Composing Path Probabilities Across Valuation Periods

Why Compose Probabilities Across Periods?

A benefit may depend on a member occupying one state at an early checkpoint and another state later. For example, a hypothetical disability benefit might require the member to be disabled at the end of one year and still be alive at the end of a later year. Between those dates, the member could recover and re-enter disability, or pass through another state. To calculate the probability of the required checkpoint events, we must link the transition probabilities for successive periods.

The central task is to account for every possible intermediate state. If the intermediate state is unrestricted, the probability of the overall event is a sum of path probabilities. Each path probability is a product of the transition probabilities along that path. This composition is the discrete-time form of the Chapman–Kolmogorov rule.

Definition: Let $p_{ij}^{(r)}$ be the probability that a person in state $i$ at the start of period $r$ is in state $j$ at its end. The states are mutually exclusive and exhaustive at each checkpoint. The transition probabilities may vary from period to period. The Markov assumption is that, conditional on the state at a checkpoint, the future state probabilities do not depend on the earlier path by which that state was reached.

A transition probability describes the state at the end of a stated period, conditional on the state at its start. It does not, by itself, say whether a particular state was occupied at some time inside the period. The distinction matters: endpoint transitions can be composed to find endpoint probabilities, but an event such as “entered disability at least once during the year” requires a model or transition probabilities that explicitly describe that event.

The Composition Rule

Consider three successive checkpoints, numbered 0, 1, and 2. Let $i$ be the state at checkpoint 0 and $k$ the state at checkpoint 2. At checkpoint 1, the person must be in exactly one state $j$ from the model’s state set. For each possible $j$, the probability of the path $i\to j\to k$ is the product of its two conditional transition probabilities.

P(X_2=k\mid X_0=i) = \sum_j P(X_1=j\mid X_0=i) P(X_2=k\mid X_1=j) = \sum_j p_{ij}^{(1)}p_{jk}^{(2)}.

The sum follows from the law of total probability: the events $X_1=j$ partition all possible outcomes at checkpoint 1. The product follows from the Markov assumption, which makes the second-period transition probability depend on the state at checkpoint 1 rather than on the earlier history.

P(X_2=k\mid X_0=i) = \sum_j P(X_1=j\mid X_0=i) P(X_2=k\mid X_1=j).

If the intermediate state is restricted to a set $B$, sum only over states in $B$. If it is specified to be one particular state $b$, there is just one path component for that checkpoint, although later unrestricted checkpoints may still require sums. For more periods, sum over every sequence of intermediate states allowed by the event. Equivalently, multiply the successive transition matrices in time order; the entry in row $i$, column $k$ of the product gives the composed probability from $i$ to $k$.

Theorem: For checkpoints $0,1,\ldots,n$, the probability of a particular state sequence $i_0,i_1,\ldots,i_n$ is $$ P(X_1=i_1,\ldots,X_n=i_n\mid X_0=i_0) = \prod_{r=1}^{n}p_{i_{r-1}i_r}^{(r)}. $$ If one or more intermediate states are unrestricted, add these products over all state sequences satisfying the event’s restrictions.

Worked Examples

Worked Example: One Specified Checkpoint and One Unrestricted Checkpoint

A hypothetical member starts in state $A$. In period 1, the probability of moving from $A$ to $B$ is $0.40$. Conditional on being in $B$ at checkpoint 1, the period-2 probabilities of being in $A,B,C$ are $0.20,0.50,0.30$, respectively. In period 3, the probabilities of reaching $C$ from $A,B,C$ are $0.10,0.40,0.80$, respectively. Find the probability that the member is in $B$ at checkpoint 1 and $C$ at checkpoint 3; checkpoint 2 is unrestricted.

Method. First apply the period-1 requirement, $A\to B$. Then sum the three possible paths from $B$ at checkpoint 1 through the state at checkpoint 2 to $C$ at checkpoint 3.

The path through $A$ at checkpoint 2 has probability $0.40(0.20)(0.10)=0.008$. The path through $B$ has probability $0.40(0.50)(0.40)=0.080$. The path through $C$ has probability $0.40(0.30)(0.80)=0.096$.

