Poisson point process in 2D with reflecting boundaries Consider a point process $\{(X_n,T_n)\}$ on a plane $[0,1/\lambda]\times\mathbf R^+$, generated from a Poisson point process $\{T_n\}$ with rate $\lambda$ on $\mathbf R^+$ (i.e. $(T_n-T_{n-1})$ is iid exponentially distributed $\sim\lambda\exp(-\lambda T)$) in the following way:
$X_0=T_0=0$ and the boundaries at $0$ and $1/\lambda$ are reflecting. The $X_n$ step direction is unchanged until hitting the boundary (or one can consider unbiased random walk alternatively), The step size is given by


*

*$X_n = T_n-T_{n-1}$ , or

*$X_n$ is uniformly distributed on $[0,T_n-T_{n-1}]$ , or

*$X_n$ is exponentially distributed with mean $T_n-T_{n-1}$
Q: Is this 2D point process Poissonian in each of the three cases? If so, is it homogeneous?
 A: In all cases i denote the 2d random measure by $\pi$.
1)  no its not, we don't have increment independence. Take two events $A=(\pi([0,1/\lambda) \times [\frac{2}{4\lambda},\frac{3}{4\lambda}) = 1), B=(\pi([\frac{3}{4\lambda},\frac{1}{\lambda})\times [\frac{3}{4\lambda},\frac{1}{\lambda} = 1)$. Clearly $P(A),P(B)>0$. Since the two boxes are disjoint, we should have
$P(A\cap B)=P(A)P(B)>0$. But under the event $A$, the T process has a jump at $[\frac{2}{4\lambda},\frac{3}{4\lambda})$, so if $T$ had another jump in $ [\frac{3}{4\lambda},\frac{1}{\lambda})$, the increment of $T$ would be smaller than $\frac{1}{2\lambda}$, and thus its corresponding $X$ would be bounded by $\frac{1}{2\lambda}$. Therefore $P(A\cap B)=0$
2)  no its not, for the exact same reason as in 1). Take two events $A=(\pi([0,1/\lambda) \times [\frac{2}{4\lambda},\frac{3}{4\lambda}) = 1), B=(\pi([\frac{3}{4\lambda},\frac{1}{\lambda})\times [\frac{3}{4\lambda},\frac{1}{\lambda} = 1)$. Clearly $P(A),P(B)>0$. Since the two boxes are disjoint, we should have
$P(A\cap B)=P(A)P(B)>0$. But under the event $A$, the T process has a jump at $[\frac{2}{4\lambda},\frac{3}{4\lambda})$, so if $T$ had another jump in $ [\frac{3}{4\lambda},\frac{1}{\lambda})$, the increment of $T$ would be smaller than $\frac{1}{2\lambda}$, and thus its corresponding $X$ would be bounded by $\frac{1}{2\lambda}$. Therefore $P(A\cap B)=0$
3) No its not. analogous to point 2). I will leave exact calculations to you, but with same events A,B as in 2) you see that $P(B)$ will be smaller given $A$, because the $X$ drawn in event $B$ is drawn from an exponential distribution with small mean.
