Why Payment Timing Creates an Identity
A discrete whole-life insurance and a whole-life annuity can be valued on the same mortality and interest basis, but their cash flows occur at different times. An insurance benefit paid at the end of the year of death is discounted one period beyond the start of that year. An annuity-due payment, by contrast, can be made at the start of each year the insured is alive. Matching those payments path by path reveals an exact relationship between their present values.
The timing conditions matter. The identities below concern a unit insurance benefit paid at the end of the year of death, and annuity payments of one at specified integer times while the insured is alive. They do not automatically apply to an insurance benefit paid immediately on death or to payments made at different dates.
The annuity-due payment at time zero is made immediately, and a payment at an anniversary is included only if the insured is alive then. For the annuity-immediate, a payment at time \(j\) is made only if the insured has survived to that time. Thus, on a path with \(K_x=k\), the annuity-due has \(k+1\) payments and the annuity-immediate has \(k\) payments.
These definitions also give the standard survival-sum forms. Here \({}_kp_x=P(K_x\geq k)\), and \({}_kp_xq_{x+k}\) is the probability of surviving \(k\) years and then dying before the next anniversary:
The sums represent expected present values. They must be finite for the corresponding finite-value identities to be used as ordinary equalities. A usual sufficient condition is a positive effective annual interest rate, \(i>0\): then \(0<v<1\), the annuity-due present value is bounded above by \(1/(1-v)\), and the unit insurance present value is at most one. More generally, other interest and mortality combinations can be used if the expected present values are finite. A nonnegative interest rate alone is not a complete convergence argument when the expected annuity value may be infinite.
Deriving the Annuity-Due Identity
First fix one possible death path with \(K_x=k\). On this path the insurance pays \(v^{k+1}\) in present-value terms. The annuity-due pays at times \(0\) through \(k\), so its present value is \(1+v+\cdots+v^k\). Since \(d=1-v\), multiplying the annuity-due present value by \(d\) makes the sum telescope:
Adding the insurance present value gives one on every such path. Taking expectations gives the identity, provided the expected present values are finite. The argument depends on the benefit being paid at time \(k+1\) and on the annuity-due payments occurring at times \(0,\ldots,k\).
The result is not tied to a particular mortality law. It follows from the cash flows on each death path, so it holds for any curtate-lifetime distribution for which the expected present values are well-defined and finite. This qualification is important: with sufficiently negative interest and a heavy enough survival tail, both expectation sums can diverge, and one cannot conclude that the identity relates finite actuarial values.
Worked Example: Verify the Identity from a Finite Mortality Distribution
A hypothetical policy has annual effective interest rate \(i=10\%\). Its curtate future lifetime has probabilities \(P(K_x=0)=0.20\), \(P(K_x=1)=0.30\), and \(P(K_x=2)=0.50\). Find \(A_x\) and \(\ddot{a}_x\), then verify the annuity-due identity. The unit death benefit is paid at the end of the year of death.
Method. Calculate the insurance and annuity present values for each possible value of \(K_x\), weight by the stated probabilities, and then check the identity. The mortality distribution has finite support, so both expected present values are finite.
First, \(v=1/(1.10)=10/11\), and \(d=1-v=1/11\). If \(K_x=0\), the insurance pays at time 1 and the annuity-due pays at time 0. If \(K_x=1\), insurance pays at time 2 and annuity-due payments occur at times 0 and 1. If \(K_x=2\), insurance pays at time 3 and annuity-due payments occur at times 0, 1, and 2. Therefore,
The annuity-due present value for \(K_x=0\) is \(1\); for \(K_x=1\) it is \(1+10/11=21/11\); and for \(K_x=2\) it is \(1+10/11+100/121=331/121\). Hence,
The check is
The displayed decimals are rounded to six places; the exact fractions verify the identity. A common mistake is to count only \(K_x\) annuity-due payments. That would omit the time-zero payment: when \(K_x=0\), the annuity-due still pays one immediately.
The Corresponding Annuity-Immediate Identity
For the annuity-immediate, payments occur at times \(1,\ldots,k\) on a path with \(K_x=k\). Its present value is therefore \(v+v^2+\cdots+v^k\), which is zero when \(k=0\). Multiplying by \(d=1-v\) yields \(v-v^{k+1}\). Adding the end-of-year-of-death insurance present value \(v^{k+1}\) gives \(v\), not one. Taking expectations gives the matching identity when the expected present values are finite:
The same relation follows from \(\ddot{a}_x=1+a_x\): the annuity-due includes the time-zero payment of one, while the annuity-immediate does not. Substitution into \(A_x+d\ddot{a}_x=1\) gives \(A_x+d(1+a_x)=1\), or \(A_x+d a_x=1-d=v\).
Worked Example: Use a Geometric Mortality Model
A hypothetical life has a constant annual death probability \(q=0.25\), so the probability of surviving each year is \(p=0.75\). The annual effective interest rate is \(i=5\%\). Find \(A_x\) and \(\ddot{a}_x\), and verify the annuity-due identity.
Method. Under this model, \({}_kp_x=p^k\), and the probability of death in year \(k+1\) is \(p^kq\). The values form geometric series. Convergence requires \(vp<1\); here \(v=1/1.05=20/21\), so \(vp=(20/21)(3/4)=5/7<1\).
