Piecewise-Constant Intensities Across Time Intervals

Introduction

A continuous-time state model may have one set of transition intensities during an initial period and a different set later. For example, a hypothetical care model might allow recovery during the first year, then change its transition pattern after a policy milestone. The transition probability over the full horizon depends on both periods and on the order in which their intensity matrices apply.

The calculation has two parts: find the transition matrix for each constant-intensity period, then multiply those matrices in chronological order. This tutorial develops that procedure, explains why the order matters, and shows how to check the resulting probabilities.

Definition: A piecewise-constant intensity model has a generator matrix $Q(t)$ that is constant within each of a sequence of time intervals, but may change at interval boundaries. For a period of length $h$ with constant generator $Q$, its transition matrix is $e^{Qh}$. Under the row-transition convention, entry $p_{ij}$ is the probability of being in state $j$ at the end of the period given that the process started in state $i$.

From Period Intensities to Period Transition Matrices

For a finite-state continuous-time Markov model, the off-diagonal entry $q_{ij}$ of a generator $Q$ is the instantaneous transition intensity from state $i$ to state $j$, for $i\ne j$. The diagonal entry is $q_{ii}=-\sum_{j\ne i}q_{ij}$; each row of $Q$ therefore sums to zero. If $Q$ is constant for a period of length $h$, the transition matrix for that period is $e^{Qh}$.

Suppose the first period has length $h_1$ and generator $Q_1$, and the second has length $h_2$ and generator $Q_2$. Write $P_1=e^{Q_1h_1}$ and $P_2=e^{Q_2h_2}$. If $P(s,t)$ denotes the transition matrix from time $s$ to time $t$, the Chapman-Kolmogorov relationship gives:

$$ P(s,s+h_1+h_2)=P_1P_2 $$

This order follows the row-transition convention: the distribution at the start of the first period is multiplied by $P_1$, then the resulting distribution is multiplied by $P_2$. For more periods, multiply the corresponding matrices from left to right in chronological order.

Key result: For successive constant-generator intervals with durations $h_1,\ldots,h_m$, the full transition matrix is $$ P(s,t)=e^{Q_1h_1}e^{Q_2h_2}\cdots e^{Q_mh_m}, $$ where the intervals run chronologically from left to right. This product is not generally equal to $e^{Q_1h_1+\cdots+Q_mh_m}$.

Why the Matrices Are Composed in This Order

The product can be justified by conditioning on the state at the boundary between periods. To go from initial state $i$ to final state $k$, the process must be in some intermediate state $j$ at the boundary. The probability of the route through $j$ is the first-period probability of reaching $j$ multiplied by the second-period probability of going from $j$ to $k$. Summing over all possible $j$ is exactly matrix multiplication.

P_{ik}(s,t)=\sum_j (P_1)_{ij}(P_2)_{jk}=(P_1P_2)_{ik}

The argument extends one boundary at a time. In a three-period model, first condition on the state after period one and then on the state after period two; the resulting sum is the $(i,k)$ entry of $P_1P_2P_3$. This derivation also shows why changing the order changes the meaning: it changes which transition rules apply before and after the boundary.

Worked Example: Two Sequential One-Way Transitions

Consider three states, ordered $0,1,2$. In a hypothetical model, during year 1 the only possible transition is from state 0 to state 1, with intensity $\ln 2$ per year. In year 2 the only possible transition is from state 1 to state 2, also with intensity $\ln 2$ per year. State 2 is absorbing in both periods, and no other transitions occur. Find the probability of ending year 2 in each state, given an initial state of 0.

Method. Construct the generator and transition matrix for each one-year period, then multiply in chronological order. For a one-way transition with constant intensity $\lambda$ over duration $h$, the probability of no transition is $e^{-\lambda h}$, and the probability of transition is $1-e^{-\lambda h}$. Here $e^{-(\ln 2)(1)}=1/2$, so each relevant period has a one-half transition probability.

In year 1, the generator and its transition matrix are:

Q_1= \begin{pmatrix} -\ln 2&\ln 2&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}, \qquad P_1=e^{Q_1}= \begin{pmatrix} 1/2&1/2&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}.

