Reconciling State Models with Multiple-Decrement Tables

Why Reconciliation Requires a Clear State Definition

A multiple-state model describes movement among states, while a multiple-decrement table records exits from a defined group of states. The two descriptions agree only when the group treated as “still in force,” the event that counts as a decrement, and the cause labels are defined consistently. A change in benefit status may be a state transition without being a decrement; conversely, an entry into a state may count as a decrement if the table defines that entry as the first exit from its in-force group.

The key task is therefore not to match state names mechanically. First define which states remain in the table’s population. Then identify the first transitions out of that set and aggregate their probabilities by cause. The resulting probabilities can be compared with a multiple-decrement table over the same interval and for the same starting population. This mapping is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Definition: For a specified interval, let the in-force set be the states in which a life remains exposed to the multiple-decrement table. A decrement of cause \(j\) is the first transition out of that set classified as cause \(j\). The probability of that decrement is \(q^{(j)}\); the probability of remaining in the in-force set through the interval is \(p\). If these are the only outcomes, then \(p+\sum_j q^{(j)}=1\).

The probabilities are conditional on being in the table’s population at the start of the interval. Their interpretation depends on the chosen in-force set. For example, if standard and reduced-benefit states both remain covered, a transition from standard to reduced benefit is not a decrement. If reduced benefit is instead defined as an exit from the table, that same transition must be classified as a decrement. Both choices can be internally valid, but they describe different tables.

Mapping a Continuous-Time Model

Consider a time-homogeneous continuous-time Markov model over an interval of length \(h\). Let \(A\) be the set of in-force states, and let the remaining states be absorbing exit states grouped into causes \(1,\ldots,m\). “Absorbing” means that after entering an exit state the process cannot leave it. Internal transitions among states in \(A\) are allowed. Use a row-vector convention: the starting distribution over in-force states is \(\alpha\), and \(T\) is the subgenerator for transitions within \(A\). For each cause \(j\), let \(c_j\) be the column vector of transition intensities from each in-force state to an exit of cause \(j\).

The probability of still being in the in-force set at time \(h\) and the probability of exiting by cause \(j\) during the interval are

p(h)=\alpha e^{Th}\mathbf{1}, \qquad q^{(j)}(h)=\int_0^h \alpha e^{Ts}c_j\,ds.

Here \(\mathbf{1}\) is a column vector of ones with one entry for each in-force state. The matrix exponential \(e^{Ts}\) gives the probabilities of occupying each in-force state at time \(s\) without having exited. Multiplying by \(c_j\) gives the instantaneous rate of a cause-\(j\) exit at that time; integrating over the interval gives its probability.

Condition for a direct table mapping: The cause groups must be mutually exclusive and collectively cover every exit from the in-force set. A person must be counted as having at most one first decrement in the interval. If the model allows exit-state transitions or re-entry, the table must explicitly classify the first exit; simply using the endpoint state can give the wrong cause probabilities.

This formulation does not require there to be just one in-force state. It does require that the in-force set and the cause classification be fixed. In particular, internal movements among covered states affect the chance of later exits by changing the state occupied, even though those movements are not themselves decrements.

Why the Probabilities Reconcile

A useful check follows from the generator structure. Since every jump out of an in-force state is assigned to exactly one exit cause, the row sums of \(T\) satisfy

T\mathbf{1}=-\sum_{j=1}^{m}c_j.

Define \(s(t)=\alpha e^{Tt}\mathbf{1}\), the probability of still being in force at time \(t\). Differentiate:

\frac{d}{dt}s(t) =\alpha e^{Tt}T\mathbf{1} =-\sum_{j=1}^{m}\alpha e^{Tt}c_j.

Integrating from \(0\) to \(h\), and using \(s(0)=\alpha\mathbf{1}=1\) for a starting distribution entirely within the in-force set, gives

1-s(h) =\sum_{j=1}^{m}\int_0^h \alpha e^{Tt}c_j\,dt =\sum_{j=1}^{m}q^{(j)}(h).

Because \(s(h)=p(h)\), this proves \(p(h)+\sum_jq^{(j)}(h)=1\). The identity is a reconciliation check, not a substitute for correct cause definitions: probabilities can sum to one and still be assigned to the wrong causes.

Worked Mappings

Worked Example: Two Competing Exit Causes

Method: Use the constant-force competing-exits model for one year. A hypothetical policyholder begins in one covered state. The force of death is \(0.10\) per year and the force of surrender is \(0.20\) per year. Both exits are absorbing, and there are no other state changes. Find the one-year in-force probability and the two decrement probabilities.

