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Link Temporary Insurance and Annuity Values

Use a survival-contingent payment identity to move efficiently between temporary insurance and annuity values, while keeping benefit timing and the maturity value consistent.

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What You'll Learn

  • Define temporary term insurance, temporary life annuity-due, and the pure endowment factor on one mortality basis
  • Derive the identity linking those values by telescoping yearly cash flows
  • Solve for an unknown insurance or annuity value from the other quantities
  • Check an identity calculation against the direct survival-contingent payment sum
  • Distinguish annuity-due timing from annuity-immediate timing

Why Link Temporary Insurance and Annuity Values?

A temporary life insurance benefit and a temporary life annuity depend on the same sequence of survival and death events. Their payment dates differ, but a simple identity connects their present values. This is useful when a problem supplies an annuity value and asks for an insurance value, or when it supplies an insurance value and asks for an annuity value. The identity also provides an independent check on a direct sum.

The timing must be specified before using the relationship. In this tutorial, the insurance benefit is paid at the end of the year of death, if death occurs within \(n\) years. The temporary annuity is paid at the beginning of each year, at times \(0,1,\ldots,n-1\), provided the life is then alive. Thus its payment at time zero is certain.

Definition: For a life aged \(x\), \(A^{1}_{x:\overline{n}|}\) is the expected present value of a unit benefit paid at the end of the year of death, if death occurs within \(n\) years. The temporary life annuity-due value \(\ddot{a}_{x:\overline{n}|}\) is the expected present value of unit payments at times \(0,1,\ldots,n-1\), conditional on being alive at each payment time. The \(n\)-year pure endowment factor is \({}_{n}E_x=v^n\,{}_np_x\), the present value of a unit payment at time \(n\) if the life survives \(n\) years.

Assume an effective annual interest rate \(i\), with \(v=1/(1+i)\), and a discrete annual mortality model. Write \({}_kp_x\) for the probability that a life aged \(x\) survives \(k\) years, and \({}_{k-1|}q_x={}_{k-1}p_x q_{x+k-1}\) for the probability of surviving \(k-1\) years and dying in the following year. The formulas below require the stated annual payment and benefit timing; continuous benefits or payments at different times require different present-value sums.

The Temporary Insurance–Annuity Identity

The direct present-value sums make the shared mortality basis visible. Each insurance term is the discounted probability of death in a particular policy year. Each annuity-due term is the discounted probability of being alive at a payment time.

$$ A^{1}_{x:\overline{n}|} =\sum_{k=1}^{n}v^k\,{}_{k-1|}q_x, \qquad \ddot{a}_{x:\overline{n}|} =\sum_{k=0}^{n-1}v^k\,{}_kp_x. $$

With \(d=1-v\), the relationship is

Theorem: For end-of-year-of-death temporary insurance and an annuity-due with payments at times \(0\) through \(n-1\), valued at the same interest and mortality assumptions,
$$ A^{1}_{x:\overline{n}|} +d\,\ddot{a}_{x:\overline{n}|} +{}_{n}E_x =1. $$

Equivalently, if \(d>0\), any one of the three values can be isolated from the other two. For example,

$$ A^{1}_{x:\overline{n}|} =1-{}_{n}E_x-d\,\ddot{a}_{x:\overline{n}|}, \qquad \ddot{a}_{x:\overline{n}|} =\frac{1-{}_{n}E_x-A^{1}_{x:\overline{n}|}}{d}. $$

The maturity term matters: temporary insurance does not pay a death benefit if the life survives the full term, whereas the pure endowment factor represents a hypothetical unit payment on survival to time \(n\). Omitting that survival value would leave the identity incomplete.

