-transform of ${\displaystyle \xi }$ if, conditional on ${\displaystyle \{\xi =\mu \}}$, the point process ${\displaystyle \zeta }$ is a ${\displaystyle \nu }$-transform of ${\displaystyle \mu }$.^[1]

If ${\displaystyle \zeta }$ is a Cox process directed by the random measure ${\displaystyle \xi }$, then the ${\displaystyle \nu }$-transform of ${\displaystyle \zeta }$ is again a Cox-process, directed by the random measure ${\displaystyle \xi \cdot \nu }$ (see Transition kernel#Composition of kernels)^[2]

Therefore, the ${\displaystyle \nu }$-transform of a Poisson process with intensity measure ${\displaystyle \mu }$ is a Cox process directed by a random measure with distribution ${\displaystyle \mu \cdot \nu }$.

It ${\displaystyle \zeta }$ is a ${\displaystyle \nu }$-transform of ${\displaystyle \xi }$, then the Laplace transform of ${\displaystyle \zeta }$ is given by

${\displaystyle {\mathcal {L}}_{\zeta }(f)=\exp \left(\int \log \left[\int \exp(-f(t))\mu _{s}(\mathrm {d} t)\right]\xi (\mathrm {d} s)\right)}$

for all bounded, positive and measurable functions ${\displaystyle f}$.^[1]