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Exam ALTAM · State Benefits and Cash Flows · Tutorial 100 of 999

Payment Timing Within a Policy Year

Determine which state is tested on a payment date and how timing changes the probability and present value of a state-contingent payment.

Intermediate 10 min read

What You'll Learn

  • Distinguish a payment date from the date at which benefit eligibility is tested.
  • Identify the occupancy probability relevant to beginning-, midyear, and end-year payments.
  • Apply a transition matrix to an initial state distribution to find payment-date probabilities.
  • Discount payments at their actual payment times using an effective annual rate.
  • Avoid substituting end-of-year probabilities when eligibility is determined earlier.

Why Payment Timing Matters

A benefit can depend on the policyholder’s state at the beginning, midpoint, or end of a policy year. Those dates generally have different state probabilities: a person may change state during the year, so the probability of being disabled at the midpoint need not equal the probability of being disabled at year-end. The payment date also determines how long the payment is discounted.

This tutorial considers fixed-date payments conditional on the state at a specified date. It is important to distinguish two questions: When is eligibility determined? and When is the payment made? They can be the same date, but a contract can also determine eligibility at one date and pay later. A payment triggered at the instant of a transition is a different cash-flow type.

Definition: Let \(J(t)\) denote the policyholder’s state at time \(t\), and let \(\pi_j(t)=P(J(t)=j)\) be the probability of occupying state \(j\) at that time. A fixed-date benefit of amount \(B_j\), paid at time \(t\) if the policyholder is in state \(j\), has expected payment \(B_j\pi_j(t)\). If the payment is made at time \(t\), its expected present value at time 0, under effective annual interest rate \(i\), is \(v^tB_j\pi_j(t)\), where \(v=1/(1+i)\).

Which Probability Belongs to Which Date?

For a one-year policy period, the three common payment dates are \(t=0\), \(t=0.5\), and \(t=1\). A beginning-of-year payment uses the state distribution at time 0. A midyear payment uses the distribution at time 0.5. An end-of-year payment uses the distribution at time 1. These are occupancy probabilities: they describe the state at a particular time, rather than the probability of having made a particular transition at some point during the year.

If the process has a transition probability matrix \(P(s,t)\), with entry \(P_{kj}(s,t)=P(J(t)=j\mid J(s)=k)\), and the initial state distribution is the row vector \(\boldsymbol{\pi}(s)\), then

$$ \boldsymbol{\pi}(t)=\boldsymbol{\pi}(s)P(s,t). $$

Thus, for a policy starting with distribution \(\boldsymbol{\pi}(0)\), use \(P(0,0.5)\) for a midpoint state test and \(P(0,1)\) for an end-year state test. For a beginning-of-year test, use \(\boldsymbol{\pi}(0)\) directly: no future transition probability is needed.

Under a constant effective annual interest rate \(i\), a payment made at time \(t\) has discount factor \(v^t=(1+i)^{-t}\). In particular, a payment at the midpoint is discounted by \(v^{0.5}\), not by \(v\). A payment that is eligible based on the midpoint state but is actually paid at year-end uses the midpoint occupancy probability and the year-end discount factor.

A Short Derivation: A One-Way State Change

Consider a hypothetical two-state continuous-time model with Active as the initial state and Disabled as an absorbing state. The transition intensity from Active to Disabled is a constant \(\lambda\) per year. Over an interval of length \(t\), the probability of remaining Active is \(e^{-\lambda t}\). Because the only two possible states are Active and Disabled, the probability of being Disabled at time \(t\) is the complement:

$$ P(J(t)=\text{Disabled}\mid J(0)=\text{Active}) =1-e^{-\lambda t}. $$

The formula follows because the event “Disabled at time \(t\)” is exactly the complement of “still Active at time \(t\)” in this model. The payment-date probability therefore changes with \(t\). The model is not suitable if Disabled is not absorbing or if there are additional possible states; in those cases, use the relevant transition probabilities from the full model.

Worked Example: Comparing Beginning-, Midyear, and End-Year Payments

A hypothetical policyholder starts the year Active. The only possible change is a transition from Active to Disabled, after which the policyholder remains Disabled. The transition intensity is \(\lambda=\ln 2\) per year. The policy pays $1,000 at the beginning, midpoint, or end of the year if the policyholder is Disabled on that payment date. Find the expected present value for each timing, using an effective annual interest rate of 10%.