P(X_1=B,X_3=C\mid X_0=A) = 0.008+0.080+0.096 = 0.40\bigl[(0.20)(0.10)+(0.50)(0.40)+(0.30)(0.80)\bigr] = 0.1840.

Check. Conditional on reaching $B$ at checkpoint 1, the probability of being in $C$ at checkpoint 3 is $0.02+0.20+0.24=0.46$, and $0.40(0.46)=0.1840$. The answer is a joint probability and cannot exceed $P(X_1=B\mid X_0=A)=0.40$. A common mistake is to omit the path through $C$ at checkpoint 2, or to add the period-2 probabilities without weighting them by the period-3 probabilities.

Worked Example: Specified States at Two Checkpoints

A hypothetical member starts in $A$. The probability of $A\to B$ during period 1 is $0.30$. Given state $B$ at checkpoint 1, the period-2 probabilities of being in $A,B,C$ at checkpoint 2 are $0.20,0.50,0.30$. The period-3 probabilities of reaching $C$ from $A,B,C$ are $0.10,0.40,1.00$, respectively. Calculate the probability of being in $B$ at checkpoint 1 and $C$ at checkpoint 3, with checkpoint 2 unrestricted. Also find the probability of being in $C$ at checkpoint 3 conditional on being in $B$ at checkpoint 1.

Method. For the joint probability, include the first-period probability and sum the three possible second-checkpoint paths. For the conditional probability, use the same sum but omit the first-period factor, because the condition already specifies state $B$ at checkpoint 1.

The three conditional path contributions from checkpoint 1 to checkpoint 3 are $0.20(0.10)=0.02$, $0.50(0.40)=0.20$, and $0.30(1.00)=0.30$. Their sum is $0.52$.

P(X_1=B,X_3=C\mid X_0=A) = 0.30(0.02+0.20+0.30) = 0.30(0.52) = 0.1560.

Thus the conditional probability is $0.5200$, and the joint probability is $0.1560$. As a check, $0.1560/0.30=0.5200$. The common mistake is to report $0.52$ as the joint answer: that is the conditional probability given the checkpoint-1 state, not the probability from the original starting state.

Worked Example: Summing Paths Allowed by a Checkpoint Condition

A hypothetical long-term care model has states $A,B,C$, with transition probabilities specified by the following three-period matrices. Rows are the state at the start of a period and columns are the state at its end; each row is in $A,B,C$ order.

P^{(1)}= \begin{pmatrix} 0.50&0.30&0.20\\ 0.20&0.50&0.30\\ 0.10&0.20&0.70 \end{pmatrix}, \quad P^{(2)}= \begin{pmatrix} 0.30&0.30&0.40\\ 0.20&0.40&0.40\\ 0.10&0.30&0.60 \end{pmatrix}, \quad P^{(3)}= \begin{pmatrix} 0.25&0.50&0.25\\ 0.30&0.40&0.30\\ 0.25&0.25&0.50 \end{pmatrix}.

A member starts in $A$. Find the probability that the member is in either $B$ or $C$ at checkpoint 1, in $C$ at checkpoint 2, and in $A$ at checkpoint 3.

Method. The checkpoint-1 condition allows two paths: $A\to B\to C\to A$ and $A\to C\to C\to A$. Multiply transition probabilities along each path and add the results.

For the first path, the probability is $0.30(0.40)(0.25)=0.030$. For the second path, it is $0.20(0.60)(0.25)=0.030$. Therefore the required probability is $0.030+0.030=0.0600$.

P(X_1\in\{B,C\},X_2=C,X_3=A\mid X_0=A) = (0.30)(0.40)(0.25)+(0.20)(0.60)(0.25) = 0.0600.

Check. The two paths are disjoint because the checkpoint-1 state cannot be both $B$ and $C$. Adding their probabilities is therefore appropriate. The common mistake is to include the path through $A$ at checkpoint 1; it does not satisfy the stated event, even though it could later reach $C$ and then $A$.

A Reliable Calculation Process

For a multi-period path problem, make the checkpoint restrictions explicit before calculating. The following process prevents omitted paths and mismatched conditioning.

Write the event by checkpoint.
Record the required state or allowed set of states at every checkpoint. Mark unrestricted checkpoints clearly.