Why the Identity Holds

Consider a single year \(k\). At the start of that year, the probability of being alive is \({}_{k-1}p_x\). By the end of the year, that group has either died during the year or survived to time \(k\). The discounted value of those two outcomes, together with the appropriate annuity contribution, reduces to the discount-weighted value at the start of the year.

$$ v^k\,{}_{k-1|}q_x +d\,v^{k-1}{}_{k-1}p_x = v^{k-1}{}_{k-1}p_x-v^k\,{}_kp_x. $$

To verify the equality, substitute \({}_{k-1|}q_x={}_{k-1}p_x-{}_kp_x\) and \(d=1-v\). The left side becomes

$$ v^k({}_{k-1}p_x-{}_kp_x) +(1-v)v^{k-1}{}_{k-1}p_x = v^{k-1}{}_{k-1}p_x-v^k\,{}_kp_x. $$

Now sum this equality over \(k=1,\ldots,n\). The year-start and year-end terms telescope: the intermediate survival terms cancel, leaving \(1-v^n{}_np_x\). The summed death-benefit terms equal \(A^{1}_{x:\overline{n}|}\), and the summed annuity terms equal \(d\,\ddot{a}_{x:\overline{n}|}\), since the latter sum runs over \(k-1=0,\ldots,n-1\). Therefore,

$$ A^{1}_{x:\overline{n}|}+d\,\ddot{a}_{x:\overline{n}|} =1-v^n{}_np_x =1-{}_{n}E_x. $$

Applying the Relationship

A reliable calculation keeps the assumptions and timing visible. First identify the term length and the survival probabilities at each integer duration. Then calculate the pure endowment and annuity-due values, apply the identity, and, when practical, compare the result with the direct insurance sum.

1
State the timing and basis.
Confirm the annual interest rate, the mortality probabilities, the number of years, and when each payment occurs.
2
Calculate survival-contingent values.
Use the stated survival probabilities to find \({}_{n}E_x\) and \(\ddot{a}_{x:\overline{n}|}\), or the supplied values if the problem gives them.
3
Apply the identity and check it.
Use \(A^{1}_{x:\overline{n}|}=1-{}_{n}E_x-d\ddot{a}_{x:\overline{n}|}\) when solving for insurance, then compare with a direct death-year sum where feasible.

Worked Example: Find a Three-Year Insurance Value from Survival Data

A hypothetical policy covers a life for three years. The effective annual interest rate is \(i=25\%\), so \(v=0.8\) and \(d=0.2\). Assume the conditional probability of survival through each policy year is \(0.8\), independently of prior survival. Find the unit end-of-year-of-death term insurance value using the annuity identity, then check it directly.

Method. Use the common survival basis to calculate the annuity-due and pure endowment first. The survival probabilities are \({}_0p_x=1\), \({}_1p_x=0.8\), \({}_2p_x=0.8^2=0.64\), and \({}_3p_x=0.8^3=0.512\).

The three annuity-due payments are at times 0, 1, and 2. Thus

$$ \ddot{a}_{x:\overline{3}|} =1+(0.8)(0.8)+(0.8)^2(0.64) =1+0.64+0.4096 =2.0496. $$

The pure endowment factor is

$$ {}_{3}E_x=(0.8)^3(0.512)=0.512(0.512)=0.262144. $$

Apply the identity:

$$ A^{1}_{x:\overline{3}|} =1-0.262144-(0.2)(2.0496) =0.737856-0.40992 =0.327936. $$

For the direct check, the probabilities of death in years 1, 2, and 3 are \(0.20\), \(0.8(0.20)=0.16\), and \(0.8^2(0.20)=0.128\). Their discounted values are

$$ (0.8)(0.20)+(0.8)^2(0.16)+(0.8)^3(0.128) =0.16+0.1024+0.065536 =0.327936. $$

The direct sum agrees with the identity. The common mistake avoided here is applying discount factors to survival probabilities but using incorrect death-year probabilities: each death-year probability must include survival to the start of that year, exactly once.

Worked Example: Solve from a Temporary Annuity Value

A hypothetical three-year term benefit pays one unit at the end of the year of death. The effective annual interest rate is \(10\%\), and the conditional probability of survival in each year is \(0.8\). Find the insurance value using the annuity identity, and verify the result by a direct sum.