Method. Find the Disabled occupancy probability at each eligibility date, then multiply by the benefit and the discount factor for the actual payment date. The common mistake to avoid is using the year-end probability for all three dates.

At the beginning of the year, the policyholder is known to be Active. Hence the probability of being Disabled at time 0 is 0, and the expected present value is \(0\).

At the midpoint, the Disabled probability is \(1-e^{-(\ln 2)(0.5)}=1-1/\sqrt{2}=0.292893\). The effective annual discount factor for half a year is \(v^{0.5}=1/\sqrt{1.10}=0.953463\). Thus the expected present value is \(1{,}000(0.292893)(0.953463)=\$279.27\), rounded to the nearest cent.

At year-end, the Disabled probability is \(1-e^{-\ln 2}=0.5\). The discount factor is \(v=1/1.10\), so the expected present value is \(1{,}000(0.5)/1.10=\$454.55\), rounded to the nearest cent.

The results are \( \$0.00\), \( \$279.27\), and \( \$454.55\) for beginning-, midyear, and end-year payments, respectively. As a check, the Disabled probability increases from 0 to approximately 0.293 to 0.500, while the later payments receive greater discounting. Both effects are reflected in the calculations.

Initial Distributions and Reversible State Changes

A policyholder need not start in a single known state. If the initial state is uncertain, the distribution at the payment date must account for both the starting probabilities and the possible changes in state. For example, in a two-state model with transitions possible in both directions, a policyholder starting Disabled might recover before the payment date, while an initially Active policyholder might become Disabled.

For a time-homogeneous continuous-time Markov model with intensity matrix \(Q\), the transition matrix over a duration \(t\) is \(P(t)=e^{Qt}\). The key calculation for a fixed-date benefit remains \(\boldsymbol{\pi}(t)=\boldsymbol{\pi}(0)P(t)\). The following example makes the row-vector calculation explicit.

Worked Example: Combining an Initial Mix with State Changes

A hypothetical group policy has two states, Active and Disabled. The transition intensity in each direction is \(\ln 2\) per year, constant throughout the year. At time 0, 60% of members are Active and 40% are Disabled. A benefit of $800 is paid if a member is Disabled at the payment date. Find the expected present value per member for a midpoint payment and for an end-year payment, using an effective annual interest rate of 8%.

Method. Calculate the transition matrix for each payment date and multiply the initial distribution by that matrix. Then apply the benefit and discount factor. Do not treat the initial 40% Disabled as though no recovery were possible.

For equal transition intensities \(\lambda\) in both directions, the probability of being in the starting state after time \(t\) is \((1+e^{-2\lambda t})/2\), and the probability of being in the other state is \((1-e^{-2\lambda t})/2\). Here \(\lambda=\ln 2\).

At \(t=0.5\), \(e^{-2(\ln 2)(0.5)}=0.5\), so the transition matrix, with rows and columns ordered Active, Disabled, is

$$ P(0.5)= \begin{pmatrix} 0.75&0.25\\ 0.25&0.75 \end{pmatrix}. $$

The initial distribution is \((0.60,0.40)\). Therefore the midpoint distribution is \((0.60,0.40)P(0.5)=(0.55,0.45)\). The Disabled probability is 0.45, giving an expected payment of \(800(0.45)=\$360\). The half-year discount factor is \(1/\sqrt{1.08}=0.962250\), so the expected present value is \(360(0.962250)=\$346.41\), rounded to the nearest cent.

At \(t=1\), \(e^{-2\ln 2}=0.25\), giving

$$ P(1)= \begin{pmatrix} 0.625&0.375\\ 0.375&0.625 \end{pmatrix}. $$

The end-year distribution is \((0.60,0.40)P(1)=(0.525,0.475)\). The expected payment is \(800(0.475)=\$380\). Discounting one year gives \(380/1.08=\$351.85\), rounded to the nearest cent.

The end-year Disabled probability is 0.475, not 0.50: the initial mix is not exactly balanced, and the model only approaches a balanced distribution over time. This also checks the matrix calculation: each distribution sums to 1, and the results reflect transitions in both directions.

