Cutting time out of a Poisson process

Question

Suppose we have a Poisson process on the time interval $[0, \infty)$. Let $N(t)$ denote the number of arrival events up to epoch $t$, and let $S_n$ denote the epoch of the $n$th arrival.

Suppose each sequence of arrival epochs $0 \le S_1 \le S_2 \le S_3 \dots$ determines another sequence $0 < a_1 < a_2 < a_3 \dots$ which satisfies the following condition:

For each $a_k$, the sequence of epochs $\{S_j | S_j \ge a_k\}$ which occur on or after $a_k$ is a Poisson process; i.e, the wait time from $a_k$ until the next arrival epoch $S_j$ is exponentially distributed, and the interarrival times from then on are exponentially distributed.

Question: If the time spent in intervals $[a_{2k}, a_{2k+1})$ is collapsed to zero, we can consider the arrivals in the union of the remaining intervals $$\bigcup_k [a_{2k-1}, a_{2k})$$ as a counting process in $\mathbb{R}_+$.
Is this new counting process a Poisson process?
If in general not, then are there additional conditions we can impose that would make it so?

Thank you very much in advance for any help or references.

Answer 1

No. Let $T_i$ be the event time of the Poisson process. Let $a_1 = T_1, a_2 = T_2$. Let $a_4= inf \lbrace T_i,i>2, T_i - T_{i-1} > T_1 \rbrace$ and $a_3 = a_4 - T_1$. $a_4$ is a stopping time so the post $a_4$ process is a poisson process independent of the past, including $T_1$, $a_4 - a_3 = T_1$ and by construction s the first event time of the post-$a_3$ process. . Repeat, let $a_6 = inf \lbrace T_i > a_4: T_i - T_{i-1} > T_1, a_5 = a_6 - T_1$. $a_6$ is a stopping time. Continuing in this way one produces intevals $a_i$ where the odd gaps are one event at time $T_1$.