Suppose $z(t)$ is the solution of sde.
Let, $ \widetilde{z}(t) = z(t) - z(t_{2k})\textbf{1}_{k\neq 0}$ for $t_{2k}\leq t < t_{2k+1}$, $ \widetilde{z}(t) = 0$ for $t_{2k+1}\leq t < t_{2k+2}$ for $k = 0, 1, 2...$
And, time interval follows the different exponential distribution, i.e. $t_{2k+1} - t_{2k}$ ~ exp($\lambda_1$), $t_{2k+2} - t_{2k+1}$ ~ exp($\lambda_2$) iid.
My final goal is to calculate $E(\widetilde{z}(t))$ with $E({z(t)}), t_0, \lambda_1, \lambda_2$.
I'm trying to calculate like this....
$E(\widetilde{z}(t)) = \sum_{k=0}^{\infty }E\left [ z(t) - z(t_{2k})\textbf{1}_{k\neq 0} \right ]P(t_{2k}\leq t<t_{2k+1})$
$= E(z(t))\sum_{k=0}^{\infty }P(t_{2k}\leq t<t_{2k+1})-\sum_{k=1}^{\infty }E(z(t_{2k}))P(t_{2k}\leq t<t_{2k+1})$
$=E(z(t))\frac{\frac{1}{\lambda_1}}{\frac{1}{\lambda_1}+\frac{1}{\lambda_2}}-\sum_{k=1}^{\infty }E(z(t_{2k}))P(t_{2k}\leq t<t_{2k+1})$
So, i tried to calculate $P(t_{2k}\leq t<t_{2k+1})$.
$P(t_{2k}\leq t<t_{2k+1}) = P(t_{2k}-t_0\leq t - t_0<t_{2k+1}-t_0) = P(X_k\leq t - t_0<Y_k) $ where $X_k$~gamma(k, $\lambda_1$)+gamma(k, $\lambda_2$), $Y_k$~gamma(k, $\lambda_2$)+gamma(k+1, $\lambda_1$)
I don't know how to calculate more.
cf) This process is known to alternating renewal process.