Splitting a Transition Calculation at an Intermediate Time

Why Split a Transition Calculation?

In a multiple-state model, an individual can move among several states between the time of observation and the end of a period. If transition probabilities are known for two successive intervals, a probability over the combined interval can be calculated by considering the state occupied at the split time. The key is to include every possible intermediate state, not just one apparent route.

For example, a hypothetical policyholder might be active at time 0 and disabled at time 2. The policyholder may have been active, disabled, or deceased at time 1. Each possible state at time 1 contributes to the probability of being disabled at time 2, subject to the model's transition probabilities.

Definition: Let \(X_t\) denote the state occupied at time \(t\). For states \(i\) and \(j\), the interval transition probability is \(p_{ij}(s,t)=P(X_t=j\mid X_s=i)\), for \(s\leq t\). The transition matrix \(\mathbf P(s,t)\) has entry \(p_{ij}(s,t)\) in the row for starting state \(i\) and the column for ending state \(j\). The state ordering is kept the same for every interval matrix in a calculation.

The split calculation uses the Markov property: conditional on the state at the split time, the distribution of the later state does not depend on the earlier path. For a split at \(u\), where \(s\leq u\leq t\), the states at time \(u\) form mutually exclusive and exhaustive possibilities. Conditioning on them gives the Chapman–Kolmogorov relationship:

p_{ij}(s,t)=\sum_k p_{ik}(s,u)p_{kj}(u,t).

The first factor in each term is the probability of reaching intermediate state \(k\) from \(i\) during the first interval. The second is the probability of reaching \(j\) from \(k\) during the second interval. Their product is that route's contribution; summing the contributions gives the total probability of the endpoint event.

Why the Sum of Products Holds

For each possible state \(k\), consider the event that \(X_u=k\). These events partition the outcomes conditional on \(X_s=i\). The law of total probability therefore gives \(P(X_t=j\mid X_s=i)=\sum_k P(X_t=j,X_u=k\mid X_s=i)\).

For a particular \(k\), the multiplication rule and Markov property yield \(P(X_t=j,X_u=k\mid X_s=i)=P(X_u=k\mid X_s=i)P(X_t=j\mid X_u=k)\). The first conditional probability is \(p_{ik}(s,u)\), and the second is \(p_{kj}(u,t)\). Substitution gives the stated sum. This argument requires that the listed states cover all possible states at the split time and that the Markov property applies. It does not require the two intervals to have equal lengths.

Fix the start and endpoint.
Identify the starting state \(i\), ending state \(j\), and split time \(u\).

List the intermediate states.
Include every state in the model that could be occupied at time \(u\).

Multiply along each route.
For each intermediate state \(k\), multiply \(p_{ik}(s,u)\) by \(p_{kj}(u,t)\).

Add the contributions.
The sum is the required longer-interval probability. Check that the answer lies between 0 and 1.

Worked Calculations

Worked Example: A Three-State Model With Several Routes

A hypothetical model has states \(A,B,C\), in that order. The transition matrices for the intervals from time 0 to 1 and from time 1 to 3 are, respectively,

\mathbf P(0,1)= \begin{pmatrix} 0.7&0.2&0.1\\ 0.1&0.6&0.3\\ 0.2&0.2&0.6 \end{pmatrix}, \qquad \mathbf P(1,3)= \begin{pmatrix} 0.5&0.3&0.2\\ 0.2&0.5&0.3\\ 0.1&0.4&0.5 \end{pmatrix}.

Find the probability of starting in \(A\) at time 0 and being in \(C\) at time 3. The method is to split at time 1 and sum over the three possible states then.

Intermediate state \(A\): the contribution is \(0.7(0.2)=0.14\).

Intermediate state \(B\): the contribution is \(0.2(0.3)=0.06\).

Intermediate state \(C\): the contribution is \(0.1(0.5)=0.05\).

Therefore, \(p_{AC}(0,3)=0.14+0.06+0.05=0.25\), or \(0.2500\) to four decimal places. As a check, the other endpoint probabilities from starting state \(A\) are \(p_{AA}(0,3)=0.7(0.5)+0.2(0.2)+0.1(0.1)=0.40\) and \(p_{AB}(0,3)=0.7(0.3)+0.2(0.5)+0.1(0.4)=0.35\). The three endpoint probabilities sum to \(0.40+0.35+0.25=1.00\).

Common mistake avoided: using only the route \(A\to B\to C\) would give \(0.06\), not the total probability. The routes \(A\to A\to C\) and \(A\to C\to C\) also end in \(C\), so their contributions must be included.

Worked Example: A Transition Probability and a Conditional Route

For a different hypothetical three-state process, states are \(1,2,3\). The interval matrices are

\mathbf P(0,2)= \begin{pmatrix} 0.6&0.3&0.1\\ 0.2&0.5&0.3\\ 0.1&0.2&0.7 \end{pmatrix}, \qquad \mathbf P(2,5)= \begin{pmatrix} 0.4&0.4&0.2\\ 0.1&0.6&0.3\\ 0.3&0.2&0.5 \end{pmatrix}.

Suppose the process starts in state 2 at time 0. First find the probability it is in state 1 at time 5, then find the conditional probability that it was in state 3 at time 2, given that it starts in state 2 and ends in state 1.