In year 2, only the transition from state 1 to state 2 is possible. Thus:

Q_2= \begin{pmatrix} 0&0&0\\ 0&-\ln 2&\ln 2\\ 0&0&0 \end{pmatrix}, \qquad P_2=e^{Q_2}= \begin{pmatrix} 1&0&0\\ 0&1/2&1/2\\ 0&0&1 \end{pmatrix}.

Multiply in chronological order:

P_1P_2= \begin{pmatrix} 1/2&1/2&0\\ 0&1&0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&1/2&1/2\\ 0&0&1 \end{pmatrix} = \begin{pmatrix} 1/2&1/4&1/4\\ 0&1/2&1/2\\ 0&0&1 \end{pmatrix}.

The initial state-0 row is $(1/2,1/4,1/4)$. Therefore the probabilities of ending in states 0, 1, and 2 are respectively $1/2$, $1/4$, and $1/4$. As a check, they are nonnegative and sum to $1$. The final state-2 probability also has a direct interpretation: the process must move from 0 to 1 during year 1 and then from 1 to 2 during year 2, giving $(1/2)(1/2)=1/4$.

Common mistake avoided: The period matrices are not interchangeable. Reversing them would apply the state-1-to-state-2 rule first, when the process starts in state 0, and would describe a different sequence of rules.

Different Periods Can Have Different Transition Patterns

A change in the generator need not mean only that an intensity becomes larger or smaller. Transitions may become possible or impossible, or rates in opposite directions may change differently. Each period matrix must reflect its own generator and duration. Once the matrices have been found, the composition rule is unchanged.

Worked Example: A Two-State Model With a Later Change in Rates

A hypothetical two-state model has states 0 and 1. During the first year, the transition intensity is $q_{01}=q_{10}=(\ln 2)/2$ per year. During the second year, the intensity from 0 to 1 is $\ln 2$ per year and the intensity from 1 to 0 is zero. Find the probability of ending in state 1 after two years, given that the process starts in state 0.

Method. Calculate each one-year transition matrix from its generator and multiply the first-year matrix by the second-year matrix. In the first year, the total rate out of either state is $\ln 2$. For this symmetric two-state model, the probability of being in the starting state after time $h$ is $(1+e^{-2\lambda h})/2$, and the probability of being in the other state is $(1-e^{-2\lambda h})/2$. Here $\lambda=(\ln 2)/2$ and $h=1$, so $e^{-2\lambda h}=1/2$.

Q_1= \begin{pmatrix} -(\ln 2)/2&(\ln 2)/2\\ (\ln 2)/2&-(\ln 2)/2 \end{pmatrix}, \qquad P_1= \begin{pmatrix} 3/4&1/4\\ 1/4&3/4 \end{pmatrix}.

In the second year, state 0 changes to state 1 with intensity $\ln 2$, and state 1 is absorbing. The probability of changing during this one-year period is $1-e^{-\ln 2}=1/2$, giving:

Q_2= \begin{pmatrix} -\ln 2&\ln 2\\ 0&0 \end{pmatrix}, \qquad P_2= \begin{pmatrix} 1/2&1/2\\ 0&1 \end{pmatrix}.

The full transition matrix is:

P_1P_2= \begin{pmatrix} 3/4&1/4\\ 1/4&3/4 \end{pmatrix} \begin{pmatrix} 1/2&1/2\\ 0&1 \end{pmatrix} = \begin{pmatrix} 3/8&5/8\\ 1/8&7/8 \end{pmatrix}.

Starting from state 0, the probability of ending in state 1 is the entry in row 0, column 1: $5/8=0.625$. It can also be checked by conditioning on the state after year 1. The process is in state 0 with probability $3/4$, then reaches state 1 with probability $1/2$; or it is already in state 1 with probability $1/4$, then stays there with probability 1. Thus $(3/4)(1/2)+(1/4)(1)=3/8+1/4=5/8$. Each row of the product sums to 1.

Common mistake avoided: The second-year probability of ending in state 1 is not $1/2$ regardless of the state at the year boundary. A process already in state 1 stays there with probability 1, so the intermediate state must be accounted for.

Three Periods and a Return Path

With three or more intervals, every boundary state matters. A process may move through several states and later return to a state it occupied earlier. Matrix multiplication includes all such possible paths without requiring them to be listed separately.

