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edited Sep 18, 2017 at 18:27

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Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$$\mathbb{P}$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$$\mathbb{E}[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$$\mathbb{P}^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$$\mathbb{E}^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$$$ \mathbb{E}^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb{E}\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb{E}[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$$\mathbb{E}[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs $B$ bounded and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$$\mathbb{P}^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs $B$ bounded and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $\mathbb{P}$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb{E}[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $\mathbb{P}^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb{E}^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb{E}^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb{E}\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb{E}[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $\mathbb{E}[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs $B$ bounded and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $\mathbb{P}^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

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edited Sep 18, 2017 at 18:10

nullUser

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Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs "bounded"$B$ bounded and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs "bounded" and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs $B$ bounded and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

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edited Sep 18, 2017 at 16:00

nullUser

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Does the Palm version of a stationary point process of finite intensity have boundedly finite first moment measure?

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs "bounded" and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.

Does the Palm version of a stationary point process of finite intensity have finite first moment measure?

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs "bounded" and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely.

Does the Palm version of a stationary point process of finite intensity have boundedly finite first moment measure?

Suppose $\Phi$ is a stationary point process on $\mathbb{R}^d$ under $P$ with intensity $0<\lambda<\infty$. In particular the first moment measure $\mathbb E[\Phi(B)] = \lambda |B|$ where $|B|$ denotes volume. Denote $P^0$ the Palm measure of $\Phi$. Does it hold that $\mathbb E^0[\Phi(B)] < \infty$ for all $B$ bounded? Taking $B:=B(0,r)$, for any $\epsilon > 0$ $$ \mathbb E^0[\Phi(B)] = \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E\int_{B(0,\epsilon)} \Phi(B(t,r))\,\Phi(dt) \leq \frac{1}{\lambda |B(0,\epsilon)|}\mathbb E[\Phi(B(0,\epsilon))\Phi(B(0,r+\epsilon))]. $$

Certainly if $E[\Phi(B(0,r +\epsilon))^2]<\infty$ then the previous expectation would be finite, but I would like to avoid this assumption if possible.

Note that if this is true for an arbitrary $B$, one needs "bounded" and not just $|B|<\infty$. E.g. if $\Phi$ is a stationary copy of $\mathbb{Z}^2$ in $\mathbb{R}^2$, then the $x$-axis is a set of measure zero that under $P^0$ contains infinitely many points almost surely. Essentially, the question reduces to the question just for balls.