Tutorials › Exam ALTAM - Advanced Long-Term Actuarial Mathematics › Initial State Distributions at Policy Issue

Exam ALTAM · Long-Term Care Models · Tutorial 171 of 999

Initial State Distributions at Policy Issue

Represent the health and care conditions present at issue with an initial-state distribution, and calculate how different issue conditions change later state probabilities.

Intermediate 10 min read

What You'll Learn

  • Define an initial-state distribution for a long-term care model.
  • Distinguish issue-state probabilities from transition probabilities.
  • Calculate projected care-state probabilities using a row-vector convention.
  • Compare projections for different underwriting groups and known issue states.
  • Check that each distribution is valid and interpret the results.

Why the Starting State Matters

A long-term care model describes how a policyholder may move among health and care states over time. The first projection question is not only which transitions are possible, but also where the policyholder is at issue. A model for applicants who are all healthy at issue starts differently from one for a hypothetical group that includes people already receiving home care. Even if both groups use the same subsequent transition assumptions, their projected care-state probabilities need not be the same.

An initial-state distribution records the probabilities assigned to the possible states at the time the policy is issued. These probabilities can represent uncertainty about an individual’s starting state, or the composition of a group of policies. They must be based on the model’s state definitions and the contract’s issue conditions. For example, a state representing a person already in facility care should not receive positive issue probability if the contract is assumed to be issued only to people who are not receiving care.

Definition: For a model with \(m\) states, an initial-state distribution is a vector \(\boldsymbol{\alpha}=(\alpha_1,\ldots,\alpha_m)\) in which \(\alpha_j\) is the probability of being in state \(j\) at issue. Each component is nonnegative and \(\sum_{j=1}^{m}\alpha_j=1\). We use a row-vector convention: if \(P(0,t)\) is the transition-probability matrix from issue to time \(t\), the state-probability vector at time \(t\) is \(\boldsymbol{\alpha}P(0,t)\).

Throughout the worked examples, the state order is \(A,H,F,D\): active and not receiving care; receiving home care; receiving facility care; and dead. Death is absorbing, so a person in state \(D\) remains there. The examples use hypothetical transition probabilities for one-year periods. A transition matrix is assumed to have nonnegative entries and each row must sum to one. Its row for a given state describes probabilities conditional on starting that period in that state.

Important distinction: The initial-state distribution describes the state at issue. A transition matrix describes changes in state after issue, conditional on the state at the start of a period. They are different objects: \(\boldsymbol{\alpha}\) is a probability vector, while each row of \(P\) is a conditional probability distribution.

From Issue Conditions to a Projection

Suppose the transition assumptions are represented by a one-year matrix \(P\). If the same matrix applies in each year, then the state-probability vector after \(n\) years is \(\boldsymbol{\alpha}P^n\). If transition assumptions depend on age or duration, use the appropriate matrix for each period in chronological order; for example, the two-year projection is \(\boldsymbol{\alpha}P_1P_2\). These formulas require that the matrix entries are valid transition probabilities for the specified period and that the state definitions remain consistent.

In a continuous-time model, the analogous transition matrix from issue to time \(t\) is written \(P(0,t)\). If the intensity matrix is constant, \(P(0,t)=\exp(Qt)\), where \(Q\) is the generator with nonnegative off-diagonal intensities and rows summing to zero. If intensities vary with age or duration, the transition matrix must reflect that variation over the interval. In either setting, the initial distribution enters on the left, as a row vector.

1
Set the state order.
Use the same order for the initial vector and every transition matrix. State clearly what each state means at issue.
2
Specify issue probabilities.
Assign each state its issue probability, using the underwriting and eligibility assumptions. Check that the probabilities are nonnegative and sum to one.
3
Apply the transition assumptions.
Multiply the initial row vector by the relevant matrix or sequence of matrices, in time order.
4
Check and interpret.
Confirm that the resulting probabilities sum to one and explain them as probabilities at the stated future time, not as probabilities conditional on a different issue group.

Why the Matrix Product Works

The row-vector formula follows from conditioning on the state at issue. Let \(I_j\) mean that the policyholder is in state \(j\) at issue, and let \(K_k(t)\) mean that the policyholder is in state \(k\) at time \(t\). For each issue state \(j\), \(P_{jk}(0,t)=P(K_k(t)\mid I_j)\). The events \(I_1,\ldots,I_m\) partition the possible issue states, so the law of total probability gives

$$ P(K_k(t))=\sum_{j=1}^{m}P(I_j)P(K_k(t)\mid I_j) =\sum_{j=1}^{m}\alpha_jP_{jk}(0,t). $$

The right-hand side is exactly component \(k\) of the row-vector product \(\boldsymbol{\alpha}P(0,t)\). Thus, each projected state probability is a weighted average of the conditional probabilities associated with the possible issue states. This also explains why a change in the initial distribution can change the projection even when the transition assumptions stay fixed.

