Consider the following setup. We have two Poisson processes, the primary process $P_1$, and the secondary
process $P_2$. The secondary process $P_2$ is a standard Poisson process with rate $\lambda$. Unfortunately, the primary
process $P_1$ is not. $P_1$ is a modified non-homogenous Poisson process, where the rate function itself is random.
In particular, the rate function $\lambda_1(t)$ for $P_1$ is the value of $P_2(t)$, namely the number of events that
have occured on $[0, t]$ in the secondary Poisson process $P_2$. Note that conditional on the times of events in
$P_2$ on $[0, t]$, one has that $P_1$ is a non-homogenous Poisson process with piece-wise linear rate function on
$[0, t].$ Compute the m.g.f. of $P_1(t)$, the number of events to have occured in $P_1$ on $[0, t]$.
My idea is $E[e^{xP_1(t)}]=\sum_{k=0}^\infty E[e^{xP_1(t)}|P_2(t)=k]P(P_2(t)=k),$ and since $P_1(t)$ is a Poisson distribution with rate $\int_0^t\lambda_1(s)ds,$ then what remains is to find out the value of this integral. Recall $E[e^{x\cdot\text{expo($\gamma$)}}]=e^{\gamma(e^x-1)}.$ Since conditional on k arrivals of a standard Poisson process, these k arrivals arrive independently and uniformly over $[0,t],$ say $U_1[0,t],...,U_k[0,t]$. Then I think $\int_0^t\lambda_1(s)ds=\int_0^{x_1}0ds+\int_{x_1}^{x_2}1ds+...+\int_{x_k}^tkds=kt-(x_1+...+x_k),$ if $\{U_1,...,U_k\}=\{x_1,...,x_k\}$ and $x_1<...<x_k.$ Then $$E[e^{xP_1(t)}|P_2(t)=k]=\int_{0\le x_1<...<x_n\le t}e^{(kt-(x_1+...+x_k))(e^x-1)}\frac{1}{t^k}dx_1...dx_k.$$ It seems we need to find the value of integral of this kind $$\int_{0\le x_1<...<x_k\le 1}e^{-(x_1+...+x_k)}dx_1...dx_k$$ I have no idea how to deal with this integral.