Question 7.EX.18: Compute the renewal function when the interarrival distribut......
Compute the renewal function when the interarrival distribution F is such that
1−F(t) = pe^{−μ_1t} + (1 −p)e^{−μ_2t}We can imagine that a renewal corresponds to a machine failure, and each time a new machine is put in use its life distribution will be exponential with rate \mu_1 with probability p, and exponential with rate \mu_2 otherwise. Hence, if our state is the index of the exponential life distribution of the machine presently in use, then this is a two-state continuous-time Markov chain with intensity rates
q_{1,2}=\mu_1(1-p), \quad q_{2,1}=\mu_2 p
Hence,
\begin{aligned}P_{11}(t)= & \frac{\mu_1(1-p)}{\mu_1(1-p)+\mu_2 p} \exp \left\{-\left[\mu_1(1-p)+\mu_2 p\right] t\right\} \\& +\frac{\mu_2 p}{\mu_1(1-p)+\mu_2 p}\end{aligned}
with similar expressions for the other transition probabilities [P_{12}(t) = 1 − P_{11}(t), and P_{22}(t) is the same with μ_{2}p and μ_{1}(1 − p) switching places]. Conditioning on the initial machine now gives
E[Y(t)] = pE[Y(t)|X(0) = 1] + (1 – p)E[Y(t)|X(0) = 2]
=p\left[\frac{P_{11}(t)}{\mu_1}+\frac{P_{12}(t)}{\mu_2}\right]+(1-p)\left[\frac{P_{21}(t)}{\mu_1}+\frac{P_{22}(t)}{\mu_2}\right]
Finally, we can obtain m(t) from
\mu[m(t)+1]=t+E[Y(t)]
where
\mu=p / \mu_1+(1-p) / \mu_2
is the mean interarrival time.