Why Occupancy Probabilities Matter
A long-term care model can describe more than whether a person has entered care. At a specified future time, the person may be healthy, receiving care, recovered, or dead. Occupancy probabilities give the probability of being in each of these states at that time. They provide a direct way to summarize a multi-state model over one or more periods.
This tutorial uses transition probability matrices: the entries are probabilities of moving from one state to another over a specified period. The calculations rely on the Markov assumption that the next period’s state probabilities depend on the current state and the transition probabilities for that period, rather than on the earlier path used to reach the current state. The matrices below are hypothetical. No interest rate is needed because the quantities being calculated are probabilities, not present values.
Moving an Occupancy Distribution Forward
Use a row vector for the current occupancy probabilities and multiply it on the left by the transition matrix. If the person is known to start in state \(H\), the initial vector is \((1,0,0,0)\). If the person starts in a mixture of states, the initial vector records the probability of starting in each one.
This formula follows from the law of total probability. For any destination state \(j\), the probability of being in \(j\) next period is the sum, over all possible current states \(i\), of the probability of being in \(i\) now multiplied by the probability of moving from \(i\) to \(j\). Thus \(\pi_{t+1,j}=\sum_i \pi_{t,i}p_{ij}\), which is exactly the \(j\)th entry of the row-vector product \(\boldsymbol{\pi}_tP\).
For a time-homogeneous model, the same matrix applies in every period. Repeating the one-period calculation gives \(\boldsymbol{\pi}_n=\boldsymbol{\pi}_0P^n\). If transition probabilities change by period, use the matrix for each period in chronological order: \(\boldsymbol{\pi}_n=\boldsymbol{\pi}_0P_1P_2\cdots P_n\). The matrices generally cannot be rearranged.
An absorbing death state is represented by a death row with probability 1 of remaining in death and probability 0 of moving elsewhere. An absorbing row is appropriate when death is permanent and the model includes no transitions out of death. The other states need not be absorbing: for example, a recovery state may allow relapse to care if the model specifies that transition.
Worked Example: Two Years from a Healthy Starting State
A hypothetical policyholder is healthy at issue. The model uses the following same annual transition matrix in both years. Calculate the occupancy probabilities at the end of year 1 and year 2.
Method. Start with the healthy-state vector and multiply by \(P\) once for year 1. Multiply the resulting occupancy vector by \(P\) again for year 2.
At issue, \(\boldsymbol{\pi}_0=(1,0,0,0)\). Therefore, the year-1 vector is the first row of \(P\):
For year 2, calculate each destination probability using all four possible states at the end of year 1:
Thus \(\boldsymbol{\pi}_2=(0.655,0.231,0.054,0.060)\). The check is \(0.655+0.231+0.054+0.060=1.000\). The death probability includes people who died in year 1 and remained dead, as well as people who entered death during year 2.
Common mistake avoided: Using only the healthy-to-destination probabilities for year 2 would ignore people who were in care or recovery after year 1. The second multiplication must include every state with nonzero occupancy.
Starting from a Distribution Across States
A policy cohort need not begin in one known state. For example, issue records or a model specification may assign probabilities to several possible initial states. The same matrix multiplication applies: each starting-state probability weights that state’s transition row. After propagation, probabilities of sets of states can be found by adding the relevant occupancy probabilities. The sum is appropriate when the states are mutually exclusive, as they are in a state model.
Worked Example: A Mixed Initial Distribution and Care Probability
For a hypothetical insured group, the initial probabilities are 0.70 healthy, 0.20 in care, and 0.10 in recovery, with no one in death. Use the annual matrix from the preceding example. Find the year-1 occupancy vector and the probability of being alive and in care or recovery at that time.
Method. Multiply the initial distribution by \(P\), then add the care and recovery entries for the requested combined event.
The healthy probability at year 1 is \(0.70(0.80)+0.20(0.05)+0.10(0.25)=0.560+0.010+0.025=0.595\). The care probability is \(0.70(0.15)+0.20(0.70)+0.10(0.20)=0.105+0.140+0.020=0.265\).