Eligibility Date and Payment Date Can Differ

A contract may test eligibility at one date but pay at another. In that case, use the occupancy probability for the eligibility date, but discount to the payment date. For instance, if a benefit is fixed by whether the insured is Disabled at the midpoint and is paid at year-end, later recovery does not undo eligibility unless the contract says it does. Conversely, if the contract requires the insured to be Disabled at year-end, the midpoint probability is not the correct probability.

This distinction is not a technicality. Replacing the eligibility-date probability with the payment-date probability changes the event being valued, while discounting to the wrong date changes the present value even if the probability is correct.

Worked Example: Midpoint Eligibility with Year-End Payment

A hypothetical contract begins with the insured Active. Active and Disabled each have a transition intensity of 0.5 per year to the other state, constant over the year. The contract pays $2,000 at year-end if the insured was Disabled at the midpoint. Compare this with a contract paying $2,000 at year-end only if the insured is Disabled at year-end. Use an effective annual interest rate of 5%.

Method. Calculate each probability at its stated eligibility date, then discount both year-end payments for one year. The mistake to avoid is using the year-end state probability for the midpoint eligibility condition.

For equal transition intensities of 0.5, the probability of being Disabled at time \(t\), given an Active initial state, is \((1-e^{-2(0.5)t})/2\). At the midpoint this is \((1-e^{-0.5})/2=0.196735\). The expected year-end payment for midpoint eligibility is \(2{,}000(0.196735)=\$393.47\), before discounting.

The payment is made at year-end, so its discount factor is \(1/1.05\), not \(1/\sqrt{1.05}\). Its expected present value is \(393.47/1.05=\$374.73\), rounded to the nearest cent.

For the year-end eligibility condition, the Disabled probability is \((1-e^{-1})/2=0.316060\). The expected payment is \(2{,}000(0.316060)=\$632.12\), and its expected present value is \(632.12/1.05=\$602.02\), rounded to the nearest cent.

The two values differ because the eligibility events differ, even though both benefits are paid at the same time. As a check, the year-end Disabled probability exceeds the midpoint probability in this model, which starts Active and moves toward a balanced distribution.

Common Timing Traps

  • Using the wrong state date. A midpoint eligibility condition calls for \(\pi_j(0.5)\), even if payment is delayed until year-end.
  • Confusing occupancy with transition. Being Disabled at a specified date is not the same as having entered the Disabled state during the year. A person may recover before the specified date.
  • Using a full-year discount for a midpoint payment. Under effective annual interest \(i\), the midpoint discount factor is \((1+i)^{-0.5}\).
  • Ignoring the initial distribution. For an uncertain initial state, calculate \(\boldsymbol{\pi}(0)P(0,t)\); do not assume every member starts Active.
  • Assuming the payment date determines eligibility. Read the condition carefully. A contract can test state at one date and pay at another.
Key takeaway: First identify the date at which the state condition is tested. Use the state occupancy probability at that date. Separately identify when money is paid, and discount to that payment date. Beginning-, midyear, and end-year payments can therefore require different probabilities, different discount factors, or both.

Check Your Understanding

For each problem, identify the relevant state probability and payment date before calculating.

  1. A policyholder starts Active and can only move from Active to Disabled, with constant intensity \(\ln 2\) per year; Disabled is absorbing. A $600 benefit is paid at the midpoint if Disabled then. Find its expected present value at effective annual interest of 4%. (Answer: $172.32)
  2. At time 0, a member is Disabled with probability 0.30 and Active otherwise. Active can move to Disabled at constant intensity \(\ln 2\) per year, and Disabled is absorbing. A $900 benefit is paid at year-end if Disabled then. Find its expected present value at effective annual interest of 5%. (Answer: $557.14)
  3. A member starts Active in a two-state model with equal transition intensities \(\ln 2\) per year in both directions. A $400 benefit is paid at year-end if the member was Disabled at the midpoint; eligibility is fixed at the midpoint. Find the expected present value at effective annual interest of 10%. (Answer: $90.91)
  4. A member starts Active. Active changes to Disabled at intensity 0.4 per year, and Disabled changes to Active at intensity 0.2 per year; both rates are constant. A $1,500 benefit is paid at year-end if the member is Disabled then. Find its expected present value at effective annual interest of 3%. (Answer: $438.05)
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