Why the Payment Date Matters
A benefit amount alone is not enough to determine its present value: the payment date matters too. With annual effective interest, a payment one year from now is discounted by one annual factor, while a payment several years away is discounted by the corresponding power of that factor. A benefit paid immediately when a decrement occurs may fall between annual dates, so its discount factor requires care.
In a competing-decrements setting, the cause and timing of an exit determine when a benefit is payable. This tutorial focuses on translating that payment date into a discount factor. The probabilities of the competing exits are separate from the timing calculation: a discount factor values a payment at a specified date; it does not, by itself, account for the chance that the payment occurs.
Place Each Payment on a Timeline
Take time 0 to be the valuation date and number years forward from there. Year 1 runs from time 0 to time 1; year 2 runs from time 1 to time 2. Consequently, the start of year \(k\) is at time \(k-1\), and the end of year \(k\) is at time \(k\).
This means that the end of year \(k\) and the start of year \(k+1\) are the same instant, time \(k\). A payment described as being made at the start of the year following a decrement in year \(k\) is therefore discounted for \(k\) years, not \(k+1\), provided no additional delay is specified. The contract’s wording should be checked for any genuine delay beyond that common date.
| Payment description | Payment time from time 0 | Discount factor |
|---|---|---|
| End of year \(k\) | \(k\) | \(v^k\) |
| Start of year \(k\) | \(k-1\) | \(v^{k-1}\) |
| Start of year \(k+1\), following year \(k\) | \(k\) | \(v^k\) |
| Immediately at decrement time \(T\) | \(T\) | Depends on the stated within-year assumption if \(T\) is noninteger |
Worked Example: Converting Payment Dates to Discount Factors
A hypothetical pension plan pays a lump sum of $7,000 at the end of year 4. It also specifies that a separate $2,000 benefit, if triggered by a decrement in year 4, is paid at the start of the following year. The annual effective interest rate is \(i=3\%\). Assume the payment dates are certain for purposes of this timing calculation; no mortality or other decrement probability is being valued. Find the present value of each payment and their total.
Method. First put each payment date on the timeline. The $7,000 is paid at time 4. The $2,000 is paid at the start of year 5, which is also time 4. Both payments use \(v^4\), where \(v=1/1.03\).
The annual discount factor and its fourth power are
Thus the present values are
The total is \(9{,}000v^4=\$7{,}996.38\). The values are rounded to the nearest cent only at the final step. As a check, adding the unrounded present values gives the same total as discounting the combined $9,000 at time 4. The common error here is to treat the start of year 5 as time 5; it is time 4.
Integer-Year Discounting and Its Justification
For annual effective interest \(i\), one dollar invested at time 0 accumulates to \(1+i\) dollars at time 1. Under the same annual accumulation basis, it accumulates to \((1+i)^k\) dollars at time \(k\), for an integer \(k\). The amount that has present value one dollar at time 0 is therefore the reciprocal, \(v^k=(1+i)^{-k}\). By linearity, the present value of a payment \(B\) at time \(k\) is \(Bv^k\).
This argument establishes the standard annual discount factor at integer-year dates. It does not uniquely determine the accumulation between those dates. In particular, an annual effective rate alone does not say how to discount a benefit paid at time 2.5. A within-year convention is needed.
To justify the fractional-time formula, let the accumulation function under a constant force \(\delta\) be \(a(t)=e^{\delta t}\). Consistency with an annual effective rate requires \(a(1)=1+i\). Hence \(e^\delta=1+i\), so \(\delta=\ln(1+i)\). The discount factor to time \(t\) is the reciprocal of accumulation to that time:
This derivation depends on the constant-force assumption. It is not a claim that every interest model with annual effective rate \(i\) must have this particular within-year accumulation pattern.
Worked Example: A Benefit Paid at the Start of a Year
A hypothetical plan pays $4,000 at the start of year 3 and another $4,000 at the end of year 3. The annual effective interest rate is \(i=10\%\). Treat both payment dates as certain and assume no mortality or decrement probability is being valued. Find the present value of the two payments together.
Method. The start of year 3 is time 2, so the first payment uses \(v^2\). The end of year 3 is time 3, so the second payment uses \(v^3\). Here \(v=1/1.10\).
The present values are
Adding the unrounded amounts gives \(4{,}000(v^2+v^3)=\$6{,}311.04\), to the nearest cent. The check is to compare each date against the timeline: the start-of-year payment is one year earlier than the end-of-year payment, so its present value must be larger. The common mistake is to discount the start-of-year-3 payment by \(v^3\) as if it occurred at the end of year 3.
