A3.1 B3.1

Markov Chains | Part II, 2002

(i) Consider the continuous-time Markov chain (Xt)t0\left(X_{t}\right)_{t \geqslant 0} on {1,2,3,4,5,6,7}\{1,2,3,4,5,6,7\} with generator matrix

Q=(6200040230001001512010000000022060212000300010102)Q=\left(\begin{array}{rrrrrrr} -6 & 2 & 0 & 0 & 0 & 4 & 0 \\ 2 & -3 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & -5 & 1 & 2 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 & -6 & 0 & 2 \\ 1 & 2 & 0 & 0 & 0 & -3 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & -2 \end{array}\right)

Compute the probability, starting from state 3 , that XtX_{t} hits state 2 eventually.

Deduce that

limtP(Xt=2X0=3)=415\lim _{t \rightarrow \infty} \mathbb{P}\left(X_{t}=2 \mid X_{0}=3\right)=\frac{4}{15}

[Justification of standard arguments is not expected.]

(ii) A colony of cells contains immature and mature cells. Each immature cell, after an exponential time of parameter 2, becomes a mature cell. Each mature cell, after an exponential time of parameter 3, divides into two immature cells. Suppose we begin with one immature cell and let n(t)n(t) denote the expected number of immature cells at time tt. Show that

n(t)=(4et+3e6t)/7n(t)=\left(4 e^{t}+3 e^{-6 t}\right) / 7

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