The total exit force is \(0.10+0.20=0.30\) per year. The probability of no exit during one year is

p=e^{-0.30}=0.7408182.

For constant competing forces, the probability of exit by a particular cause is its share of the total force multiplied by the probability of any exit. Thus the death probability is

q^{(D)} =\frac{0.10}{0.30}\left(1-e^{-0.30}\right) =\frac{1}{3}(0.2591818) =0.0863939.

The surrender probability is

q^{(S)} =\frac{0.20}{0.30}\left(1-e^{-0.30}\right) =\frac{2}{3}(0.2591818) =0.1727879.

Rounded to four decimal places, \(p=0.7408\), \(q^{(D)}=0.0864\), and \(q^{(S)}=0.1728\). The rounded values sum to \(1.0000\), and the unrounded values also sum to one. Check: the two decrement probabilities add to \(0.2591818=1-p\). The common mistake avoided is treating annual forces as annual probabilities; the forces \(0.10\) and \(0.20\) do not themselves give the two one-year decrement probabilities.

The method in the first example generalizes through the occupancy probabilities of multiple in-force states. In a discrete-time state model, let \(P\) be the transition-probability matrix, with absorbing exit states grouped by cause. Let \(P_{AA}\) be the block of transitions within the in-force set and \(P_{A,j}\) the block of transitions from that set to cause \(j\). For a row-vector starting distribution \(\alpha\),

p=\alpha P_{AA}\mathbf{1}, \qquad q^{(j)}=\alpha P_{A,j}\mathbf{1}.

These expressions apply when the matrix gives transitions over the table interval and the exit states are absorbing for the purpose of classification. Internal transitions in \(A\) are included in the probability of remaining in force.

Worked Example: Separating Covered-State Changes from Decrements

Method: Aggregate the row of a one-year transition matrix over the defined in-force set. A hypothetical product has two covered states, Standard (\(S\)) and Reduced Benefit (\(R\)), and two absorbing exits, Death (\(D\)) and Lapse (\(L\)). The one-year transition probabilities are:

\begin{array}{c|rrrr} & S & R & D & L\\ S & 0.68 & 0.12 & 0.13 & 0.07\\ R & 0.08 & 0.72 & 0.14 & 0.06\\ D & 0 & 0 & 1 & 0\\ L & 0 & 0 & 0 & 1 \end{array}

Both \(S\) and \(R\) are in force. A group of 2,500 policies begins in \(S\). From the \(S\) row, the probability of remaining in force is \(0.68+0.12=0.80\). The transition from \(S\) to \(R\) is an internal covered-state change, not a decrement. The cause probabilities are \(q^{(D)}=0.13\) and \(q^{(L)}=0.07\).

The expected one-year counts are \(2{,}500(0.68)=1{,}700\) in \(S\), \(2{,}500(0.12)=300\) in \(R\), \(2{,}500(0.13)=325\) deaths, and \(2{,}500(0.07)=175\) lapses. Hence \(1{,}700+300=2{,}000\) policies remain in force, which is also \(2{,}500(0.80)\). The exits total \(325+175=500\), or \(2{,}500(0.20)\). Check: \(0.80+0.13+0.07=1\), and the four expected counts total 2,500. The common mistake avoided is counting the \(S\)-to-\(R\) transition as a decrement even though both states were defined as in force.

Worked Example: Reconciling a Table for a Mixed Starting Population

Method: Weight state-conditional probabilities by the starting-state distribution. A hypothetical closed group has two in-force states, Active (\(A\)) and Modified (\(M\)), and absorbing Death and Withdrawal exits. The one-year probabilities conditional on starting state are:

\begin{array}{c|rrrr} & A & M & D & W\\ A & 0.70 & 0.14 & 0.10 & 0.06\\ M & 0.10 & 0.60 & 0.18 & 0.12 \end{array}

At the start of the year, 75% of the group is in \(A\) and 25% is in \(M\). The in-force probability conditional on \(A\) is \(0.70+0.14=0.84\); conditional on \(M\), it is \(0.10+0.60=0.70\). Therefore the pooled in-force probability is

$$ p=0.75(0.84)+0.25(0.70)=0.63+0.175=0.805. $$

The pooled death and withdrawal probabilities are

q^{(D)}=0.75(0.10)+0.25(0.18)=0.075+0.045=0.120, $$ $$ q^{(W)}=0.75(0.06)+0.25(0.12)=0.045+0.030=0.075.