The recovery probability is \(0.70(0.03)+0.20(0.10)+0.10(0.50)=0.021+0.020+0.050=0.091\). The death probability is \(0.70(0.02)+0.20(0.15)+0.10(0.05)=0.014+0.030+0.005=0.049\). Therefore:
The probability of being alive and in care or recovery is \(0.265+0.091=0.356\). As a check, all four entries sum to \(0.595+0.265+0.091+0.049=1.000\), and the probability of being alive is \(0.595+0.265+0.091=0.951=1-0.049\).
Common mistake avoided: The initial probabilities are not transition probabilities. In particular, the initial 0.10 in recovery must be multiplied by the recovery row of \(P\), not added unchanged to the year-1 recovery probability.
When Transition Probabilities Change by Period
A model may specify different transition matrices for successive periods, for example because transition probabilities depend on age or duration. The current occupancy vector must first be advanced using the first period’s matrix, then the resulting vector using the next period’s matrix. Each matrix’s rows and columns must use the same state ordering.
Worked Example: Two Different Annual Matrices
A hypothetical policyholder starts healthy. The first-year and second-year transition matrices are:
Find the occupancy probabilities at the end of year 2. Each matrix has nonnegative entries and each row sums to 1, including the absorbing-death row.
Method. Apply \(P_1\) to the initial healthy vector, then apply \(P_2\) to the year-1 vector. The intermediate vector is \((1,0,0,0)P_1=(0.60,0.30,0.10,0)\).
Using that vector, the year-2 healthy probability is \(0.60(0.50)+0.30(0)+0.10(0.30)+0(0)=0.330\). The care probability is \(0.60(0.30)+0.30(0.50)+0.10(0.20)+0(0)=0.350\). The recovery probability is \(0.60(0.10)+0.30(0.20)+0.10(0.40)+0(0)=0.160\). The death probability is \(0.60(0.10)+0.30(0.30)+0.10(0.10)+0(1)=0.160\). Hence:
The entries sum to \(0.330+0.350+0.160+0.160=1.000\), as required. Common mistake avoided: Applying \(P_2\) first would use the wrong period’s transitions for the initial healthy state. The chronological product is \(\boldsymbol{\pi}_0P_1P_2\), not \(\boldsymbol{\pi}_0P_2P_1\).
Checks and Common Interpretation Traps
An occupancy calculation is only as meaningful as the period and state definitions behind its matrix. Before multiplying, confirm that each matrix entry is a probability over the intended interval and that rows represent current states while columns represent destination states. Transition intensities are rates, not probabilities; an intensity matrix cannot be substituted directly for \(P\). Converting intensities into transition probabilities requires a separate model calculation.
- Check each row. A transition probability row must sum to 1. A row that does not usually signals an omitted destination or an arithmetic or transcription error.
- Check the occupancy vector. Its entries must be nonnegative and sum to 1, subject to rounding. The death entry should not decrease in a model with absorbing death.
- Keep state order fixed. If rows use \(H,C,R,D\), the initial vector and columns must use \(H,C,R,D\) too.
- Distinguish occupancy from entry. The probability of being in care at a given time is not necessarily the probability of ever having entered care. A person may have entered and then recovered, so occupancy at the endpoint does not record every prior transition.
- Combine events only after propagation. To find the probability of being in either care or recovery at the endpoint, add those endpoint probabilities; do not add transition probabilities from different starting states without weighting them by current occupancy.
Check Your Understanding
Use the state order \(H,C,R,D\). Unless stated otherwise, use the annual matrix from the first worked example. Each problem is hypothetical.
- A person is healthy at the start of the year. Find the one-year occupancy vector. (Answer: \((0.8000,0.1500,0.0300,0.0200)\))
- A person starts in care. Find the one-year probability of being in recovery. (Answer: \(0.1000\))
- An initial distribution is \((0.50,0.30,0.20,0)\). Find the one-year occupancy vector. (Answer: \((0.4650,0.3250,0.1450,0.0650)\))
- A person starts healthy and follows the same annual matrix for two years. Find the two-year death probability. (Answer: \(0.0600\))
- An initial distribution is \((0.40,0.40,0.20,0)\). Find the probability of being in care or recovery after one year. (Answer: \(0.5320\))