Closed-Form Kolmogorov Solution for Two States

Why Two-State Models Have a Closed-Form Solution

Many long-term insurance models begin with a simple distinction: a person is either in one state or in another. For example, a hypothetical disability-income model might distinguish an able state from a disabled state, with transitions possible in both directions. If the transition intensities are constant, the forward equations for this model can be solved explicitly. The result gives the probability of occupying either state at any elapsed time, for either starting state or any initial mixture.

The key feature is that the probability of entering a state depends on the probability of currently occupying the other state, while the probability of leaving depends on the probability of currently occupying that state. The solution therefore depends on both transition intensities. This tutorial derives the formulas and shows how to use and check them. The treatment is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: Consider a continuous-time Markov chain with states 0 and 1. Let $\alpha$ be the constant intensity of transition from state 0 to state 1, and let $\beta$ be the constant intensity of transition from state 1 to state 0. The generator, using the row convention, is $$ Q=\begin{pmatrix}-\alpha&\alpha\\ \beta&-\beta\end{pmatrix}, \qquad \alpha,\beta\geq 0. $$ The intensity $\alpha$ is the instantaneous transition rate out of state 0, conditional on being in state 0; $\beta$ has the corresponding meaning for state 1.

The model assumes time-homogeneous transition intensities: the rates do not vary with elapsed time, and the process has the Markov property. Thus, conditional on the current state, future transitions do not depend on the earlier path. Time is measured in the same units used for the intensities. The model describes transitions between the two states only; any additional states or causes of exit would require a different model.

Write $P_{ij}(t)$ for the probability of being in state $j$ at time $t$, given that the process was in state $i$ at time 0. The transition probability matrix is $P(t)=(P_{ij}(t))$, and, under the row convention, the forward equation is $P'(t)=P(t)Q$, with $P(0)$ equal to the identity matrix. Equivalently, if the initial state probabilities are the row vector $\boldsymbol{\pi}(0)=(\pi_0(0),\pi_1(0))$, then $\boldsymbol{\pi}'(t)=\boldsymbol{\pi}(t)Q$.

Deriving the Occupancy Formula

Let $p_1(t)$ be the probability of occupying state 1 at time $t$, and let $p_0(t)=1-p_1(t)$. The forward equation for state 1 balances probability entering from state 0 against probability leaving for state 0. This gives $p_1'(t)=\alpha p_0(t)-\beta p_1(t)$. Substituting $p_0(t)=1-p_1(t)$ reduces the two-state system to one scalar differential equation.

p_1'(t)=\alpha-(\alpha+\beta)p_1(t).

Set $s=\alpha+\beta$. If $s>0$, multiply the differential equation by the integrating factor $e^{st}$. The left side then becomes the derivative of $e^{st}p_1(t)$. Integrating from 0 to $t$ gives the solution below. The initial state enters through $p_1(0)$; no assumption that the process starts in state 0 is needed.

\begin{aligned} e^{st}p_1(t)-p_1(0) &=\int_0^t \alpha e^{su}\,du\\ &=\frac{\alpha}{s}(e^{st}-1),\\ p_1(t) &=\frac{\alpha}{s}+\left(p_1(0)-\frac{\alpha}{s}\right)e^{-st}. \end{aligned}

Since $p_0(t)=1-p_1(t)$, the probability of state 0 follows directly. In particular, if the process starts in state 0, then $p_1(0)=0$; if it starts in state 1, then $p_1(0)=1$. Applying those initial conditions gives all four transition probabilities.

Theorem: For a two-state homogeneous continuous-time Markov chain with constant intensities $\alpha,\beta$ and $s=\alpha+\beta>0$, $$ P(t)= \begin{pmatrix} \frac{\beta}{s}+\frac{\alpha}{s}e^{-st} & \frac{\alpha}{s}(1-e^{-st})\\[4pt] \frac{\beta}{s}(1-e^{-st}) & \frac{\alpha}{s}+\frac{\beta}{s}e^{-st} \end{pmatrix}. $$ The first row is conditional on starting in state 0; the second row is conditional on starting in state 1. For an arbitrary initial distribution with probability $q$ in state 1, the state-1 occupancy probability is $$ p_1(t)=\frac{\alpha}{s}+\left(q-\frac{\alpha}{s}\right)e^{-st}, \qquad p_0(t)=1-p_1(t). $$

The coefficients $\alpha/s$ and $\beta/s$ are the long-run probabilities of states 1 and 0, respectively, when both intensities are positive. The factor $e^{-st}$ measures how quickly the initial occupancy probability gives way to that long-run balance. If $\alpha=0$ or $\beta=0$, the formula remains valid when $s>0$, with the state having no outgoing transition treated as absorbing. If both intensities are zero, there are no transitions and $P(t)$ is the identity matrix for all $t$.