Thus the compatible pooled table has \(p=0.805\), \(q^{(D)}=0.120\), and \(q^{(W)}=0.075\). For 4,000 starting policies, there are 3,000 in \(A\) and 1,000 in \(M\). Expected deaths are \(3{,}000(0.10)+1{,}000(0.18)=300+180=480\); expected withdrawals are \(3{,}000(0.06)+1{,}000(0.12)=180+120=300\). Expected policies remaining in force are \(4{,}000-480-300=3{,}220\), also \(4{,}000(0.805)\). Check: \(0.805+0.120+0.075=1\). The common mistake avoided is using the probabilities for one starting state as if they described the whole group.

Compatibility Conditions and Exam Traps

A state model and a multiple-decrement table can be compared directly only if they refer to the same interval, starting population, in-force definition, and event classification. The following checks expose many apparent disagreements:

Check the interval. A one-year table probability cannot be compared directly with an intensity, a monthly probability, or a probability over a different duration.
Check the in-force set. Decide whether each state represents continued exposure to the table. Internal transitions count toward remaining in force if both states are included.
Check the cause definition. Causes must be mutually exclusive. If “disability” means first entry into disability, a later recovery does not undo that first decrement for a table based on first exits. If coverage continues through recovery, however, that table may not describe continued exposure under the product’s actual rules.
Check the starting mix. If table probabilities are pooled across several starting states, weight the state-conditional probabilities by the actual starting distribution. A different mix can produce different pooled decrement probabilities even when the transition model is unchanged.
Check what probabilities describe. A transition matrix’s endpoint probabilities identify exit causes directly when exits are absorbing. If exit states can be left, endpoint occupancy may not identify the cause of the first exit; use a first-exit definition and its corresponding probabilities instead.

A particularly common error is to add cause-specific forces and treat the result as the probability of any decrement. In the first worked example, the incorrect calculation would be \(q^{(D)}=0.10\), \(q^{(S)}=0.20\), and \(p=1-0.10-0.20=0.70\). Those are not the one-year probabilities under the stated continuous-time model. The correct total decrement probability is \(1-e^{-0.30}=0.2591818\), allocated by the force proportions, giving \(q^{(D)}=0.0863939\), \(q^{(S)}=0.1727879\), and \(p=0.7408182\). Forces are rates; the interval probabilities account for the possibility that no exit occurs until later in the interval.

Key takeaway: Reconcile a state model with a multiple-decrement table by defining the in-force set first, classifying each first exit exactly once, and aggregating over internal state changes and the correct starting-state mix. Then verify that continued-in-force probability plus all cause probabilities equals one.

Check Your Understanding

For each problem, state or use the in-force set and cause definitions given. Probabilities in final answers are rounded to four decimal places where needed.

A single covered state has constant annual forces of death \(0.05\) and surrender \(0.15\), with both exits absorbing. Find the one-year in-force, death, and surrender probabilities. (Answer: \(p=0.8187\), \(q^{(D)}=0.0453\), \(q^{(S)}=0.1360\).)
A one-year transition row from covered state \(C\) gives probabilities \(0.50\) to \(C\), \(0.20\) to another covered state \(R\), \(0.20\) to death, and \(0.10\) to lapse. If both covered states are in force, find the three multiple-decrement probabilities. (Answer: \(p=0.7000\), \(q^{(D)}=0.2000\), \(q^{(L)}=0.1000\).)
A population starts 40% in state \(A\) and 60% in state \(B\). Conditional on \(A\), the one-year probabilities are \(0.90\) remaining in force, \(0.06\) death, and \(0.04\) withdrawal. Conditional on \(B\), they are \(0.75\) remaining in force, \(0.15\) death, and \(0.10\) withdrawal. Find the pooled probabilities and verify the total. (Answer: \(p=0.8100\), \(q^{(D)}=0.1140\), \(q^{(W)}=0.0760\); the total is \(1.0000\).)
A table defines disability as a decrement at first entry, but the state model allows recovery and continued coverage. Can a table that treats disabled lives as still in force after recovery be reconciled by simply adding the one-year probability of being disabled at year-end to the decrement probability? Explain. (Answer: No. First entry and year-end occupancy are different events; use a first-exit/first-entry classification consistent with the table, or redefine the in-force set and table to match continued coverage.)
A group of 2,000 begins in covered state \(A\). During the year, each policy has probabilities \(0.72\) of ending in \(A\), \(0.08\) of ending in covered state \(B\), \(0.12\) of death, and \(0.08\) of lapse; the exits are absorbing. Find expected counts by outcome and the table probabilities when both \(A\) and \(B\) are in force. (Answer: 1,440 in \(A\), 160 in \(B\), 240 deaths, and 160 lapses; \(p=0.8000\), \(q^{(D)}=0.1200\), \(q^{(L)}=0.0800\).)

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