Worked Example: A Mixed Issue Population

A hypothetical policy group is issued with 80% of policyholders active, 15% receiving home care, and 5% receiving facility care. No policyholder is dead at issue. For each one-year period, assume the transition matrix below, with rows and columns ordered \(A,H,F,D\). Find the state probabilities after one year.

$$ \boldsymbol{\alpha}=(0.80,0.15,0.05,0),\qquad P= \begin{pmatrix} 0.86&0.10&0.03&0.01\\ 0.08&0.72&0.15&0.05\\ 0.02&0.08&0.82&0.08\\ 0&0&0&1 \end{pmatrix}. $$

Method. Multiply the issue distribution by the one-year matrix. Each resulting component is the sum of the issue-state probability times the corresponding one-year transition probability.

$$ \begin{aligned} P(A\text{ at 1})&=0.80(0.86)+0.15(0.08)+0.05(0.02)+0(0)\\ &=0.688+0.012+0.001=0.701,\\ P(H\text{ at 1})&=0.80(0.10)+0.15(0.72)+0.05(0.08)+0(0)\\ &=0.080+0.108+0.004=0.192,\\ P(F\text{ at 1})&=0.80(0.03)+0.15(0.15)+0.05(0.82)+0(0)\\ &=0.024+0.0225+0.041=0.0875,\\ P(D\text{ at 1})&=0.80(0.01)+0.15(0.05)+0.05(0.08)+0(1)\\ &=0.008+0.0075+0.004=0.0195. \end{aligned} $$

The projected vector is \((0.7010,0.1920,0.0875,0.0195)\). As a check, \(0.7010+0.1920+0.0875+0.0195=1.0000\). The group’s one-year probability of being in either care state is \(0.1920+0.0875=0.2795\). This is an unconditional probability for the stated issue population.

Common mistake avoided: The home-care probability is not just the home-care entry from the active row. Policyholders who began in home care may remain there, and policyholders in other states may move into home care. The full matrix product includes all of these paths.

Underwriting Groups and Alternative Starting Conditions

Underwriting can affect the model in two distinct ways. First, different groups may have different initial-state distributions. Second, the groups may have different transition assumptions after issue. These effects should not be conflated. If the groups share the same transition matrix, their projections differ only because they start with different state mixes. If they have group-specific transition matrices as well, each group must be projected using its own assumptions.

Worked Example: Comparing Two Issue Classes

Consider two hypothetical underwriting classes. Class P consists of policyholders all active at issue. Class S has issue distribution \((0.75,0.20,0.05,0)\). Assume both classes use the same one-year transition matrix shown in the previous example. Calculate and compare their one-year probabilities of being in care or dead.

For Class P, \(\boldsymbol{\alpha}_P=(1,0,0,0)\). Multiplying by \(P\) selects its active-state row:

$$ \boldsymbol{\alpha}_PP=(0.86,0.10,0.03,0.01). $$

Therefore, the Class P probability of being in care is \(0.10+0.03=0.13\), and its probability of being dead is \(0.01\).

For Class S, calculate each component using its initial distribution:

$$ \begin{aligned} P_S(A\text{ at 1})&=0.75(0.86)+0.20(0.08)+0.05(0.02)=0.645+0.016+0.001=0.662,\\ P_S(H\text{ at 1})&=0.75(0.10)+0.20(0.72)+0.05(0.08)=0.075+0.144+0.004=0.223,\\ P_S(F\text{ at 1})&=0.75(0.03)+0.20(0.15)+0.05(0.82)=0.0225+0.030+0.041=0.0935,\\ P_S(D\text{ at 1})&=0.75(0.01)+0.20(0.05)+0.05(0.08)=0.0075+0.010+0.004=0.0215. \end{aligned} $$

The Class S vector is \((0.6620,0.2230,0.0935,0.0215)\), which sums to \(1.0000\). Its probability of being in care is \(0.2230+0.0935=0.3165\), and its probability of being dead is \(0.0215\). Compared with Class P, Class S has a higher projected care probability because its issue distribution includes policyholders already in care. The calculation does not establish that the underwriting class itself caused the difference; the specified initial distributions explain it under the shared transition assumptions.

Common mistake avoided: Do not apply a transition row for a single starting state to a group that begins in several states. The Class S result requires weighting all relevant rows by their issue probabilities.

