Stationary Distribution

Study guide: Exam MAS-I

The Stationary Distribution $\boldsymbol{\pi}$ of a Markov Chain is a probability vector that one transition leaves unchanged: a chain started in $\boldsymbol{\pi}$ stays in $\boldsymbol{\pi}$ . For an irreducible chain, $\pi_j$ is the long-run proportion of periods spent in state $j$ ; if the chain is also aperiodic, $\pi_j$ is the limit of $P^n_{ij}$ whatever the starting state $i$ .

$\pi_j = \sum_i \pi_i\, P_{ij} \quad \text{for every } j$

$\sum_j \pi_j = 1$

Solving: write the balance equations $\boldsymbol{\pi}\mathbf{P} = \boldsymbol{\pi}$ , drop any one of them (they are linearly dependent because every row of the Transition Probability Matrix sums to one) and use $\sum_j \pi_j = 1$ in its place.
A finite irreducible chain always has a unique stationary distribution. If it is also aperiodic (ergodic), $\lim_{n\to\infty} P^n_{ij} = \pi_j$ . A periodic chain — one that alternates between two states, say — has a $\boldsymbol{\pi}$ , but $\mathbf{P}^n$ never settles.
Mean return time to state $j$ is $1/\pi_j$ . Long-run average reward: if state $j$ earns $r(j)$ per period, the long-run average is $\sum_j r(j)\,\pi_j$ — the average premium in a Bonus-Malus System, or the average expected loss of a portfolio whose risk mix migrates.
Shortcuts: a two-state chain with $P_{01} = a$ and $P_{10} = b$ has $\boldsymbol{\pi} = \big(b/(a+b),\ a/(a+b)\big)$ . A chain that moves at most one state at a time satisfies $\pi_i P_{i,i+1} = \pi_{i+1} P_{i+1,i}$ . A doubly stochastic matrix has a uniform $\boldsymbol{\pi}$ .
With an absorbing state that every other state can reach (death in a Multi-State Model, default in Credit Rating Migration) the stationary distribution just piles all probability into that state; those models are worked over finite horizons with the Chapman-Kolmogorov Equations instead.

Long-Run Risk Mix and Average Loss Policyholders move among Low, Medium and High risk tiers each year with $\mathbf{P} = \begin{array}{c|ccc} & L & M & H \\ \hline L & 0.85 & 0.15 & 0 \\ M & 0.20 & 0.70 & 0.10 \\ H & 0 & 0.30 & 0.70 \end{array}$ Expected annual losses are $\$400$ , $\$800$ and $\$2{,}000$ by tier. Find the long-run tier mix, the long-run average expected loss per policy, and the mean time between years spent in H.
Answer The chain moves one tier at a time, so balance the flows between neighbours:
 $\begin{align*} 0.15\,\pi_L &= 0.20\,\pi_M \\ \pi_M &= 0.75\,\pi_L \\ 0.10\,\pi_M &= 0.30\,\pi_H \\ \pi_H &= 0.25\,\pi_L \end{align*}$ 
Normalising, $\pi_L(1 + 0.75 + 0.25) = 1$ , so $\boldsymbol{\pi} = (0.500,\ 0.375,\ 0.125)$ . Then
 $\begin{align*} \text{Average loss} &= 0.500(400) + 0.375(800) + 0.125(2{,}000) \\ &= 200 + 300 + 250 \\ &= \$750 \end{align*}$ 
The mean return time to H is $1/0.125 = 8$ years. High-risk policies are one-eighth of the book but a third of its expected loss.

Solving the Full Balance Equations A reinsurer classes each quarter's catastrophe activity as Quiet, Normal or Active, with rows $\mathbf{P} = \begin{array}{c|ccc} & Q & N & A \\ \hline Q & 0.5 & 0.4 & 0.1 \\ N & 0.3 & 0.4 & 0.3 \\ A & 0.2 & 0.3 & 0.5 \end{array}$ Find the long-run proportion of Active quarters.
Answer Use the Q and A balance equations and drop the N one:
 $\begin{align*} 0.5\,\pi_Q &= 0.3\,\pi_N + 0.2\,\pi_A \\ 0.5\,\pi_A &= 0.1\,\pi_Q + 0.3\,\pi_N \end{align*}$ 
The first gives $\pi_Q = 0.6\,\pi_N + 0.4\,\pi_A$ . Substituting into the second:
 $\begin{align*} 0.5\,\pi_A &= 0.06\,\pi_N + 0.04\,\pi_A + 0.3\,\pi_N \\ 0.46\,\pi_A &= 0.36\,\pi_N \\ \pi_A &= \tfrac{18}{23}\,\pi_N \end{align*}$ 
so $\pi_Q = \tfrac{21}{23}\pi_N$ . Normalising, $\pi_N(21 + 23 + 18)/23 = 1$ , giving $\boldsymbol{\pi} = \big(\tfrac{21}{62},\ \tfrac{23}{62},\ \tfrac{18}{62}\big) = (0.339,\ 0.371,\ 0.290)$ .

The dropped N equation checks it: $0.4(21) + 0.4(23) + 0.3(18) = 23$ . About $29\%$ of quarters are Active in the long run.