Renewal process problem, where $X_i$'s are i.i.d. with exponential distribution. A room is lit by $2$ bulbs. Bulbs are replaced only when both bulbs burn out. Lifetimes of bulb's are i.i.d exponentially distributed with parameter $λ=1$. What fraction of the time is the room only half-lit?
 A: The immediate thought (which we will correct later) is to describe this situation as a continuous time Markov Chain with state space $$\Bbb X=\{0,1,2\}$$ where $X(t)$ denotes the number of bulbs that are lit at time $t\ge 0$. The initial state is $X(0)=2$ and the possible transitions are the following: 


*

*$2\to 0$: Is not possible, as two events cannot occur simultaneously.

*$2\to 1$: Is the minimum of two exponentials with parameter $1$ and hence is again exponentially distributed with parameter $2\times 1=2$.

*$1\to 0$: Due to the memoryless property of the exponential distribution this time is again exponential with parameter $λ=1$.

*$1\to 2$: Is not possible (due to the rules).

*$0\to 1$: Is not possible (due to the rules).

*$0\to 2$: This change is immediate. So, state $0$ is called instantaneous, that is when the process enters state $0$ it immediately leaves for state $2$. 


This last observation, allows us to modify the above as follows: We ommit state $0$ since the chain spends $0$ time there and consider instead the following state space $$\Bbb X=\{1,2\}$$ and the following transitions:


*

*$1\to 2$: This is an exponential time with parameter $λ=1$. Specifically, when the time expires (and the single bulb dies) we immediately replace both bulbs, hence the transition $1\to2$ instead of $1\to0\to2$.

*$2\to 1$: Is the minimum of two exponentials with parameter $1$ and hence is again exponentially distributed with parameter $2\times 1=2$.


So, the transition rate matrix $Q$ of the chain is given by $$Q=\begin{pmatrix}-1 &1\\2&-2\end{pmatrix}$$
You are looking for the stationary probability $π_1$. It can be found by solving the system \begin{align}\vec πQ&=0\\\vec π\mathbf{1}&=1\end{align} where $\vec p=(π_1, π_2)$. This gives $π_1=\frac23$, i.e. the room is half-lit $2/3$ of the time and fully lit $1/3$ of the time.
A: Let $Z(t)=1$ if the room is fully lit at time $t$ and $0$ if the room is half-lit at time $t$. Let $U_n\stackrel{\mathsf{iid}}\sim\mathsf{Exp}(2)$ and $D_n\stackrel{\mathsf{iid}}\sim\mathsf{Exp}(1)$, then $\{(U_n,D_n):n\geqslant 1\}$ is the alternating renewal process associated with $Z(t)$. Let $X_n=U_n+D_n$, then $X_n$ is a $\mathsf{iid}$ sequence with density
$$f_{X}(t) = f_U\star f_D(t) = \int_0^t f_U(s)f_D(t-s)\ \mathsf ds = 2(e^{-t}-e^{-2t}).$$
It follows that $\{(X_n,U_n):n\geqslant 1\}$ is a renewal reward process and by the renewal reward theorem, we have
$$\lim_{t\to\infty}\frac1t\int_0^t Z(s)\ \mathsf ds = \frac{\mathbb E[U_1]}{\mathbb E[U_1]+\mathbb E[D_1]}= \frac{\frac12}{\frac12+1}=\frac13. $$
Therefore the long-run fraction of time that the room is half lit is $1-\frac13=\frac23$.
