Introduction
A disability income valuation can depend on more than the probability of becoming disabled. A claim may have to satisfy an elimination period before benefits begin; recovery or death can interrupt the claim; and a maximum benefit duration can stop payments even if disability continues. Premiums may also stop while the insured is disabled and resume after recovery. The valuation is reliable only when the state history and each cash-flow date are explicit.
This tutorial uses an annual, discrete-time model to work through these features together. The numbers are hypothetical. The model is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus; it is not a description of an actual policy or insurer.
Specify the States and Cash-Flow Timing
Use the states \(A\) (active and not disabled), \(W\) (disabled and completing a one-year elimination period), \(B_1\) (in the first payable benefit year), \(B_2\) (in the second and final payable benefit year), \(E\) (still disabled but benefits exhausted), and \(D\) (dead). A recovery from any disabled state returns the insured to \(A\). Death is absorbing.
The annual transition probabilities are conditional on the state at the start of the year. The cash-flow convention is just as important: premiums are paid at the start of each policy year if the insured is in \(A\); a benefit is paid at a year-end only if the insured completes a payable benefit year. Thus a transition \(B_1\to B_2\) produces the first annual benefit, and \(B_2\to E\) produces the second. Recovery or death during the year produces no year-end benefit under this discrete model.
| State at start of year | Possible next states and probabilities | Interpretation |
|---|---|---|
| \(A\) | \(A: 0.90,\ W: 0.08,\ D: 0.02\) | Active continuation, incidence, or death |
| \(W\) | \(A: 0.20,\ B_1: 0.75,\ D: 0.05\) | Recovery, completion of the waiting year, or death |
| \(B_1\) | \(A: 0.25,\ B_2: 0.70,\ D: 0.05\) | Recovery, continuation to the second payable year, or death |
| \(B_2\) | \(A: 0.20,\ E: 0.75,\ D: 0.05\) | Recovery, continued disability after the benefit limit, or death |
| \(E\) | \(A: 0.20,\ E: 0.75,\ D: 0.05\) | Recovery, continued exhausted disability, or death |
| \(D\) | \(D: 1.00\) | Death is absorbing |
Each row sums to 1. The probabilities are annual transition assumptions, not force-of-transition rates. In particular, do not treat an annual incidence probability such as 0.08 as an annual force or convert it to a rate without specifying a separate model. The one-year waiting period here means a full annual interval in \(W\); a contract with a different elimination period needs a time scale and states that represent that period.
Why State Probabilities Give Expected Cash Flows
Let \(\mathbf{p}_k\) be the row vector of state probabilities at time \(k\), and let \(P\) be the matrix whose rows contain the annual transition probabilities. Conditioning on the state at time \(k\) gives the probability of each state at \(k+1\):
This recursion follows by adding the probabilities of all paths that enter a given next state. For example, if the only ways to be in \(B_1\) at time 2 are a transition \(W\to B_1\), then the probability of \(B_1\) at time 2 is the probability of \(W\) at time 1 multiplied by the conditional probability of that transition.
Expected cash flow at a date is the sum of each state- or transition-specific payment multiplied by the probability of that state or transition. Discount each expected cash flow from its actual payment date. This is the same probability-weighting principle used in state and decrement valuations; the main care here is assigning benefits and premiums to the correct transitions and dates.
Start with the known initial state and apply the transition probabilities one year at a time.
Identify which transitions complete a payable benefit year and which year-start states permit a premium.
Multiply each expected payment at time \(k\) by \(v^k\), where \(v=1/(1+i)\).
Confirm that the waiting period is completed before any benefit, no more than the allowed benefit years are paid, and the probabilities sum to 1.
Worked Case: Two-Year Maximum Benefit and Active-Only Premiums
The first example combines incidence, recovery, death, a one-year elimination period, a two-year maximum benefit duration, and premiums payable only while active. It uses the transition table above. The policy is issued to an insured initially in \(A\), premiums are payable at times 0, 1, and 2, and benefits can continue after the three-year premium period. Assume an annual effective interest rate \(i=5\%\), an annual benefit of $10,000, and the stated transition probabilities.