Using the Closed-Form Solution

A reliable calculation starts by identifying the direction of each intensity, the time unit, and the initial state or distribution. Then calculate $s=\alpha+\beta$, evaluate the exponential $e^{-st}$, and substitute into the appropriate row or occupancy formula. The rows of the transition matrix are conditional distributions, so each row must sum to 1.

Worked Example: Starting in the Able State

Method: Use the first row of the transition matrix. Consider a hypothetical disability model with state 0 meaning able and state 1 meaning disabled. The intensity of moving from able to disabled is $\alpha=0.20$ per year, and the recovery intensity is $\beta=0.30$ per year. A person starts able. Find both state probabilities after 2 years. There is no additional mortality or other state in this model.

First calculate the total rate appearing in the exponential:

s=\alpha+\beta=0.20+0.30=0.50\text{ per year}, \qquad st=0.50(2)=1.

Thus $e^{-st}=e^{-1}=0.3678794$. Starting in state 0, the probability of being disabled is

P_{01}(2)=\frac{\alpha}{s}(1-e^{-st}) =\frac{0.20}{0.50}(1-0.3678794) =0.40(0.6321206) =0.2528482.

The probability of being able is

P_{00}(2)=\frac{\beta}{s}+\frac{\alpha}{s}e^{-st} =\frac{0.30}{0.50}+\frac{0.20}{0.50}(0.3678794) =0.60+0.1471518 =0.7471518.

Rounded to four decimal places, the probabilities are $P_{00}(2)=0.7472$ and $P_{01}(2)=0.2528$. Check: $0.7471518+0.2528482=1.0000000$, and the probability of disability is greater than zero but below its long-run value $\alpha/s=0.40$, as expected when starting able. The common mistake to avoid is using $\alpha$, rather than $\alpha+\beta$, in the exponential.

Worked Example: Starting in the Second State

Method: Use the second row of the transition matrix, not the first row. In a hypothetical two-state care model, state 0 means not receiving care and state 1 means receiving care. The transition intensity into care is $\alpha=0.40$ per year, and the transition intensity out of care is $\beta=0.10$ per year. A person starts in state 1. Find the probabilities of each state after 3 years. No other state transitions are included.

Here $s=0.40+0.10=0.50$ per year, so $st=0.50(3)=1.50$, and $e^{-1.50}=0.2231302$. Starting in state 1, the probability of moving to state 0 by year 3 is

P_{10}(3)=\frac{\beta}{s}(1-e^{-st}) =\frac{0.10}{0.50}(1-0.2231302) =0.20(0.7768698) =0.1553740.

The probability of remaining in state 1 is

P_{11}(3)=\frac{\alpha}{s}+\frac{\beta}{s}e^{-st} =\frac{0.40}{0.50}+\frac{0.10}{0.50}(0.2231302) =0.80+0.0446260 =0.8446260.

To four decimal places, $P_{10}(3)=0.1554$ and $P_{11}(3)=0.8446$. Check: the probabilities sum to 1.0000. The probability of state 1 remains high because the initial state is state 1 and the return intensity $\beta$ is smaller than the entry intensity $\alpha$. A frequent error is to calculate $P_{01}$ from the first row despite the stated initial state; the starting state determines which row applies.

Worked Example: Starting With a Mixed Initial Distribution

Method: Apply the occupancy formula to the initial state-1 probability. Suppose a hypothetical two-state benefit model starts with 70% of the population in state 1 and 30% in state 0. The constant transition intensities are $\alpha=0.25$ per year from state 0 to state 1 and $\beta=0.15$ per year from state 1 to state 0. Find the state occupancy probabilities after 4 years. There are no other states or exits.

