Can anyone give me some hint on the following problem? Thanks a lot!
Let $\{T_n:n\ge 0\}$ be a point process and $\{N_t: t\ge 0\}$ be the corresponding counting process which admits a bounded intensity $\lambda_t$. From this point process we define a new point process $\{\hat T_n: n\ge 0\}$ as follows: an epoch $T_n$ of the original process is also an epoch for the new process if and only if $n\ge g(T_n)$, where $g(t)$ is a given increasing and continuous function. Furthermore, let $\{\hat N_t: t\ge 0\}$ be the counting process associated to $\hat T_n$.
Prove: $\hat N_t$ also admits an intensity $\hat\lambda_t$. Express $\hat \lambda_t$ as a function of $\lambda_t$, $N_t$, and $g$.
My current Proof: (not quite sure)
Let $C_t$ be an arbitrary nonnegative $\mathcal F_t$-predictable process. Notice that we always have $n=N_{T_n}$ for all $n\in\mathbb N$, and hence $\mathbb 1_{\{n\ge g(T_n)\}}=\mathbb 1_{\{N_{T_n}\ge g(T_n)\}}$. Therefore, by definition it follows that \begin{align*} \mathbb E\bigg[\int_0^\infty C_s\,d \tilde N_s\bigg]&=\mathbb E\bigg[\sum_{T_n\ge 0} C_{T_n}\cdot \mathbb 1_{\{N_{T_n}\ge g(T_n)\}}\bigg]=\mathbb E\bigg[\int_0^\infty C_s\cdot \mathbb 1_{\{N_{s^-}+1\ge g(s)\}}\,d N_s\bigg]\\ &=\mathbb E \bigg[\int_0^\infty C_s\cdot \mathbb 1_{\{N_{s}+1\ge g(s)\}}\cdot \lambda_s\,d s\bigg]. \end{align*} As the function $\mathbb 1_{\{N_{s+1}\ge g(s)\}}$ is left-continuous, it is also $\mathcal F_t$-predictable, which implies that the product $\mathbb 1_{\{N_{s+1}\ge g(s)\}}\,\lambda_s$ is $\mathcal F_t$-predictable. Hence, it is safe to conclude that the new process $\{\tilde N_t\,|\, t\ge 0\}$ admits an intensity $\tilde\lambda_s=\lambda_s\cdot \mathbb 1_{\{N_{s+1}\ge g(s)\}}$.
(The question is horribly asked (no personal input and all that) but the problem is interesting and my soul is weak hence here we go...)
Let $\Lambda_t=\displaystyle\int_0^t\lambda_s\mathrm ds$, then the number of events in $[0,t)$ is Poisson with parameter $\Lambda_t$ hence, for every $n\geqslant0$, the distribution of $T_{n+1}$ has density $$ \lambda_t\cdot\mathrm e^{-\Lambda_t}\frac{\Lambda_t^n}{n!}. $$ This is an event of the new process if and only if $g(t)\leqslant n+1$. Summing on $n\geqslant0$, one gets the intensity of the new process when $(\lambda_t)$ is deterministic, as $$ \lambda_t\cdot\mathrm e^{-\Lambda_t}\sum_{n=0}^\infty\frac{\Lambda_t^n}{n!}\mathbf 1_{n+1\geqslant g(t)}. $$ When $(\lambda_t)$ is random and predictable with respect to the natural filtration of $(N_t)$, the intensity of the new process conditionally on $(N_t)$ is $$ \lambda_t\cdot\mathbf 1_{N_t+1\geqslant g(t)}. $$