Benefits Paid Immediately on Decrement
Suppose an exit occurs at time \(T\). If the benefit is paid immediately at that instant, its payment time is \(T\), not automatically the end of the policy year in which \(T\) falls. For a fractional \(T\), use the fractional-time discount factor only after specifying the within-year interest convention. Under constant force consistent with \(i\), the present value of amount \(B\) is \(Bv^T\).
The event time and the interest time are measured from the same valuation date. A benefit paid at time 2.5 is discounted for 2.5 years, not for 2 years merely because it occurs during the third year, and not for 3 years unless the contract delays payment to the end of that year. If the decrement time is random, this timing rule identifies the factor for each possible payment date; assessing the probability-weighted value is a separate step.
Worked Example: Immediate Payment Versus Year-End Payment
A hypothetical exit benefit is $20,000 and is payable immediately upon a decrement at time \(T=2.5\) years. For comparison, suppose the contract instead paid the same amount at the end of year 3. The annual effective interest rate is \(i=4\%\), and fractional-year discounting uses a constant force \(\delta=\ln(1.04)\). Treat the decrement time as deterministic at 2.5 years (a point-mass distribution); no mortality or other decrement assumption is required for this timing comparison.
Method. Apply \(v^t=(1.04)^{-t}\) to each stated payment time. For the immediate payment, \(t=2.5\); for the year-end payment, \(t=3\).
Therefore,
The immediate-payment value exceeds the year-end value by \(20{,}000(v^{2.5}-v^3)=\$352.11\), to the nearest cent. This difference is positive because the immediate payment arrives half a year earlier. Rounding is to the nearest cent only at the final step. The common mistake is to use the incorrect fractional factor or to round the factors too early; retain sufficient precision until the dollar amounts are calculated.
A Reliable Timing Procedure
For each benefit, write down the valuation date, identify the actual payment date, and convert it to elapsed time. Only then select the discount factor. This sequence prevents timing language from being replaced by an assumed convention.
Measure every payment date from the valuation date, not from the beginning of the decrement year unless those dates coincide.
End of year \(k\) is time \(k\); start of year \(k\) is time \(k-1\); start of the year after year \(k\) is time \(k\).
At integer times use \(v^k\). At a noninteger time, state the within-year interest assumption before using a fractional power.
Use the amount multiplied by the factor for its own payment date, then add present values when appropriate.
Common Errors and Why They Matter
- Adding one to a start-of-year date. Start of year 3 is time 2, not time 3. The extra year understates present value.
- Separating two labels for the same instant. End of year 2 and start of year 3 are both time 2 unless the contract states a delay.
- Substituting year-end for immediate payment. A decrement at time 2.5 followed by immediate payment is not a time-3 payment. Discounting to time 3 understates the value under positive interest.
- Assuming fractional discounting is automatic. The annual effective rate determines annual accumulation, but fractional-time discounting also needs a within-year convention.
- Mixing probability and discounting. Discounting determines the value conditional on a payment date. It does not establish the probability of a decrement or the probability that the benefit is paid.
Getting these dates right is essential when valuing benefits associated with competing exits. The same benefit amount can have different present values under different payment provisions, and even a one-year timing error changes the discount factor materially. This timing discipline is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each hypothetical benefit, identify the time and apply the stated interest assumption. Round dollar amounts to the nearest cent.
- A $12,000 benefit is payable at the end of year 5. The annual effective rate is \(5\%\); the payment date is certain, with no mortality or decrement probability to value. Find its present value. (Answer: $9,402.31)
- A $9,000 benefit is payable at the start of year 4. The annual effective rate is \(6\%\); treat the payment date as certain. Find its present value and state the payment time. (Answer: Time 3; $7,556.57)
- A plan pays $3,000 at the start of year 2 and a separate $3,000 at the end of year 2. At \(i=8\%\), both dates are certain and no mortality assumption is needed. Find the combined present value. (Answer: $5,349.79)
- A $10,000 benefit is paid immediately when a hypothetical decrement occurs at deterministic time \(T=1.75\). The annual effective rate is \(3\%\), with constant force \(\delta=\ln(1.03)\). Find the present value. (Answer: $9,495.87)
- A $6,000 benefit is paid either immediately at deterministic time \(T=2.4\), or, under an alternative provision, at the end of year 3. The annual effective rate is \(2\%\), and use constant force \(\delta=\ln(1.02)\) for the fractional date. Find each present value and the difference. (Answer: Immediate $5,721.51; year-end $5,653.93; immediate value exceeds it by $67.58)