15.2 Absorbing chains
A state is absorbing if \(p_{ii} = 1\). If every state can reach an absorbing one, order the states with the transient ones first: \[ P = \begin {pmatrix} Q & R \\ 0 & I \end {pmatrix} . \]
Theorem 15.2.1. The fundamental matrix \(N = (I - Q)^{-1} = I + Q + Q^2 + \cdots \) exists; \(N_{ij}\) is the expected number of visits to transient state \(j\) starting from transient state \(i\); \(N\mathbf 1\) is the vector of expected times to absorption; and \(B = NR\) gives the probabilities of ending in each absorbing state.
Proof. \(Q^n \to 0\), since absorption occurs eventually with probability one from every transient state, so the series converges and equals \((I - Q)^{-1}\). \((Q^n)_{ij}\) is the probability of being in \(j\) at step \(n\), so the sum is the expected number of visits. The probability of absorption in \(k\) is \(\sum _n\sum _j(Q^n)_{ij}R_{jk} = (NR)_{ik}\), summing over the last transient state \(j\) visited and the step \(n\) at which it is visited. □
Example 15.2.2. A firm’s accounts are \(0\)–\(30\) days old, \(31\)–\(60\) days old, paid, or written off. Each month, \(40\%\) of \(0\)–\(30\) day accounts are paid and \(60\%\) age; of \(31\)–\(60\) day accounts, \(70\%\) are paid and \(30\%\) are written off. Then \(N = \begin {pmatrix} 1 & 0.6 \\ 0 & 1 \end {pmatrix}\) and \(B = NR = \begin {pmatrix} 0.82 & 0.18 \\ 0.70 & 0.30 \end {pmatrix}\): of new receivables of Rs. 10 lakh, Rs. 8.2 lakh will be collected and Rs. 1.8 lakh lost, a provision for bad debts computed from the chain.
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