Worked Example: Value the Benefits and Find the Level Premium
Method. Propagate state probabilities for the benefit transitions, then equate the present values of expected benefits and active-only premiums. This avoids counting a benefit during the elimination period or charging a premium while disabled.
At time 0, the insured is active with probability 1. At time 1, the probabilities are \(P(A)=0.90\), \(P(W)=0.08\), and \(P(D)=0.02\).
At time 2, the probability of \(B_1\) is \(0.08(0.75)=0.0600\). The active and waiting-state probabilities are \(P(A)=0.90(0.90)+0.08(0.20)=0.8260\) and \(P(W)=0.90(0.08)=0.0720\). The death probability is \(0.02+0.90(0.02)+0.08(0.05)=0.0420\). These state probabilities sum to \(0.8260+0.0720+0.0600+0.0420=1.0000\).
At time 3, the probability of \(B_2\) is \(0.0600(0.70)=0.0420\). The first benefit is paid at time 3 only on the transition \(B_1\to B_2\). At time 4, the probability of the transition \(B_2\to E\), and hence the second benefit payment, is \(0.0420(0.75)=0.0315\). There are no further benefit payments because the maximum of two payable years has then been reached.
The discount factor is \(v=1/1.05=0.9523809524\). The benefit present value is
For premiums, the probability of being active at the three premium dates is 1 at time 0, 0.90 at time 1, and 0.8260 at time 2. If \(P\) is the annual premium, its expected present value is
Equating the expected present values gives \(P=621.9631/2.6063492063=\$238.63\), rounded to the nearest cent. The unrounded benefit present value is $621.9631, and the unrounded premium factor is 2.6063492063.
Check. The first payment is at time 3, after the one-year elimination period and a completed payable benefit year; the second is at time 4, and no third payment is included. Premiums are weighted by active-state probabilities only. The wrong shortcut \(10{,}000(0.08)\) would treat every incidence as immediately benefit-eligible and ignore the required survival through the waiting and payable periods.
A Second Valuation with Different Transition Assumptions
The same method applies when the annual transition probabilities and interest rate change. The state definitions and timing convention must remain explicit; otherwise a numerical comparison can mix different product designs with different assumptions.
Worked Example: Premium for a Three-Year Premium Term
A hypothetical policy has the same states and cash-flow timing, but the active-state transitions are \(A\to A:0.88\), \(A\to W:0.10\), and \(A\to D:0.02\). From \(W\), the probabilities of recovery, entry to \(B_1\), and death are \(0.25, 0.70, 0.05\). From \(B_1\), they are \(0.20, 0.75, 0.05\) for recovery, transition to \(B_2\), and death. From \(B_2\), they are \(0.15, 0.80, 0.05\) for recovery, transition to \(E\), and death. Assume \(i=4\%\), a $8,000 annual benefit, and premiums at times 0, 1, and 2 only while active. Find the level premium by equivalence.
Method. Follow the claim path to each benefit transition, and separately find active probabilities at premium dates. From an initial \(A\), the probability of \(W\) at time 1 is 0.10. At time 2, \(P(B_1)=0.10(0.70)=0.0700\). Therefore the probability of the first benefit transition \(B_1\to B_2\) at time 3 is \(0.0700(0.75)=0.0525\). The probability of the second benefit transition \(B_2\to E\) at time 4 is \(0.0525(0.80)=0.0420\).
The annual discount factor is \(v=1/1.04=0.9615384615\), so \(v^2=0.9245562130\), \(v^3=0.8889963587\), and \(v^4=0.8548041911\). The expected benefit present value is
The active probabilities at the premium dates are \(1\), \(0.88\), and \(0.88(0.88)+0.10(0.25)=0.7744+0.0250=0.7994\). The expected present value of premiums is consequently
Equivalence gives \(P=660.5926784/2.5852440829=\$255.52\), rounded to the nearest cent. The result is consistent with the cash-flow dates: the first benefit is not paid at time 2, when the claim has only entered \(B_1\); it is paid at time 3 after completing the first payable year.
