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(This is to answer a question about Poisson processes asked by the OP in the comments.)

Here is an analog of Alekk's argument about the mutual singularity of the probability distributions of diffusions on bounded intervals, but for Poisson processes on $[0,+\infty[$. For every positive $\lambda$, call $Q_\lambda$ the probability distribution of the homogenoushomogeneous Poisson process with intensity $\lambda$ on $[0,+\infty[$, and $D_\lambda$ the set of locally finite subsets $S$ of $[0,+\infty[$ with asymptotic density $\lambda$: these are the sets $S$ such that $x^{-1}\\#(S\cap[0,x])\to\lambda$$x^{-1}\#(S\cap[0,x])\to\lambda$ when $x\to+\infty$.

Then, if $\lambda\ne\mu$, sets $D_\lambda$ and $D_\mu$ are disjoint by definition and $Q_\lambda(D_\lambda)=Q_\mu(D_\mu)=1$ by the law of large numbers. Hence $Q_\lambda$ and $Q_\mu$ are mutually singular.

This is in contrast to Poisson processes of finite total intensity. Let $I$ denote any measurable subset of $[0,+\infty[$ of finite Lebesgue measure $|I|$, and $Q^I_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $I$. Then all the measures $Q^I_\lambda$ are mutually absolutely continuous and the density of $Q^I_\lambda$ with respect to $Q^I_\mu$ at a finite subset $S$ of $I$ is $\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\\#S}$$\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\#S}$.

(This is to answer a question about Poisson processes asked by the OP in the comments.)

Here is an analog of Alekk's argument about the mutual singularity of the probability distributions of diffusions on bounded intervals, but for Poisson processes on $[0,+\infty[$. For every positive $\lambda$, call $Q_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $[0,+\infty[$, and $D_\lambda$ the set of locally finite subsets $S$ of $[0,+\infty[$ with asymptotic density $\lambda$: these are the sets $S$ such that $x^{-1}\\#(S\cap[0,x])\to\lambda$ when $x\to+\infty$.

Then, if $\lambda\ne\mu$, $D_\lambda$ and $D_\mu$ are disjoint by definition and $Q_\lambda(D_\lambda)=Q_\mu(D_\mu)=1$ by the law of large numbers. Hence $Q_\lambda$ and $Q_\mu$ are mutually singular.

This is in contrast to Poisson processes of finite total intensity. Let $I$ denote any measurable subset of $[0,+\infty[$ of finite Lebesgue measure $|I|$, and $Q^I_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $I$. Then all the measures $Q^I_\lambda$ are mutually absolutely continuous and the density of $Q^I_\lambda$ with respect to $Q^I_\mu$ at a finite subset $S$ of $I$ is $\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\\#S}$.

(This is to answer a question about Poisson processes asked by the OP in the comments.)

Here is an analog of Alekk's argument about the mutual singularity of the probability distributions of diffusions on bounded intervals, but for Poisson processes on $[0,+\infty[$. For every positive $\lambda$, call $Q_\lambda$ the probability distribution of the homogeneous Poisson process with intensity $\lambda$ on $[0,+\infty[$, and $D_\lambda$ the set of locally finite subsets $S$ of $[0,+\infty[$ with asymptotic density $\lambda$: these are the sets $S$ such that $x^{-1}\#(S\cap[0,x])\to\lambda$ when $x\to+\infty$.

Then, if $\lambda\ne\mu$, sets $D_\lambda$ and $D_\mu$ are disjoint by definition and $Q_\lambda(D_\lambda)=Q_\mu(D_\mu)=1$ by the law of large numbers. Hence $Q_\lambda$ and $Q_\mu$ are mutually singular.

This is in contrast to Poisson processes of finite total intensity. Let $I$ denote any measurable subset of $[0,+\infty[$ of finite Lebesgue measure $|I|$, and $Q^I_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $I$. Then all the measures $Q^I_\lambda$ are mutually absolutely continuous and the density of $Q^I_\lambda$ with respect to $Q^I_\mu$ at a finite subset $S$ of $I$ is $\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\#S}$.

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(This is to answer a question about Poisson processes asked by the OP in the comments.)

Here is an analog of Alekk's argument about the mutual singularity of the probability distributions of diffusions on bounded intervals, but for Poisson processes on $[0,+\infty[$. For every positive $\lambda$, call $Q_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $[0,+\infty[$, and $D_\lambda$ the set of locally finite subsets $S$ of $[0,+\infty[$ with asymptotic density $\lambda$: these are the sets $S$ such that $x^{-1}\\#(S\cap[0,x])\to\lambda$ when $x\to+\infty$.

Then, if $\lambda\ne\mu$, $D_\lambda$ and $D_\mu$ are disjoint by definition and $Q_\lambda(D_\lambda)=Q_\mu(D_\mu)=1$ by the law of large numbers. Hence $Q_\lambda$ and $Q_\mu$ are mutually singular.

This is in contrast to Poisson processes of finite total intensity. Let $I$ denote any measurable subset of $[0,+\infty[$ of finite Lebesgue measure $|I|$, and $Q^I_\lambda$ the probability distribution of the homogenous Poisson process with intensity $\lambda$ on $I$. Then all the measures $Q^I_\lambda$ are mutually absolutely continuous and the density of $Q^I_\lambda$ with respect to $Q^I_\mu$ at a finite subset $S$ of $I$ is $\mathrm{e}^{(\mu-\lambda)|I|}(\lambda/\mu)^{\\#S}$.