Multi-State Model

A Multi-State Model describes an individual moving among a finite set of states — for example healthy, disabled and dead — with transitions governed by a Markov Chain; benefits and premiums attach to being in a state or to moving between states. The two-state alive–dead Survival Model is its simplest case.

$\text{APV} = \sum_{k=1}^{n} b\, v^{k}\, P_{ij}^{\,k}$

$P(X_1 = \cdots = X_n = i \mid X_0 = i) = (P_{ii})^{n}$

The APV is for a benefit $b$ payable at the end of each period $k$ in which a life that starts in state $i$ is in state $j$ ; $v$ is the one-period discount factor (Present Value) and $P^k_{ij}$ comes from the Chapman-Kolmogorov Equations.
A disability model typically has states H (healthy), D (disabled) and X (dead, absorbing), with transitions H $\to$ D (disablement), D $\to$ H (recovery) and H, D $\to$ X.
Being in a state versus staying in it: $P^n_{ii}$ counts every path that is back in $i$ at time $n$ , including those that left and returned; $(P_{ii})^n$ counts only paths that never left. Waiting periods and "continuously disabled" conditions need the second.
Transition probabilities change with age, so realistic models are non-homogeneous — a different Transition Probability Matrix each year, multiplied in order.
In continuous time the matrix is replaced by transition intensities, and the force of mortality (Hazard Rate) is the intensity of the alive $\to$ dead transition. A pair of lives $(x)$ and $(y)$ is another multi-state model, with four states: both alive, only $(x)$ alive, only $(y)$ alive, and both dead. The joint-life status fails at the first death and the last-survivor status at the last.

Two-Year Disability Probabilities Annual transitions among Healthy, Disabled and Dead are $\mathbf{P} = \begin{array}{c|ccc} & H & D & X \\ \hline H & 0.85 & 0.10 & 0.05 \\ D & 0.20 & 0.65 & 0.15 \\ X & 0 & 0 & 1 \end{array}$ For a healthy life, find the state probabilities two years from now and the probability of staying healthy throughout.
Answer Row H of $\mathbf{P}^2$ :
 $\begin{align*} P^2_{HH} &= 0.85(0.85) + 0.10(0.20) \\ &= 0.7425 \\ P^2_{HD} &= 0.85(0.10) + 0.10(0.65) \\ &= 0.1500 \\ P^2_{HX} &= 0.85(0.05) + 0.10(0.15) + 0.05(1) \\ &= 0.1075 \end{align*}$ 
Staying healthy throughout has probability $0.85^2 = 0.7225$ . The gap $0.7425 - 0.7225 = 0.02$ is the path H $\to$ D $\to$ H, $(0.10)(0.20)$ : disabled in year 1, recovered by year 2.

Valuing a Disability Income Benefit Using the same matrix, a policy on a healthy life pays $\$10{,}000$ at the end of each of the next two years if the insured is then disabled. At $5\%$ annual interest, find the actuarial present value.
Answer The disabled probabilities are $P^1_{HD} = 0.10$ and $P^2_{HD} = 0.15$ :
 $\begin{align*} \text{APV} &= 10{,}000\left(\frac{0.10}{1.05} + \frac{0.15}{1.05^2}\right) \\ &= 10{,}000\,(0.095238 + 0.136054) \\ &= \$2{,}312.93 \end{align*}$ 
The year-2 payment is worth more despite the extra discounting, because disablement accumulates faster than recovery removes it.