Holding time

The holding time of a continuous-time Markov chain is the time spent in its current state before the next jump. For $q_i>0$ it has an exponential distribution of total outgoing rate $q_i$; at an absorbing state with $q_i=0$ it is infinite.

The holding rate at state $i$ of a continuous-time Markov chain with generator matrix $Q=(q_{ij})$ isFor a conservative chain with finite $q_i>0$, its holding time $T_i$ has exponential distribution: $\mathrm{Pr}(T_i>t)=e^{-q_{i}t}$ and $\mathbb{E}{T}_i=1/q_i$. Conditional on a jump, the next state is $j$ with probability $q_{ij}/q_i$. Thus the rate determines when departure occurs, while normalized outgoing rates determine where it goes.

If the outgoing rates are $q_{i1}=2$ and $q_{i2}=3$ for two distinct other states, the total rate is five, the mean holding time is $1/5$, and departure chooses them with probabilities $2/5$ and $3/5$. A zero rate gives an absorbing state with infinite holding time. Statewise finite rates alone do not rule out explosion of a continuous-time Markov chain.

A state's holding time in a time-homogeneous continuous-time Markov chain is exponentially distributed with rate $-q_{rr}>0$. Its mean is $-1/q_{rr}$; the probability that its next jump is to state $s$ is $q_{rs}/(-q_{rr})$. An absorbing state has infinite holding time.

For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval $[{\lambda }_L,{\lambda }_U]$ with $0<{\lambda }_L\le {\lambda }_U$ for the rate transforms to one for the mean by reversing and inverting its endpoints. The monotonic transformation preserves the coverage event; it does not give a prediction interval for one realized holding time. A zero lower rate endpoint corresponds to an infinite upper mean endpoint.