Why Payment Timing Determines Present Value
A death benefit payable at the end of the year of death is not paid at the instant of death. If death occurs during the first policy year, payment is made one year after issue; if death occurs during the second policy year, payment is made two years after issue. That timing determines the number of annual discount periods and therefore the present value.
The central object here is a present-value random variable: its value depends on the random year in which death occurs. The benefit amount may be fixed, but the payment date—and hence the present value—is not known in advance. We will express this random variable using curtate future lifetime and an annual discount factor.
Let \(B\) be a fixed benefit payable at the end of the year of death. Let \(i\) be the effective annual interest rate, assumed nonnegative, and let \(v=1/(1+i)\) be the annual discount factor. The present value at issue of a payment made at time \(n\) is its amount multiplied by \(v^n\). Since the payment time is \(K_x+1\), the present-value random variable is:
This expression describes the value for each possible death year. It does not say that the benefit is paid immediately at death: the extra \(1\) in the exponent accounts for payment at the end of the year of death. The result assumes a level benefit paid once, at that specified time, and no other cash flows are included.
From Curtate Lifetime to Present-Value Outcomes
For each nonnegative integer \(k\), the event \(K_x=k\) is the event that the life survives \(k\) complete years and then dies in the following year. Using \({}_kp_x\) for the probability of surviving \(k\) years and \(q_{x+k}\) for the probability of death during the next year, conditional on reaching age \(x+k\), we have:
Here \({}_0p_x=1\), so \(P(K_x=0)=q_x\). These probabilities identify how likely each present-value outcome is. Because \(Z=Bv^{K_x+1}\), the outcome associated with \(K_x=k\) is \(Bv^{k+1}\).
Why the Exponent Is \(K_x+1\)
The formula follows directly from the definition of curtate future lifetime and the stated payment timing. When \(K_x=k\), death occurs after \(k\) complete years but before \(k+1\) complete years have elapsed. The contract pays at the end of the year of death, which is time \(k+1\). Discounting a payment at time \(k+1\) back to time 0 gives \(Bv^{k+1}\).
If \(K_x=k\), exactly \(k\) full years have elapsed before the year in which death occurs.
The end of that year is time \(k+1\), not time \(k\).
A payment \(B\) at time \(k+1\) has present value \(Bv^{k+1}\), giving \(Z=Bv^{K_x+1}\).
The same reasoning gives a useful expression for the mean present value when the possible death years are listed. Multiply each present-value outcome by its probability and sum. For a discrete distribution of \(K_x\), this is:
This is an expectation of the random present value, not a statement that the actual payment is made at that expected value. The actual present value is one of the outcomes \(Bv^{k+1}\), determined by the realized death year. The sum is valid when the probabilities describe the lifetime model and the benefit and interest assumptions remain as stated.
Worked Example: Listing Present-Value Outcomes From Annual Mortality
A hypothetical model has annual mortality probabilities \(q_x=0.20\), \(q_{x+1}=0.25\), and \(q_{x+2}=1\). A level benefit of $10,000 is paid at the end of the year of death. The effective annual interest rate is \(i=25\%\). Find the distribution of the present-value random variable and its expected value.
Method. First find the probabilities of each curtate lifetime value. Then discount the benefit for the corresponding payment time. This avoids confusing the year of death with the number of discount periods.
The first possible outcome is \(K_x=0\), meaning death in the first year. Its probability is \(q_x=0.20\). For \(K_x=1\), the life must survive year 1 and die in year 2:
For \(K_x=2\), the life must survive the first two years and then die in year 3. The probability of surviving year 2 is \(1-q_{x+1}=0.75\), so:
The probabilities check: \(0.20+0.20+0.60=1.00\). The annual discount factor is \(v=1/(1.25)=0.80\). The payment dates and present values are therefore:
Thus \(Z\) is $8,000 with probability 0.20, $6,400 with probability 0.20, and $5,120 with probability 0.60. Its expected value is:
The probability total is 1.00, and the expected value lies between the smallest and largest possible present values. The common mistake avoided is using \(v^k\): for \(K_x=0\), payment is at time 1, so the present value is $8,000, not $10,000.
Reading and Checking the Random Variable
A reliable way to check the exponent is to test the first few lifetime outcomes. If \(K_x=0\), death occurs during the first year and the end-of-year payment is discounted one year. If \(K_x=1\), death occurs during the second year and the payment is discounted two years. If \(K_x=2\), the payment is discounted three years. These cases must produce exponents 1, 2, and 3, respectively.
