Finding the distribution of the total time spent in a transient state of a continuous time Markov chain

Question

Let $i$ be a transient state of a continuous-time Markov chain $X$ with $X(0) = i$. Furthermore $X$ has right continuous paths and the $Q$-matrix is stable and conservative. How do I show, that the total time spent in state $i$ has an exponential distribution and how do I find it´s rate? I suspect, that showing the memoryless property holds is sufficient, however I do not even know where to start. Any hints would be appreciated.

Answer 1

Keep in mind the following statement:

"Let $T_i$ be the holding time at state $i$, that is, the length of time that a continuous-time Markov chain started in $i$ stays in $i$ before transitioning to a new state. Then, $T_i$ has an exponential distribution."

As correctly you suggest in your question you have to prove that the holding time is exponentially distributed. Doing so we have to prove in the statement that $T_i$ is memoryless.

Let $s, t \geq 0$. For the chain started in $i$, the event$\{T_i >s\}$ is equal to the event that $\{X_u =i,\quad \text{for} \quad 0 \leq u \leq s\}$.Since$\{T_i >s+t\}$ implies $\{T_i > s\}$,

\begin{align*} P(T_i >s+t|X_0 =i)&=P(T_i >s+t,T_i >s|X_0 =i) \\ &=P(T_i >s+t|X_0 =i,T_i >s)P(T_i >s|X_0 =i)\\ &=P(T_i >s+t|X_u =i,\quad \text{for} \quad 0\leq u\leq s)P(T_i >s|X0 =i)\\ &=P(T_i >s+t|X_s =i)P(T_i >s|X_0 =i) \\ &=P(T_i >t|X_0 =i)P(T_i >s|X_0 =i), \end{align*} where the next to last equality is because of the Markov property, and the last equality is because of homogeneity. This gives that $T_i$ is memoryless. The result follows since the exponential distribution is the only continuous distribution that is memoryless.