List the admissible paths.
Include only state sequences that satisfy every restriction. For an unrestricted checkpoint, include each possible state that can contribute.

Multiply along each path.
Use the transition probability for each step, conditional on that step’s starting state and period.

Add disjoint path probabilities.
Sum the products for the admissible paths. If the requested probability is conditional on a checkpoint state, do not include the probability of reaching that conditioned state.

Check the result.
Confirm that it is between zero and one, that it does not exceed the probability of any required event, and that each transition row used sums to one.

Common Errors and Why the Rule Matters

The composition rule is useful whenever a benefit depends on state occupancy or state changes at multiple dates. It lets the actuary calculate the probability of the qualifying state path before using that probability in a benefit calculation. This tutorial concerns the probability composition itself; it does not determine payment amounts or their present values.

Confusing joint and conditional probabilities. A probability conditional on being in a state at checkpoint 1 does not include the probability of reaching that state. A joint probability from the original starting state does.
Forgetting an unrestricted state. “Unrestricted at checkpoint 2” means sum over all possible states there, not ignore the checkpoint’s transitions.
Adding transition probabilities directly. A path probability is a product of successive conditional transition probabilities. Add only after calculating the separate path products.
Using the wrong period’s probabilities. If rates or transition probabilities change over time, use the matrix for the particular interval being traversed. Do not assume a time-homogeneous model unless it is stated.
Interpreting endpoint probabilities as within-period events. Knowing the state at the end of an interval does not establish whether the person entered or occupied another state during the interval.

The method assumes that the transition probabilities are defined on the same checkpoint structure and that the states form a complete partition. It also relies on the Markov property for the period-by-period products. If future transitions depend on duration in a state, or on a relevant earlier history not captured by the current state, the state description must be expanded or the conditional probabilities must explicitly incorporate that dependence.

This composition approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In an exam calculation, writing the checkpoint event and admissible paths before doing arithmetic is often the simplest way to expose a missing factor or an incorrectly included path.

Check Your Understanding

For each problem, identify which probabilities are joint and which are conditional before calculating.

A member starts in $X$, with $P(X\to U)=0.25$ in period 1. Given $U$ at checkpoint 1, the period-2 probabilities to $M,V$ are $0.40,0.60$. In period 3, the probabilities of reaching $Z$ from $M,V$ are $0.20,0.70$. Find the probability of being in $U$ at checkpoint 1 and $Z$ at checkpoint 3. (Answer: $0.1250$)
A member starts in $R$, with $P(R\to S)=0.40$ in period 1. Given $S$ at checkpoint 1, the probabilities to $T,U$ in period 2 are $0.30,0.70$. In period 3, the probabilities of reaching $K$ from $T,U$ are $0.50,0.10$. Find the joint probability of $S$ at checkpoint 1 and $K$ at checkpoint 3. (Answer: $0.0880$)
The initial state is $A$ or $B$, each with probability $0.50$. In period 1, the probabilities of reaching $C$ from $A,B$ are $0.20,0.60$. Given $C$ at checkpoint 1, the period-2 probability of reaching $D$ is $0.50$. Find the probability of $C$ at checkpoint 1 and $D$ at checkpoint 2. (Answer: $0.2000$)
A member starts in $A$. In period 1, the probabilities of reaching $B,C$ are $0.30,0.20$. In period 2, the probabilities of reaching $D$ from $B,C$ are $0.40,0.50$. In period 3, the probability of reaching $E$ from $D$ is $0.60$. Find the probability of being in $B$ or $C$ at checkpoint 1, $D$ at checkpoint 2, and $E$ at checkpoint 3. (Answer: $0.1320$)
A member starts in $A$, with $P(A\to B)=0.40$ in period 1. Given $B$ at checkpoint 1, the period-2 probabilities of reaching $C,D$ are $0.30,0.70$. In period 3, the probabilities of reaching $E$ from $C,D$ are $0.50,0.20$. Find both the probability of reaching $E$ conditional on being in $B$ at checkpoint 1 and the joint probability of $B$ at checkpoint 1 and $E$ at checkpoint 3. (Answer: Conditional probability $0.2900$; joint probability $0.1160$)

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