The random variable also depends on the benefit and interest assumptions. A larger benefit multiplies every outcome by the same factor. A higher nonnegative interest rate reduces \(v\), so for a fixed death year it reduces the present value. However, the probability of each outcome comes from the mortality model, not from the interest rate.
| Lifetime outcome | Payment time | Present value |
|---|---|---|
| \(K_x=0\) | End of year 1 | \(Bv\) |
| \(K_x=1\) | End of year 2 | \(Bv^2\) |
| \(K_x=2\) | End of year 3 | \(Bv^3\) |
| \(K_x=k\) | End of year \(k+1\) | \(Bv^{k+1}\) |
Worked Example: Finding a Probability From Present-Value Outcomes
A hypothetical contract pays $50,000 at the end of the year of death. The effective annual interest rate is \(i=25\%\). A mortality model assigns probabilities \(P(K_x=0)=0.10\), \(P(K_x=1)=0.20\), \(P(K_x=2)=0.30\), and \(P(K_x=3)=0.40\). Find the present value for each outcome, the probability that the present value is less than $25,000, and the expected present value.
Method. Convert each curtate lifetime value into its payment time, then discount. For the probability question, compare each resulting present value with the specified threshold.
Since \(v=1/(1.25)=0.80\), the outcomes are:
Only the outcome for \(K_x=3\) is less than $25,000. Therefore \(P(Z<\$25{,}000)=0.40\). The expected present value is:
The expected value is consistent with the outcomes: $26,272 lies between $20,480 and $40,000. The common mistake avoided is treating “less than $25,000” as including the $25,600 outcome; that value exceeds the threshold.
Worked Example: A Complete Distribution With Three Death Years
A hypothetical mortality model has \(q_x=0.25\), \(q_{x+1}=0.40\), and \(q_{x+2}=1\). A level benefit of $21,600 is paid at the end of the year of death. The effective annual interest rate is \(i=20\%\). Find the distribution of the present value, its expected value, and the probability that it is at most $12,500.
Method. Use conditional annual mortality to find the death-year probabilities, and use the payment time \(K_x+1\) to find each present value.
Death in year 1 has probability \(0.25\). Death in year 2 requires survival of year 1 followed by death in year 2:
Death in year 3 requires survival of years 1 and 2, followed by death under \(q_{x+2}=1\):
The probabilities sum to \(0.25+0.30+0.45=1.00\). The discount factor is \(v=1/(1.20)=5/6\). Thus:
The expected present value is:
Only the third outcome is at most $12,500, so \(P(Z\leq \$12{,}500)=0.45\). The equality is included because the question says “at most.” The probabilities sum to 1.00, and the computed expected value is between the minimum and maximum present-value outcomes.
Common Timing Errors
- Using \(K_x\) instead of \(K_x+1\). \(K_x\) counts complete years lived before the year of death; it is not the payment time. The end-of-year payment occurs at time \(K_x+1\).
- Discounting to the instant of death. The benefit in this model is paid at the end of the year of death, not at the exact time of death. The present-value expression therefore uses annual periods.
- Confusing event probability with present value. The probability \({}_kp_xq_{x+k}\) tells how likely \(K_x=k\) is. It is not itself a payment amount or discount factor.
- Leaving out a possible lifetime outcome. When calculating a full distribution, check that the death-year probabilities sum to 1 if the model specifies that death eventually occurs within the listed years. If they do not, determine whether the model includes later death years rather than silently treating the listed outcomes as exhaustive.
- Applying the interest rate to the wrong number of years. Work out the payment time first, then use that time as the exponent of \(v\).
Check Your Understanding
For each problem, identify the payment time before calculating or interpreting the present value.
- A hypothetical benefit of $8,000 is paid at the end of the year of death. The effective annual interest rate is \(25\%\). Find the present values when \(K_x=0\) and \(K_x=2\). (Answer: $6,400 and $4,096, respectively.)
- A life has \(q_x=0.20\) and \(q_{x+1}=0.25\). Find the probability that \(K_x=1\). (Answer: 0.2000.)
- A hypothetical benefit of $30,000 is paid at the end of the year of death, with an effective annual interest rate of \(20\%\). Find its present value if \(K_x=2\), rounded to the nearest cent. (Answer: $17,361.11.)
- A hypothetical mortality model has \(q_x=0.10\), \(q_{x+1}=0.20\), and \(q_{x+2}=0.25\). Find the probability of death during the first three years. (Answer: 0.4600.)
- A hypothetical benefit of $20,000 is paid at the end of the year of death. The effective annual interest rate is \(25\%\), and \(P(K_x=0)=0.20\), \(P(K_x=1)=0.30\), and \(P(K_x=2)=0.50\). Find the expected present value. (Answer: $12,160.)