Given jobs completed in succession, and time to complete the job follows a distribution $F$. Assume the shocks occur follow $\ PP(\lambda)$, and the jobs restarted at shocks. Find the rate at which jobs are completed using the renewals defined as when shocks occur.
My attempt: Let $Z$ = random time to complete the job, so $Z$ has distribution $F$, $Y = $ time to the next shock, so $Y$ follows $\exp(\lambda)$, and $X_j = $ time of the $j$th cycle.
Now, as our renewals = shocks occur, we have: $E(X_j) = \frac{1}{\lambda}$ because $X_j$ follows $PP(\lambda)$. We now need to find $E(R_j)$ and then apply the renewal reward theorem and compute $\frac{E(R_j)}{E(X_j)}$. Now, $R_j = \sum_{i=1}^{N} min(Z_j, Y_j)$ where $N = $ first index such that $Z_j<Y_j$. So $N$ follows geometric distribution with $p = P(Z_j<Y_j)$. Then $E(R_j) = E(N)E(min(Z_j, Y_j)) = \frac{E(min(Z_j, Y_j))}{P(Z_j<Y_j)}$ since $E(N) = \frac{1}{p}$. Thus, $\frac{E(R_j)}{E(X_j)} = \lambda\frac{E(min(Z_j, Y_j))}{P(Z_j<Y_j)}$.
My question: Could someone help let me know if my formula above for $R_j$ correct? If not, please help correct it if possible. Otherwise, any inputs would be welcomed as well.