Common mistake avoided. Using \(0.10(0.70)\) as the probability of the first benefit payment omits the additional \(B_1\to B_2\) continuation probability of 0.75. That product is the probability of being in \(B_1\) at time 2, not the probability of a benefit payment.
Shorter Benefit Limit: A One-Year Maximum
A benefit limit changes which transition creates a payment. It does not change the elimination-period requirement. The next case has a one-year maximum benefit and illustrates why the benefit vector must be rebuilt when the contract changes.
Worked Example: One-Year Benefit Limit and Two Active Premiums
A hypothetical plan has \(P(A\to A)=0.92\), \(P(A\to W)=0.06\), and \(P(A\to D)=0.02\). From \(W\), the probabilities of recovery, entry to the single payable state \(B_1\), and death are \(0.30, 0.65, 0.05\). From \(B_1\), the probabilities of recovery, transition to exhausted disability \(E\), and death are \(0.25, 0.70, 0.05\). Assume a one-year elimination period, a one-year maximum benefit, annual benefit $15,000, \(i=5\%\), and premiums at times 0 and 1 only while active. Find the equivalence premium.
Method. The probability of entering \(W\) at time 1 is 0.06. The probability of entering \(B_1\) at time 2 is \(0.06(0.65)=0.0390\). The single payable year is completed on the transition \(B_1\to E\) at time 3, so the probability of a benefit payment is \(0.0390(0.70)=0.0273\). The benefit present value is
The active probabilities at times 0 and 1 are 1 and 0.92. Thus the premium present-value factor is \(1+0.92v=1+0.92(0.9523809524)=1.8761904762\). The equivalence premium is \(353.7415/1.8761904762=\$188.54\), rounded to the nearest cent.
Check. The benefit is paid at time 3, not time 2: entering \(B_1\) at time 2 begins the payable year but does not complete it. Also, this one-year limit permits no second benefit payment, even if the insured remains disabled.
Common Exam Traps and Checks
These valuations are especially sensitive to state definitions and the payment convention. Before calculating, write down the event that triggers each cash flow. In the examples, a benefit is triggered by completing a payable benefit year, while a premium is triggered by being active at a premium date.
- Paying during the elimination period. Incidence alone does not mean a benefit is payable. Under the stated model, a claim must pass through \(W\) and then complete a payable benefit year.
- Confusing state probability with transition probability. Being in \(B_1\) at a year boundary is not itself a benefit payment. The payment depends on the next transition that completes the payable year.
- Exceeding the benefit limit. Continued disability in \(E\) does not produce further benefits. State \(E\) is still needed if premiums, recovery, or other benefits depend on continued disability.
- Charging premiums in the wrong state. A recovered claimant returns to \(A\) and may owe a premium at a later premium date. A disabled claimant does not owe one under the active-only premium assumption.
- Discounting from the wrong date. A benefit paid at time 3 is discounted by \(v^3\), not \(v^2\). State entry and payment dates need not coincide.
The transition-probability method is a compact way to check the complete claim path. At every time boundary, the probabilities across all states, including death, must sum to 1. Every benefit payment must correspond to a permitted payable transition, and every premium must correspond to an active premium date.
Check Your Understanding
Use the annual timing convention stated in each question. Each problem is hypothetical.
- Using the first transition table, find the probability of being in \(B_1\) at time 2 for an insured initially active. (Answer: 0.0600)
- Under the first example’s assumptions, find the present value at time 0 of only the second $10,000 benefit payment, using \(i=5\%\). (Answer: $259.15)
- In the one-year maximum-benefit example, what is the probability of the benefit payment at time 3? (Answer: 0.0273)
- For a hypothetical policy with \(i=5\%\), active probabilities 1.00 and 0.92 at its only two premium dates, and benefit present value $353.74, find the equivalence premium to the nearest cent. (Answer: $188.54)
- Under the first example’s convention, does a transition \(W\to B_1\) at time 2 trigger a benefit at time 2? State why. (Answer: No; it enters the payable state, but the first payable benefit year must be completed before the benefit is paid.)