Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 37 / 5 / ii / c

The transition matrix over five years is $P(5)=e^{5Q}$. Its first row conditions on starting in state 1. The displayed approximate probabilities mean that, five years later, about 65% remain disease-free, 6% have local recurrence only, 10% have distant metastasis only, 2% have both, and 16% have died. They are probabilities of the states occupied at five years; they are not the probabilities of the next jump destinations or annual transition intensities. In particular the positive probability of state 4 does not require a direct $1\to 4$ arrow: the paths $1\to 2\to 4$ and $1\to 3\to 4$ contribute to it. The probability of death includes deaths after any allowed sequence of intermediate states.

The printed row sums to $0.99$ rather than one because the entries are approximate; an exact stochastic transition row sums to one. Exponentiating the displayed three-decimal generator gives approximately $(0.65377,0.06490,0.09775,0.02532,0.15826)$, illustrating that the edited, low-precision table should not be treated as an exact matrix exponential of its separately rounded rates.

State 5 is an absorbing state, so once entered it can never be left. Its generator row is zero, and the last row of every positive-time transition matrix is thereforeThis explains all four leading zeroes in the printed last row.