Question: Let $\tau_i\sim \text{Exp}(\lambda_0)$ iid and $\gamma_i\sim \text{Exp}(\lambda_1)$ iid and independent of each other and set $N_t=max\{k\geq 0: \sum_{i=1}^k{(\tau_i+\gamma_i)}\leq t\}$. What is $\mathbb{E}(N_t)$?
My attempt:(inspired by section 3 of this paper) Let $(\sigma_t)_{t\geq 0}$ be a two-state continuous-time Markov chain on $\\{0,1\\}$ with jumps $0\rightarrow 1$ at rate $\lambda_0$ and $1\rightarrow 0$ at rate $\lambda_1$, i.e. the q-matrix is given by $$\begin{pmatrix}-\lambda_0&\lambda_0\\\ \lambda_1&-\lambda_1\end{pmatrix}$$
Starting from $\sigma_0=0$ the time $\rho_1$ to the first jump of $(\sigma_t)_{t\geq 0}$ is $\text{Exp}(\lambda_0)$ distributed, the time $\rho_2$ to the second jump is $\text{Exp}(\lambda_1)$-distributed, $\rho_3\sim \text{Exp}(\lambda_0)$ again, and so on.
Thus, $\rho_{2i-1}$ mimicks $\tau_i$ and $\rho_{2i}$ mimicks $\gamma_i$. Now $M_t=max\{k\geq 0: \sum_{i=1}^k{\rho_i}\leq t\}$ is a Poisson process with stochastic rate $(\lambda_{\sigma_t})_{t\geq 0}$ (a so-called Cox process) and its expectation is simply
$$\mathbb{E}(M_t)=\mathbb{E}(\mathbb{E}(M_t|\sigma_t))=\mathbb{E}(\lambda_{\sigma_t}t)=(\lambda_0\mathbb{P}(\sigma_t=0|\sigma_0=0)+\lambda_1\mathbb{P}(\sigma_t=1|\sigma_0=0))t\\ =\left(2\lambda_1+(\lambda_0-\lambda_1)e^{-(\lambda_0+\lambda_1)t}\right)\frac{\lambda_0}{\lambda_0+\lambda_1}t$$ (where we used the probabilities from Wikipedia)
$M_t$ counts every jump, $N_t$ only pairs of jumps, so (roughly) $N_t=\lfloor M_t/2\rfloor$ and I suspect that $\mathbb{E}(N_t)\approx\mathbb{E}(M_t)/2$. But I am unable to see, how one could get an exact expression for $\mathbb{E}(N_t)$.
Edit: The last claim can be refined: Since $N_t=\lfloor M_t/2\rfloor$ we have $\mathbb{P}(N_t=n)=\mathbb{P}(M_t=2n)+\mathbb{P}(M_t=2n+1)$ and thus $$\mathbb{E}(N_t)=\sum_{n=0}^\infty{n\mathbb{P}(M_t=2n)+n\mathbb{P}(M_t=2n+1)}$$ and hence $$\mathbb{E}(M_t)=\sum_{n=0}^\infty{2n\mathbb{P}(M_t=2n)+(2n+1)\mathbb{P}(M_t=2n+1)}\\ =2\sum_{n=0}^\infty{n(\mathbb{P}(M_t=2n)+\mathbb{P}(M_t=2n+1))}+\sum_{n=0}^\infty{\mathbb{P}(M_t=2n+1)}\\ =2\mathbb{E}(N_t)+\sum_{n=0}^\infty{\mathbb{P}(M_t=2n+1)}$$
From equation (3.6) in section 3 of the aforementioned paper, we know that $$\mathbb{P}(M_t=2n+1)=\frac{\lambda_0^{n+1}\lambda_1^n t^{2n+1}}{(2n+1)!}e^{-\lambda_0t}\Phi(n+1,2(n+1),(\lambda_0-\lambda_1)t)$$ where (see Wikipedia, they call it $M(a,b,z)$): $$\Phi(a,b,z)=\sum_{k=0}^\infty{\frac{a^{(k)}}{b^{(k)}}\frac{z^k}{k!}}$$ and $a^{(k)}=a(a+1)\cdot...\cdot(a+k-1)$ is the rising factorial. Since $\frac{1}{2}=\frac{n+1}{2(n+1)}\leq \frac{n+1+m}{2(n+1)+m}\quad\forall m\in\mathbb{N}_0$ we have $\frac{(n+1)^{(k)}}{(2(n+1))^{(k)}}\leq \frac{1}{2^k}$ and therefore $$\Phi(n+1,2(n+1),(\lambda_0-\lambda_1)t)=\sum_{k=0}^\infty{\frac{(n+1)^{(k)}}{(2(n+1))^{(k)}}\frac{((\lambda_0-\lambda_1)t)^k}{k!}}\leq \sum_{k=0}^\infty{\frac{((\lambda_0-\lambda_1)t/2)^k}{k!}}=e^{(\lambda_0-\lambda_1)t/2}$$ Consequently $$\mathbb{P}(M_t=2n+1)\leq \frac{\lambda_0^{n+1}\lambda_1^n t^{2n+1}}{(2n+1)!}e^{-\lambda_0t}e^{(\lambda_0-\lambda_1)t/2}=\sqrt{\frac{\lambda_0}{\lambda_1}} \frac{(\sqrt{\lambda_0\lambda_1}t)^{2n+1}}{(2n+1)!}e^{-(\lambda_0+\lambda_1)t/2}$$ and using $\sinh(x)=\sum_{n=0}^\infty{\frac{x^{2n+1}}{(2n+1)!}}$: $$\sum_{n=0}^\infty{\mathbb{P}(M_t=2n+1)}\leq \sqrt{\frac{\lambda_0}{\lambda_1}}\sinh(\sqrt{\lambda_0\lambda_1}t)e^{-(\lambda_0+\lambda_1)t}$$
Thus we get the following bounds on $\mathbb{E}(N_t)$: $$\frac{1}{2}\mathbb{E}(M_t)\geq \mathbb{E}(N_t)\geq \frac{1}{2}\mathbb{E}(M_t)-\frac{1}{2}\sqrt{\frac{\lambda_0}{\lambda_1}}\sinh(\sqrt{\lambda_0\lambda_1}t)e^{-(\lambda_0+\lambda_1)t}$$ In particular, as $t\rightarrow \infty$ we find $\mathbb{E}(N_t)\sim\frac{1}{2}\mathbb{E}(M_t)\sim \frac{\lambda_0\lambda_1}{\lambda_0+\lambda_1}t$
If possible, I would love to receive feedback on this edit. Is everything correct?
Any help or comment is greatly appreciated!