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

Question

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.

Answer 1

The answer is no. One can construct examples where $\Phi(B)$ has infinite variance. This violates the claim as you calculated. Example: Let $(m_n)_{n\in\mathbb Z}$ be i.i.d. random variables in $\mathbb N\cup \{0\}$ with mean 1 but infinite variance. For each $n\in\mathbb Z$, choose $m_n$ points in the interval $[n,n+1)$ with the uniform distribution independently. Let $\Phi_0$ denote this point process. Finally, let $\Phi:=\Phi_0-U$, where $U\in [0,1)$ is a random number with the uniform distribution.