Introduction
A benefit may be payable on a particular date only if the policyholder is in a specified state on that date. For example, a hypothetical contract might pay a lump sum at the end of year three if the insured is then disabled, but pay nothing if the insured is active or has died. The relevant probability is the probability of occupying the eligible state at the payment time—not the probability of ever having entered that state.
The calculation has two parts: determine the probability of the required state at the specified time, then discount the benefit from that time to the valuation date. This tutorial focuses on one fixed payment date and one eligible state. It does not value a stream of payments or benefits triggered by a state transition.
State Probability and Discounting
Suppose time is measured in years, the valuation date is time 0, and the insured is known to be in state \(i\) at time 0. If the benefit \(B\) is payable at time \(t\) only when the insured is in state \(j\), its present value is \(Bv^t\) when payable and zero otherwise. Here \(v=1/(1+i)\), where \(i\) is an effective annual interest rate. Under the usual compound-interest convention, the time-\(t\) discount factor is \(v^t=(1+i)^{-t}\).
If interest is specified as a constant force \(\delta\), the corresponding discount factor is \(e^{-\delta t}\), so the expected present value is \(Be^{-\delta t}p_{ij}(0,t)\). Use one interest convention consistently; do not combine an effective annual rate discount factor with a force-of-interest formula.
The model assumes the payment amount and payment date are fixed, the payment rule depends on the state at that date, and the interest basis is deterministic. The state probability must come from the stated transition model and initial state. If payment is instead triggered by entering state \(j\), the event is about a transition and its timing, not simply the state occupied at a fixed date.
Why the Formula Holds
Let \(I_j(t)\) equal 1 if the insured is in state \(j\) at time \(t\), and 0 otherwise. The present value of the benefit is the random variable \(Bv^t I_j(t)\): it equals \(Bv^t\) when the benefit is payable and zero when it is not. Since \(B\) and \(v^t\) are fixed, they can be taken outside the expectation:
The indicator's expectation equals the probability of the event it indicates. This establishes why the fixed-time payment amount is multiplied by the probability of occupying the eligible state, then by the discount factor.
Worked Example: A Benefit at the End of Year Three
A hypothetical contract begins with the insured active. At time 3, a benefit of $15,000 is payable if the insured is disabled. The supplied three-year transition matrix, with rows as initial states and columns as states at time 3, is:
Here \(A\) denotes active, \(D\) disabled, and \(X\) dead. The annual effective interest rate is \(i=4\%\). Find the expected present value at time 0.
Method. Read the probability from the row for the known initial state \(A\) and the column for the eligible state \(D\). Thus \(p_{AD}(0,3)=0.18\). The payment is made at time 3, so its discount factor is \(v^3=(1.04)^{-3}\).
The expected payment at time 3 is \(15{,}000(0.18)=\$2{,}700\). The accumulated value of $1 over three years is \((1.04)^3=1.124864\). Therefore:
Check. The probability \(0.18\) is between 0 and 1, and the present value is less than the expected payment of $2,700 because the payment is deferred. The result, rounded to the nearest cent, is $2,400.29.
Common mistake avoided: Using the correct probability but reporting $2,700 as the present value omits discounting. That is the expected payment at time 3, not its value at time 0.
Obtaining the State Probability at the Payment Time
A transition matrix may give the probability directly for the required time interval, as in the preceding example. If the supplied matrices cover successive intervals, multiply them in chronological order. With row-vector convention, a distribution at time 0 is multiplied on the right by the first transition matrix, then by the next matrix. The entry for the eligible state in the resulting distribution is the probability needed for the fixed-time benefit.
Worked Example: Combining Two Annual Transition Matrices
A hypothetical insured begins active. The states are active \(A\), disabled \(D\), and dead \(X\). The first-year and second-year transition matrices are, respectively:
A benefit of $6,000 is payable at time 2 if the insured is disabled then. The annual effective interest rate is \(i=5\%\). Find the expected present value at time 0.
Method. The initial row distribution is \((1,0,0)\). After year one, it is \((1,0,0)P_1=(0.80,0.15,0.05)\). The probability of being disabled at time 2 is the year-one probability of each living state multiplied by its second-year probability of moving to \(D\):
Thus the expected payment at time 2 is \(0.25(\$6{,}000)=\$1{,}500\). The two-year accumulation factor is \((1.05)^2=1.1025\), giving:
Check. Each row of each matrix sums to 1, so the matrices are valid transition matrices. The calculated disabled probability \(0.25\) is a valid probability, and the discounted value is below the expected payment of $1,500. The answer, rounded to the nearest cent, is $1,360.54.
