# infinitesimal generator of a Markov process

Problem: There are $$n$$ identical components in a system that operate independently. When a component fails, it undergoes repair, and after repair is placed back into the system. Assume that for a component the operating times between successive failures are i.i.d. exponential with mean $$\frac{1}{\lambda}$$, and that these are independent of the successive repair times, which are i.i.d. exponential with mean $$\frac{1}{\mu}$$. The state of the system is the number of components in operation. Determine the infinitesimal generator of this Markov process.

I try to find the transition probabilities $$\mathbb{P}(X_t = j|X_s = i)$$. First, consider the cases when $$i>j$$, which means more components failed in the time period than repaired. Let $$k$$ be the number of components that fails during $$[s,t]$$, then the rest of $$i$$ components didn't fail. Also $$k - (i-j)$$ components were repaired. I got this equation: \begin{align} \mathbb{P}(X_t = j|X_s = i) &= p_{ij}(t-s)\\ &=\sum_{k=i-j}^{i}{i \choose k}(\int_0^{t-s}\lambda e^{-\lambda u}du)^k(\int_{t-s}^{\infty}\lambda e^{-\lambda u}du)^{i-k}{n-i+k \choose n-j}(\int_0^{t-s}\mu e^{-\mu u}du)^{k-(i-j)}(\int_{t-s}^{\infty}\mu e^{-\mu u}du)^{n-j} \end{align} But this doesn't seem right, because then I get $$q_{ij} = p'_{ij}(0) = 0$$. Help is much appreciated.