The insurance value is
The annuity-due value is
Here \(d=1-v=1/21\). Thus \(A_x+d\ddot{a}_x=5/6+(1/21)(7/2)=5/6+1/6=1\). As a further timing check, \(a_x=\ddot{a}_x-1=2.5\), and \(A_x+d a_x=5/6+2.5/21=20/21=v\). This confirms both identities under the stated payment timings.
A common mistake is to use \(q\) as the probability of surviving \(k\) years in the annuity series. Survival through \(k\) years has probability \(p^k\); the factor \(q\) applies only to the year in which death occurs in the insurance sum.
Timing Checks and Convergence
The two identities differ because the annuities have different payment dates. A useful way to remember the result is to first write the cash flows on a path where \(K_x=k\), rather than trying to recall a formula from memory.
| Contract value | Payments on a path with \(K_x=k\) | Identity with end-of-year-of-death insurance |
|---|---|---|
| Whole-life annuity-due | At times \(0,1,\ldots,k\) | \(A_x+d\ddot{a}_x=1\) |
| Whole-life annuity-immediate | At times \(1,2,\ldots,k\) | \(A_x+d a_x=v\) |
These are not identities for every possible benefit timing. If the insurance benefit is paid at the moment of death rather than at the end of the year of death, its discount factor depends on the fractional part of the lifetime. The discrete value \(A_x=E[v^{K_x+1}]\) no longer describes that benefit, so the pathwise derivation above cannot be reused without additional assumptions and a value appropriate to the new timing.
Convergence also deserves an explicit check. In the geometric example, the insurance and annuity series converge because \(vp<1\). For a general mortality distribution, the relevant requirement is that the expected present values being related are finite. With \(i>0\), finiteness is assured for these unit benefits and annuities. With negative interest, \(v>1\), so survival probabilities must decline quickly enough for the weighted sums to converge; otherwise the formula is not an equality between finite values.
Worked Example: Infer Mortality and Verify the Annuity Value
A hypothetical life has a constant annual death probability \(q\), but \(q\) is unknown. At an annual effective interest rate of \(25\%\), the unit discrete whole-life insurance value is \(A_x=0.50\). Find \(q\) and \(\ddot{a}_x\), and verify the identity.
Method. Set \(p=1-q\), use the geometric insurance series, and solve for \(p\). Then calculate the annuity-due using the same mortality and interest assumptions. The geometric sums must converge; the resulting value of \(vp\) will be checked.
The discount factor is \(v=1/1.25=0.8\). From the geometric insurance formula,
Multiplying by \(1-0.8p\) gives \(0.50-0.40p=0.8-0.8p\). Thus \(0.40p=0.30\), so \(p=0.75\) and \(q=1-p=0.25\). The convergence check is \(vp=(0.8)(0.75)=0.60<1\). Therefore,
The discount rate is \(d=1-v=0.20\), so \(A_x+d\ddot{a}_x=0.50+(0.20)(2.5)=1\). The solution is consistent with the given insurance value, and the expected present values are finite. A common mistake is to solve the insurance equation while treating \(p\) as the death probability; the model specifies \(p\) as survival and \(q=1-p\) as death.
Common Exam Traps
- Using the wrong annuity identity. The annuity-due includes time zero, giving \(A_x+d\ddot{a}_x=1\). The annuity-immediate does not, giving \(A_x+d a_x=v\).
- Shifting the death benefit by one year. Under the stated discrete insurance definition, a death in year \(k+1\) produces a payment at time \(k+1\), with present value \(v^{k+1}\), not \(v^k\).
- Ignoring finiteness. The pathwise relation does not make divergent expectation sums finite. Check convergence or use a sufficient condition such as \(i>0\).
- Changing the payment condition. An annuity payment at an anniversary requires survival to that anniversary for an immediate annuity; the due annuity instead has its first payment at time zero.
For aligned work on the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus, write down the payment dates and the event that triggers each payment before substituting values. This short timing check often identifies an off-by-one error before it affects the calculation.
Check Your Understanding
For each problem, identify the payment timing before using an identity. Answers are given for checking.
- A hypothetical policy has \(i=25\%\) and \(P(K_x=0)=P(K_x=1)=0.50\). The unit benefit is paid at the end of the year of death. Calculate \(A_x\) and \(\ddot{a}_x\). (Answer: \(A_x=0.72,\ \ddot{a}_x=1.4\); \(A_x+0.20\ddot{a}_x=1\).)
- Use the same interest rate and curtate-lifetime distribution as in Question 1. Calculate \(a_x\) and verify the annuity-immediate identity. (Answer: \(a_x=0.40\); \(A_x+0.20a_x=0.80=v\).)
- A life has constant annual death probability \(q=0.20\) and \(i=25\%\). Find \(A_x\) and \(\ddot{a}_x\), checking convergence. (Answer: \(vp=0.64<1,\ A_x=4/9,\ \ddot{a}_x=25/9\); \(A_x+0.20\ddot{a}_x=1\).)
- At \(i=25\%\), a geometric mortality model has \(A_x=0.50\). Infer \(q\), then calculate \(\ddot{a}_x\). (Answer: \(p=0.75,\ q=0.25,\ \ddot{a}_x=2.5\); the geometric series converges because \(vp=0.60<1\).)
- A hypothetical life has constant annual death probability \(q=0.20\) and \(i=-10\%\). Determine whether the geometric insurance and annuity-due values are finite, and, if they are, verify the identity. (Answer: \(v=10/9,\ vp=8/9<1,\ A_x=2,\ \ddot{a}_x=9\); \(d=-1/9\) and \(A_x+d\ddot{a}_x=1\).)