Known Issue State and Multiple Periods

Sometimes the issue state is known rather than uncertain. If all policies start in home care, the initial vector is \((0,1,0,0)\), not a population-wide average. Multiplication then selects the home-care row for the first period. For later periods, the resulting distribution must be multiplied by the next period’s transition matrix. This ordering matters: the first matrix represents the first interval after issue, and the second represents the following interval.

Worked Example: Two Years Starting in Home Care

A hypothetical policyholder is known to be receiving home care at issue. Use the same transition matrix \(P\) in each of the next two one-year periods. Find the probability of each state after two years.

The issue vector is \((0,1,0,0)\), so after one year it equals the home-care row of \(P\):

$$ \boldsymbol{\alpha}P=(0.08,0.72,0.15,0.05). $$

For the second year, multiply this intermediate distribution by \(P\). The active-state probability is

$$ 0.08(0.86)+0.72(0.08)+0.15(0.02)+0.05(0)=0.0688+0.0576+0.0030=0.1294. $$

The remaining components are

$$ \begin{aligned} P(H\text{ at 2})&=0.08(0.10)+0.72(0.72)+0.15(0.08)+0.05(0)\\ &=0.0080+0.5184+0.0120=0.5384,\\ P(F\text{ at 2})&=0.08(0.03)+0.72(0.15)+0.15(0.82)+0.05(0)\\ &=0.0024+0.1080+0.1230=0.2334,\\ P(D\text{ at 2})&=0.08(0.01)+0.72(0.05)+0.15(0.08)+0.05(1)\\ &=0.0008+0.0360+0.0120+0.0500=0.0988. \end{aligned} $$

The two-year vector is \((0.1294,0.5384,0.2334,0.0988)\). Its components sum to \(1.0000\). The death probability includes both those who died during the first year and those who died during the second year; the \(0.05(1)\) term accounts for first-year deaths remaining in the absorbing death state.

Common mistake avoided: Using \(P^2\) alone does not give the two-year distribution for a specific policyholder unless it is also combined with the initial vector. The required calculation is \(\boldsymbol{\alpha}P^2\), or equivalently \((\boldsymbol{\alpha}P)P\).

Checks and Interpretation

Before using an initial distribution in a valuation or projection, confirm that the issue-state probabilities represent the population actually being modeled. A group of newly issued policies may have a different starting mix from a group observed after a period of selection, and a model for in-force policies should not silently be given the issue distribution of new business. If the issue condition rules out a state, assign it probability zero unless the model explicitly includes a different interpretation.

Several checks help catch errors:

  • The components of the initial vector are nonnegative and sum to one.
  • The state order in the vector matches the order of rows and columns in every matrix.
  • Every transition matrix has rows that sum to one, including the absorbing death row.
  • The future-state vector is calculated by multiplying a row vector on the left.
  • The projected components are nonnegative and sum to one, apart from any stated rounding.

A particularly common exam error is to use a conditional probability as if it were unconditional. For example, an entry in the active-state row is the probability of a destination state given active status at the start; it is not automatically the probability for a mixed issue group. The correction is to weight each applicable row by that state’s initial probability and add the results. Also check the time unit: a one-year transition matrix cannot be applied as though it described a month or a full five-year interval.

Key takeaway: The initial-state distribution encodes issue conditions. Combine it with transition probabilities using \(\boldsymbol{\alpha}P(0,t)\), with consistent state ordering and time units. If issue conditions or underwriting groups change the starting mix, calculate their projections separately before comparing them.

Check Your Understanding

Use the state order \(A,H,F,D\) and the one-year transition matrix given in the worked examples unless a question states otherwise. Give probabilities to four decimal places where needed.

  1. A hypothetical issue group has distribution \((0.60,0.30,0.10,0)\). Find its one-year state-probability vector. (Answer: \((0.5420,0.2840,0.1450,0.0290)\))
  2. A policyholder is known to be in home care at issue. What is the probability vector after one year? (Answer: \((0.0800,0.7200,0.1500,0.0500)\))
  3. A group has initial distribution \((0.50,0.50,0,0)\). Find its one-year probability of being in either care state. (Answer: \(0.5000\))
  4. A policyholder starts in home care and the same transition matrix applies for two years. Find the probability of being dead after two years. (Answer: \(0.0988\))
  5. A group starts with distribution \((0.60,0.30,0.10,0)\), and the same matrix applies in both years. Find the full two-year state-probability vector. (Answer: \((0.4917,0.2703,0.1778,0.0602)\))
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