For the endpoint probability, sum over the intermediate states. The contributions to ending in state 1 are:

Through state 1: \(0.2(0.4)=0.08\). Through state 2: \(0.5(0.1)=0.05\). Through state 3: \(0.3(0.3)=0.09\).

Thus \(p_{21}(0,5)=0.08+0.05+0.09=0.22\). The joint probability of the specified start, intermediate state 3, and endpoint is the third contribution, \(0.09\). Hence the requested conditional probability is \(0.09/0.22=0.409090\ldots\), or \(0.4091\) to four decimal places.

Check: the conditional probability is between 0 and 1. The numerator is one part of the total endpoint probability, and \(0.09\leq0.22\). The common mistake is to report \(0.09\) as the conditional probability: it is the joint route contribution, not the contribution divided by the probability of the endpoint.

Worked Example: Splitting a Hypothetical Disability Model

A hypothetical disability model has three states: active \(A\), disabled \(D\), and dead \(M\), in that order. Death is absorbing. Given transition matrices over two successive intervals,

\mathbf P(0,1)= \begin{pmatrix} 0.8&0.15&0.05\\ 0.2&0.7&0.1\\ 0&0&1 \end{pmatrix}, \qquad \mathbf P(1,4)= \begin{pmatrix} 0.7&0.2&0.1\\ 0.1&0.8&0.1\\ 0&0&1 \end{pmatrix},

calculate the probability that a person active at time 0 is disabled at time 4. The split-time state can be active, disabled, or dead.

If active at time 1, the contribution is \(0.8(0.2)=0.16\). If disabled at time 1, it is \(0.15(0.8)=0.12\). If dead at time 1, the contribution is \(0.05(0)=0\), because the absorbing state cannot transition to disabled.

The total is \(p_{AD}(0,4)=0.16+0.12+0=0.28\), or \(0.2800\) to four decimal places. The intermediate-state probabilities from \(A\) sum to \(0.8+0.15+0.05=1\); the result also cannot exceed the probability \(0.95\) of being alive at the split. A common error is to treat the second-interval probability \(0.2\) as the answer. It applies only to those active at time 1, so it must be weighted by the probability \(0.8\) of reaching that state, and the disabled-at-time-1 route must also be counted.

Checks and Exam Traps

A longer-interval probability is not generally equal to either interval probability, nor is it obtained by adding the two probabilities. It accounts for the state at the split, which determines which second-interval probabilities apply. The following checks are useful:

Use the matching intermediate state. Each first-interval entry ending in \(k\) is paired with the second-interval entry starting in that same \(k\).
Include every possible intermediate state. A state with a zero probability of reaching the endpoint may contribute zero, but it should not be confused with an unexamined state.
Keep route and endpoint probabilities distinct. A single product is one route's contribution; the sum over routes is the endpoint probability.
Check the scale. Each contribution is nonnegative, and their sum cannot exceed 1. If all endpoint states are calculated from one starting state, those probabilities should sum to 1.
Use the interval-specific entries supplied. The first factor describes the first interval and the second factor the later interval; do not substitute an entry from the wrong interval.

This calculation is useful whenever a transition probability over a longer period is needed but the interval model is given in pieces. In actuarial applications, intermediate states can represent changes in health or policy status. The same probability logic applies regardless of the state labels or the lengths of the two intervals.

Key takeaway: To calculate \(p_{ij}(s,t)\) by splitting at \(u\), multiply the probability of each possible intermediate state \(k\) by the probability of reaching \(j\) from \(k\), then add across all \(k\). This is the conditioning form of the Chapman–Kolmogorov relationship.

Check Your Understanding

For each problem, use the state at the split time to organize the calculation.

For states \(A,B\), the interval matrices are \(\begin{pmatrix}0.8&0.2\\0.3&0.7\end{pmatrix}\) and \(\begin{pmatrix}0.6&0.4\\0.1&0.9\end{pmatrix}\). Find the probability of starting in \(A\) and ending in \(B\). (Answer: \(0.5000\))
For states \(A,B,C\), the first interval matrix is \(\begin{pmatrix}0.5&0.3&0.2\\0.2&0.5&0.3\\0.1&0.3&0.6\end{pmatrix}\), and the second is \(\begin{pmatrix}0.7&0.2&0.1\\0.1&0.6&0.3\\0.2&0.2&0.6\end{pmatrix}\). Find the probability of starting in \(A\) and ending in \(C\). (Answer: \(0.2600\))
For states \(A,B,C\), the first interval matrix is \(\begin{pmatrix}0.6&0.3&0.1\\0.2&0.5&0.3\\0.1&0.2&0.7\end{pmatrix}\), and the second is \(\begin{pmatrix}0.5&0.3&0.2\\0.1&0.6&0.3\\0.2&0.3&0.5\end{pmatrix}\). Given a start in \(B\) and an endpoint in \(A\), what is the conditional probability that the intermediate state was \(C\)? (Answer: \(0.2857\))
For states \(A,B,C\), the first interval matrix is \(\begin{pmatrix}0.4&0.4&0.2\\0.2&0.5&0.3\\0.1&0.2&0.7\end{pmatrix}\), and the second interval's probabilities of ending in \(A\), from starting states \(A,B,C\), are \(0.5,0.2,0.1\), respectively. Find the probability of starting in \(A\) and ending in \(A\) over both intervals. (Answer: \(0.3000\))

What You'll Learn