The total rate is $s=0.25+0.15=0.40$ per year, and $st=0.40(4)=1.60$. Therefore $e^{-st}=e^{-1.60}=0.2018965$. The long-run state-1 probability is $\alpha/s=0.25/0.40=0.625$. With $q=p_1(0)=0.70$,

\begin{aligned} p_1(4) &=\frac{\alpha}{s}+\left(q-\frac{\alpha}{s}\right)e^{-st}\\ &=0.625+(0.70-0.625)(0.2018965)\\ &=0.625+(0.075)(0.2018965)\\ &=0.625+0.0151422\\ &=0.6401422. \end{aligned}

Then $p_0(4)=1-0.6401422=0.3598578$. Rounded to four decimal places, the state-1 and state-0 probabilities are 0.6401 and 0.3599. Check: they sum to 1.0000, and the state-1 probability has moved from 0.70 toward the long-run value 0.625 without passing it. The common mistake is to treat the initial 70% as if everyone started in state 1; the initial distribution must be used as $q$, not replaced by 1.

Checks and Common Exam Traps

The formula can be checked without repeating the derivation. At $t=0$, $e^{-st}=1$, so the transition matrix becomes the identity matrix: $P_{00}(0)=P_{11}(0)=1$ and $P_{01}(0)=P_{10}(0)=0$. For any $t\geq 0$, the two entries in each row sum to 1. When both intensities are positive, as $t$ grows, $e^{-st}$ tends to zero and both rows approach the same long-run distribution $(\beta/s,\alpha/s)$.

The following incorrect calculation illustrates why the total rate matters. In Worked Example 1, using only $\alpha=0.20$ in the exponent would give

P_{01}(2)\text{ (incorrect)} =\frac{0.20}{0.50}(1-e^{-0.20(2)}) =0.40(1-0.6703200) =0.1318720.

This is not the solution of the forward equation: the rate at which the state-1 occupancy moves toward equilibrium is $\alpha+\beta=0.50$, because both entry into and exit from state 1 affect its occupancy. The correct exponent is $-(\alpha+\beta)t=-1$, giving $P_{01}(2)=0.2528482$. Always add the two transition intensities before forming the exponential.

A second practical trap is a time-unit mismatch. If an intensity is given per month, $t$ must be in months; if it is given per year, $t$ must be in years. The product $(\alpha+\beta)t$ must be dimensionless. Also distinguish the conditional transition probability $P_{ij}(t)$ from an unconditional occupancy probability: for a mixed initial distribution, combine the starting-state probabilities or use $p_1(t)$ with the given initial value $q$.

Key takeaway: With two states and constant intensities, state-1 occupancy follows $p_1(t)=\alpha/(\alpha+\beta)+(p_1(0)-\alpha/(\alpha+\beta))e^{-(\alpha+\beta)t}$. State-0 occupancy is its complement. The starting condition determines the initial value; the sum of the two intensities determines the rate of convergence.

Check Your Understanding

For each problem, use the same constant-intensity, two-state assumptions stated in the question. Answers are rounded to four decimal places unless a time is requested.

A process starts in state 0. Its transition intensities are $\alpha=0.30$ and $\beta=0.20$ per year. Find the probabilities of states 0 and 1 after 2 years. (Answer: $p_0(2)=0.6207,\ p_1(2)=0.3793$)
A process starts in state 1, with $\alpha=0.10$ and $\beta=0.40$ per year. Find the probability of state 0 after 2 years. (Answer: $P_{10}(2)=0.5057$)
An initial population has probability 0.70 in state 1. If $\alpha=0.20$ and $\beta=0.30$ per year, find both occupancy probabilities after 2 years. (Answer: $p_1(2)=0.5104,\ p_0(2)=0.4896$)
A process starts in state 0, with $\alpha=0.12$ and $\beta=0.18$ per year. Find $P_{01}(5)$ and $P_{00}(5)$. (Answer: $P_{01}(5)=0.3107,\ P_{00}(5)=0.6893$)
A process starts in state 0, with $\alpha=0.30$ and $\beta=0.20$ per year. How long does it take for the state-1 occupancy probability to reach 0.50? (Answer: $t=2\ln(6)=3.5835$ years)

What You'll Learn