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Dividing the expected occupation time of a continuous-time Markov chain by $t$ and letting $t\to \infty$ givesFor $\lambda ,\mu >0$ this is also the almost-sure long-run time fraction, not just the limit of an expectation. The two-state continuous-time Markov chain is irreducible and positive recurrent. Alternatively, a regeneration cycle consists of a mean $1/\lambda$ symptom-free holding time followed by a mean $1/\mu$ symptomatic holding time; the renewal-reward theorem gives $(1/\mu )/(1/\lambda +1/\mu )=\lambda /(\lambda +\mu )$. This agrees with the stationary distribution solving ${\pi }_1\lambda ={\pi }_2\mu$ and ${\pi }_1+{\pi }_2=1$.

Starting in state 1, if $\lambda =0$ the fraction is zero; if $\lambda >0$ and $\mu =0$ the patient eventually enters state 2 forever, and the fraction is one. If both rates vanish, the fraction remains zero from this initial state and the quotient $0/0$ is undefined.

Let $A_2(t)={\int }_0^{t}1\{X(s)=2\}ds$ be the occupation time of a continuous-time Markov chain in the symptoms state. Tonelli theorem permits exchanging the nonnegative integral and expectation. Starting from state 1, put $q=\lambda +\mu >0$. ThensoFor small $t$ the answer is $\lambda t^2/2+O(t^3)$, as expected when onset must first occur. It is between zero and $t$. If $\lambda =\mu =0$, no transition can occur and the answer is zero; the displayed quotient is interpreted separately in that degenerate case.

Let $r_i$ be the probability of returning to $i$ after its first departure. Transience means $r_i<1$; in particular $q_i>0$. By the Strong Markov property, the number $N$ of visits, including the initial visit, has the geometric distribution $\mathbb{P}(N=n)=(1-r_i)r_i^{n-1}$, $n\ge 1$. The holding times of these visits are independent $\mathrm{Exp}(q_i)$ variables and are independent of the jump-chain return decisions.

For the total occupation time of a continuous-time Markov chain $T_i$ and $s\ge 0$, summing over $N$ givesUniqueness of the Laplace transform identifiesStarting elsewhere adds an atom at zero if the state might never be hit; the purely exponential conclusion uses the specified initial state $i$.

Write $Z(u)$ for the continuous-time Markov chain and ${\mathbb{E}}_r$ for expected value conditional on $Z(0)=r$. The occupation time of a continuous-time Markov chain in state $s$ is the integral of its indicator random variable. By Tonelli theorem,The interchange is valid because the integrand is nonnegative; indeed the random integral is bounded by $t$.

Integrate the expected Markov-chain entry count density and use the occupation time of a continuous-time Markov chain formula:Each term is a transition intensity multiplied by the expected exposure time to that transition. This counts every entry, including returns, rather than only the first visit. For countably many states the same nonnegative interchange follows from Tonelli theorem, provided the resulting expected count is finite. The sum excludes $i=s$.