Expected hitting time of a state in a continuous time Markov chain

Expected time for a continuous time Markov chain started in state i to reach state j for the first time.

Usage

ExpectedTime(C, i, j, useRCpp = TRUE)

Arguments

C

A ctmc object.

i, j

Indices (positions in states(C)) of the starting and target states.

useRCpp

Logical: solve the linear system in C++ (default) or with solve()?

Value

A number, the expected hitting time of j from i (0 if i == j).

Details

The expected hitting times \(m_k\) of j from the states \(k \ne j\) solve \(\sum_{l \ne j} Q_{kl} m_l = -1\) (Norris 1997, Section 3.3). Every state other than j must have a positive exit rate, otherwise an error is given; the times are finite only if j is reachable from every state, otherwise the system is singular.

References

Norris, J. R. (1997). Markov Chains. Cambridge University Press.

See also

Other continuous time Markov chains: ctmcFit(), freq2Generator(), generatorToTransitionMatrix(), impreciseProbabilityatT(), is.CTMCirreducible(), is.TimeReversible(), probabilityatT(), rctmc(), transition2Generator()

Author

Vandit Jain

Examples

states <- c("a", "b", "c", "d")
gen <- matrix(c(-1, 1/2, 1/2, 0,
                1/4, -1/2, 0, 1/4,
                1/6, 0, -1/3, 1/6,
                0, 0, 0, 0), nrow = 4, byrow = TRUE,
              dimnames = list(states, states))
ctmc <- new("ctmc", states = states, byrow = TRUE, generator = gen,
            name = "testctmc")
ExpectedTime(ctmc, 1, 4) # from "a" to the absorbing state "d"
#> [1] 7