An ice cream machine is either operation (state 1) or being repaired (state 2). Repairs take on average 6 hours and the ice cream machine breaks down once a day on average. This follows a time-homogeneous markov process where time is measured in days.
Question: Calculate the constant transition intensities
I cannot calculate the transition intensities from state 1 to state 2. For the transition intensity from state 2 to state 1 I have used the fact that E(T2) = 0.25 and since T2 follows an exponential (-h22) where T2 is the sjourn time in state 2 however how would i calculate h12 or h11.
Any help would much be appreciated.
The transition matrix of the embedded Markov chain is given by $$ Q = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, $$ and the generator matrix by $$ A = \begin{pmatrix} -1 & 1\\ \frac14 & -\frac14 \end{pmatrix}. $$ Hence the Kolmogorov forward equation $P'(t) = P(t)A$ are \begin{align} P'_{11}(t) &= - P_{11}(t) + \frac14 P_{12}(t)\tag1\\ P'_{21}(t) &= - P_{21}(t) + \frac14 P_{22}(t). \end{align} Setting $P_{12}(t) = 1-P_{11}(t)$, equation $(1)$ becomes $$ P'_{11}(t) = -\frac54 P_{11}(t)+\frac14. $$ This is a first order linear differential equation of the form $$ y'(t) + p(t)y(t) = g(t), $$ so we compute the integrating factor $\mu(t) = Ce^{\int p(t)\ \mathsf dt}$, that is, $$ Ce^{\int \frac54\ \mathsf du} = Ce^{\frac54t} $$ to get $$ \left[Ce^{\frac34t}P_{11}(t)\right]' = \frac C4e^{\frac54t}. $$ Integrating both sides and solving for $P_{11}(t)$, we find that $$ P_{11}(t) = Ce^{-\frac{5 }{4}t}+\frac{1}{5}. $$ Since $P(0)=I$, we have $P_{11}(0)=1$ as an initial condition, and hence $C=\frac45$. So $$ P_{11}(t) = \frac15\left(1 + 4e^{-\frac 54} \right). $$ We can also find the transition probability matrix by computing the exponential of the generator matrix: $$ P(t) = e^{At} = \sum_{n=0}^\infty \frac{(At)^n}{n!} = \left( \begin{array}{cc} \frac15\left(1 + 4 e^{-\frac{5 }{4}t}\right) & \frac45\left(1- e^{-\frac{5 }{4}t}\right) \\ \frac{1}{5}\left(1-e^{-\frac{5 }{4}t}\right) & \frac{1}{5}\left(4 + e^{-\frac{5 }{4}t}\right) \\ \end{array} \right). $$