Common mistake avoided: Using only the second-year transition probability \(0.20\) ignores the possibility that the insured was disabled after year one and remained disabled. The correct time-2 probability accounts for each possible state after year one.
A Continuous-Time State Model
In a continuous-time Markov model, transition intensities determine the probability of being in a state at the specified time. As a simple case, consider two states, healthy \(H\) and sick \(S\), with constant intensity \(\alpha\) from \(H\) to \(S\) and constant intensity \(\beta\) from \(S\) to \(H\). There is no death or other state in this model. If the insured starts healthy, let \(q(t)\) be the probability of being sick at time \(t\).
Over a short interval, probability flows into \(S\) from \(H\) at rate \(\alpha(1-q(t))\), and out of \(S\) to \(H\) at rate \(\beta q(t)\). Consequently:
Solving this first-order differential equation gives \(q(t)=\frac{\alpha}{\alpha+\beta}\left(1-e^{-(\alpha+\beta)t}\right)\). This is the probability of being sick at time \(t\) given a healthy initial state; it is not the probability of ever becoming sick by time \(t\). Recovery means those probabilities need not be the same.
Worked Example: Fixed-Time Sick-State Benefit in Continuous Time
A hypothetical policyholder is healthy at time 0. The constant intensity of transition from healthy to sick is \(\alpha=0.5\) per year, and the constant recovery intensity is \(\beta=0.5\) per year. A benefit of $10,000 is payable at time 1 if the policyholder is sick then. The annual effective interest rate is \(i=5\%\). Find the expected present value.
Method. First calculate the probability of being sick at time 1:
The expected payment at time 1 is \(\$10{,}000(0.3160603)=\$3{,}160.6028\). Discount one year at the effective annual rate:
The expected present value, rounded to the nearest cent, is $3,010.10. The probability is less than 1, and discounting reduces the value below the expected payment at time 1, as expected.
Common mistake avoided: Treating the transition intensity \(0.5\) as the probability of being sick at time 1 gives an incorrect probability. An intensity is a rate, not a one-year transition probability; the probability here follows from the continuous-time model and includes the possibility of recovery.
Common Checks and Traps
The key question is always which event makes the payment due at the specified date? For this topic, that event is occupation of a named state at time \(t\). Keep that event separate from both the discount calculation and the process that produces the state probability.
- Confusing occupancy with entry. Being in a state at time \(t\) is not the same as having entered it at some point before \(t\). Recovery or other transitions can change the time-\(t\) state.
- Reversing the transition indices. With \(p_{ij}(s,t)\), \(i\) is the state at the starting time \(s\), and \(j\) is the state at the later time \(t\). Check the row and column labels in a matrix.
- Using the wrong payment time. A benefit due at time \(t\) is discounted for \(t\) years, not for the number of transitions or the number of states in the model.
- Ignoring possible paths before payment. When the payment-time probability is not provided directly, account for all possible intermediate states that can lead to the eligible state at the specified time.
Check Your Understanding
For each problem, find the expected present value at time 0. Round monetary answers to the nearest cent.
- A hypothetical insured begins active. The probability of being disabled exactly one year later is 0.18. A $5,000 benefit is payable then only if disabled. The annual effective interest rate is 5%. (Answer: $857.14)
- A hypothetical insured begins healthy. At time 2, the row of the transition matrix corresponding to healthy is \((0.60,0.30,0.10)\) for the states healthy, sick, and dead. A $12,000 benefit is payable then only if sick. The annual effective interest rate is 4%. (Answer: $3,328.40)
- A hypothetical insured begins active. The first-year probabilities from active to active, disabled, and dead are \(0.70,0.20,0.10\). In the second year, the probabilities of moving to disabled from active and disabled are \(0.10\) and \(0.60\), respectively; dead is absorbing. A $8,000 benefit is payable at time 2 only if disabled. The annual effective interest rate is 5%. (Answer: $1,378.68)
- A hypothetical policyholder begins healthy in a two-state model. The constant healthy-to-sick and sick-to-healthy intensities are both \(0.4\) per year. A $25,000 benefit is payable at time 2 only if sick. The force of interest is \(\delta=0.03\). (Answer: $9,395.32)