Worked Example: Composing Three Periods

Use states $0,1,2$, with one-year intervals. In period 1, the only transition is $0\to1$ at intensity $\ln 2$. In period 2, the only transition is $1\to2$ at intensity $\ln 2$. In period 3, the only transition is $2\to0$ at intensity $\ln 2$. All unspecified transitions have intensity zero, and transitions out of state 2 in periods 1 and 2, or out of states 0 and 1 in period 3, are unavailable. Find the ending-state distribution from initial state 0.

Method. Each period has a one-half chance of its specified transition and a one-half chance of no transition from the state where that transition is available. The first two period matrices are those calculated in the first worked example. The third-period matrix is:

P_3= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 1/2&0&1/2 \end{pmatrix}.

First combine periods 1 and 2:

P_1P_2= \begin{pmatrix} 1/2&1/4&1/4\\ 0&1/2&1/2\\ 0&0&1 \end{pmatrix}.

Then apply period 3:

P_1P_2P_3= \begin{pmatrix} 1/2&1/4&1/4\\ 0&1/2&1/2\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&1&0\\ 1/2&0&1/2 \end{pmatrix} = \begin{pmatrix} 5/8&1/4&1/8\\ 1/4&1/2&1/4\\ 1/2&0&1/2 \end{pmatrix}.

The initial state-0 row gives ending probabilities $5/8$ in state 0, $1/4$ in state 1, and $1/8$ in state 2. These add to $5/8+1/4+1/8=1$. In particular, ending in state 0 includes both the path that remained in state 0 throughout and paths that reached state 2 in period 2 and returned in period 3. Matrix multiplication aggregates those routes.

Common mistake avoided: Counting only paths that move forward through the state labels would omit the return $2\to0$ in the final period. The product includes it because the third-period transition probability from state 2 to state 0 is in $P_3$.

Common Exam Traps and Checks

A useful check is to verify that each period matrix and the final product are valid transition matrices: all entries should be nonnegative and every row should sum to 1. A row-sum failure often indicates an incorrectly constructed generator, a matrix multiplication error, or an invalid proposed period matrix. This check does not by itself prove that the model is appropriate, but it catches many arithmetic and setup errors.

Reversing the order. Under the row-transition convention, use $P_1P_2$, not $P_2P_1$, when period 1 precedes period 2.
Adding generators without justification. The general piecewise-constant result is a product of matrix exponentials. Replacing it by one exponential of a sum is not generally valid; it requires special conditions, such as the relevant generator matrices commuting.
Using the wrong duration. The exponent is $Qh$, where $h$ is the length of that specific interval in the same time units used for the intensities.
Ignoring intermediate states. An overall transition probability sums over all possible states at each boundary, not just one guessed route.
Accepting an invalid period matrix. Every row must sum to 1. For instance, a proposed row $(0,1/2)$ cannot represent a complete distribution over two states because it omits probability $1/2$.

The key is to keep the time structure visible: label each interval, attach its own generator and duration, calculate its transition matrix, and then compose the matrices from earliest to latest. This procedure is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the row-transition convention and give the requested probability or distribution.

A two-state model has state 0 moving to absorbing state 1 at intensity $\ln 2$ per year during a one-year interval. What is the probability of remaining in state 0 at the end? (Answer: $1/2$)
There are three states $0,1,2$. During year 1, only $0\to1$ is possible, at intensity $\ln 2$; during year 2, only $1\to2$ is possible, also at intensity $\ln 2$. Starting from 0, what is the probability of ending in state 2 after two years? (Answer: $1/4$)
For the two-period model in which $P_1=\begin{pmatrix}3/4&1/4\\1/4&3/4\end{pmatrix}$ and $P_2=\begin{pmatrix}1/2&1/2\\0&1\end{pmatrix}$, what is the probability of ending in state 0 from initial state 1? (Answer: $1/8$)
In the three-period model of the worked example, starting from state 0, what is the probability of ending in state 0 after the final $2\to0$ transition opportunity? (Answer: $5/8$)
Let $P_1=\begin{pmatrix}1/2&1/2\\0&1\end{pmatrix}$ and $P_2=\begin{pmatrix}1&0\\0&1/2\end{pmatrix}$. Is this a valid pair of transition matrices for two successive periods? If not, identify the defect. (Answer: No. The second row of $P_2$ sums to $1/2$, not 1, so $P_2$ is not a transition matrix.)

What You'll Learn