Given a birth process $\{B_t:t\geqslant0\}$ with $\lambda >0$, define $$K_t=\int_{0}^{t}B_s ds=\sum_{i=1}^{n}B_{t_{i}}(t_{i+1}-t_i)$$ if there were $n$ births in $[0,t]$ and let $t_{i}$ be the times of birth. Note that the soujorn time $$S_{i}:=(t_{i+1}-t_i)\sim\mathrm{ Exp}(\lambda) .$$ Define $$ \sigma_{t}=\inf\{s\geq 0:K_{s}>t\} $$ Can we think of the time scaling $t\rightarrow \sigma_t$ that $B_{\sigma_t}$ is Poisson?

The reason behind this:

It is well known, that $$ B_{t}-\underbrace{\lambda\int_{0}^{t}B_{s}ds}_{A_t} $$ is a Martingale, where $A_{t}$ is a strictly increasing continuous process . Hence $A_t$ is the compensator of $B_t$, which is a counting process, since $A_0=0$ the representation is unique. Since $A_t$ is continuous, $B_t$ is quasi-left-continuous. For a quasi-left-continuous counting process $B_t$ with compensator $A_t$ define the stopping time $$ \tau_t:=\inf\{s\geq 0:A_s>t\} $$ Then $B_{\tau_t}$ is poisson process with rate 1, see Theorem 5 here. The relation is $\lambda K_t=A_t$. As Did mentions, it holds $$ \sigma_{t}=\tau_{\lambda t}. $$

There are some problems considering the starting value $B_{0}=x$, since a poisson-prozess starts at 0. So can we think of $B_{\sigma_t}$ as a Poisson-Prozess starting at $x$?

If this is not valid, then we just consider $\tilde{B_t}=B_{t}-B_{0}$ (which is the classical counting process starting at 0, which counts the births) and the martingale is $$ B_{t}-B_{0}-\lambda \int_{0}^{t}B_sds $$ With the same argumentation as above and same defined stopping time $\tau_{t}$ then least $\tilde{B}_{\tau_{t}}$ is a poisson process with rate 1. Since $\tilde{B}_{\lambda \tau_t}=\tilde{B}_{\sigma_{t}}$, our process $(\tilde{B}_{\sigma_{t}})_{t\geq 0}$ is a poisson-process with rate $\lambda$.

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    $\begingroup$ Pathwise fact: the process $X_t=B_{\sigma_t}$ starts from $X_0=1$, jumps from $i$ to $i+1$, for every $i\geqslant1$, and its sojourn time at $i$ is $iS_{i+1}$. Stochastic consequence: $(X_t)$ is not a Poisson process. $\endgroup$ – Did Jun 3 '16 at 5:57
  • $\begingroup$ @Did i edited some theory behind it. I guess it has to be poisson. Maybe if we watch the counting process $\tilde{B}_t=B_{t}-B_{0}$ for given population. $\endgroup$ – ziT Jun 4 '16 at 9:26
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    $\begingroup$ "And if im right $\lambda \sigma_t=\tau_t$" Actually, $$\sigma_t= \tau_{\lambda t}.$$ $\endgroup$ – Did Jun 4 '16 at 16:44
  • $\begingroup$ @Did thank you so much for your support. Then $\tilde{B}_{\sigma_{t}}$ is a poisson process with rate $\lambda$. This coincides with the expectation. I dont know how i can reward you :) $\endgroup$ – ziT Jun 4 '16 at 17:47

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