Finding limiting probability for continuous-time Markov chain

Question

Assume that an individual only has two possible states: susceptible (S) and infected (I). Further, assume that the individuals in the population are independent, and that for each susceptible individual the time until the next infection follows an exponential distribution with expected value 1/λ = 100 days and that the durations of the infections follow independent exponential distributions with expected values 1/μ = 7 days.

Consider a single individual, and let X(t) be the state (S or I) of the individual at time t measured in days.

Specify the transition rates for this continuous-time Markov chain, and calculate the long-run mean fraction of time per year that an individual has a cold.

So the transition rates I believe must be $\lambda = \frac{1}{100}$ and $\mu = \frac{1}{7}$. Since there are only two possible states, intuitively the long-run mean fraction of time per year that an individual has a cold should be around 0.07 (around 7 days every 100 days on average). But I don't know how to calculate the limiting probability of this rigorously

Answer 1

The transition rates are $\lambda^{-1}=100$ and $\mu^{-1}=7$ (recall that the rate is the reciprocal of the mean). Since this Markov chain only has two states, there is but one balance equation: $$ \lambda\pi_S = \mu\pi_I, $$ from which $\pi_I = \frac\lambda\mu \pi_S$. Now, $\pi_S+\pi_I=1$, and so $$ 1 = \pi_S\left(1 + \frac\lambda\mu\right)\implies \pi_s = \frac{1}{1+\frac\lambda\mu} = \frac{\mu}{\lambda+\mu}. $$ It follows readily that $\pi_I = 1 - \pi_S = \frac\lambda{\lambda+\mu}$. The long-run mean fraction of time per year that an individual has a cold is simply $\pi_S$. Substituting $\lambda = \frac1{100}$ and $\mu=\frac17$, this is $$ \pi_S = \frac{\frac17}{\frac1{100}+\frac17} = \frac{100}{107}. $$