Here \(v=10/11\), \(d=1/11\), and the survival probabilities at times 1, 2, and 3 are \(0.8\), \(0.64\), and \(0.512\). The annuity-due value is

$$ \ddot{a}_{x:\overline{3}|} =1+\frac{10}{11}(0.8)+\left(\frac{10}{11}\right)^2(0.64) =1+\frac{8}{11}+\frac{64}{121} =\frac{273}{121}. $$

The three-year pure endowment factor is

$$ {}_{3}E_x =\left(\frac{10}{11}\right)^3(0.512) =\frac{512}{1331}. $$

Therefore,

$$ A^{1}_{x:\overline{3}|} =1-\frac{512}{1331}-\frac{1}{11}\left(\frac{273}{121}\right) =\frac{1331-512-273}{1331} =\frac{546}{1331} \approx 0.4102. $$

To check, the death probabilities in years 1, 2, and 3 are \(0.20\), \(0.16\), and \(0.128\). The direct discounted sum is

$$ \frac{10}{11}(0.20) +\left(\frac{10}{11}\right)^2(0.16) +\left(\frac{10}{11}\right)^3(0.128) = \frac{242}{1331}+\frac{176}{1331}+\frac{128}{1331} =\frac{546}{1331} \approx 0.4102. $$

The identity and direct calculation agree. The method avoids mixing up the conditional survival probability \(0.8\) with the unconditional probability of death in year 2, which is \(0.8(0.2)=0.16\).

Worked Example: Recover an Annuity-Due Value

For a hypothetical two-year contract, the effective annual interest rate is \(5\%\). The conditional survival probability in each year is \(0.9\). Suppose the unit end-of-year-of-death term insurance value is known to be \(0.17687075\). Recover the temporary annuity-due value using the identity, and check the inputs from the stated mortality model.

The discount factor and discount rate are \(v=1/1.05=20/21\) and \(d=1/21\). The two-year survival probability is \(0.9^2=0.81\), so

$$ {}_{2}E_x =\left(\frac{20}{21}\right)^2(0.81) =\frac{400}{441}\frac{81}{100} =\frac{36}{49} \approx 0.73469388. $$

Solve for the annuity-due value:

$$ \ddot{a}_{x:\overline{2}|} =\frac{1-{}_{2}E_x-A^{1}_{x:\overline{2}|}}{d} =\frac{1-0.73469388-0.17687075}{1/21} \approx 1.85714277 \approx 1.8571. $$

Check from the survival-contingent annuity sum: \(\ddot{a}_{x:\overline{2}|}=1+v(0.9)=1+(20/21)(0.9)=1+6/7=13/7\approx1.85714286\). The small difference before rounding is due to the stated insurance value being rounded. Directly, the insurance value is \((20/21)(0.1)+(20/21)^2(0.09)=0.09523810+0.08163265=0.17687075\), confirming consistency to the displayed precision.

The common mistake avoided is dividing by \(i\) instead of \(d\). The identity contains \(d\,\ddot{a}_{x:\overline{n}|}\), so the annuity is isolated by division by \(d=1-v\).

Timing Traps and Practical Checks

The identity is compact, but it is not a timing-free formula. An annuity-due has an initial payment at time zero; an annuity-immediate does not. If \(a_{x:\overline{n}|}\) denotes payments at times 1 through \(n\) while alive, then \(\ddot{a}_{x:\overline{n}|}=1+a_{x:\overline{n}|}\). Substituting the immediate value directly for the annuity-due value loses the certain initial payment.

For example, in the first worked example, the annuity-immediate value is \(2.0496-1=1.0496\). Incorrectly using that value in the identity would give

$$ 1-0.262144-(0.2)(1.0496)=0.527936, $$

rather than the correct insurance value \(0.327936\). The discrepancy is exactly \(d\), the omitted contribution of the time-zero annuity payment multiplied by \(d\). A second check is to ensure that the pure endowment uses survival through all \(n\) years and is discounted \(n\) years: \({}_{n}E_x=v^n{}_np_x\).

Key takeaway: For end-of-year-of-death temporary insurance and an annuity-due paid at times \(0\) through \(n-1\), use \(A^{1}_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}+{}_{n}E_x=1\). Keep the term, payment timing, and pure endowment on the same mortality and interest basis.

Check Your Understanding

In each problem, identify the relevant timing and use the relationship or a direct sum to check your work. Answers are rounded only where indicated.

  1. A hypothetical two-year policy has \(i=4\%\), \({}_1p_x=0.95\), and \({}_2p_x=0.90\). Find the unit end-of-year-of-death term insurance value. Give the answer to four decimals. (Answer: 0.0943)
  2. A hypothetical two-year contract has \(i=25\%\), with conditional survival probability \(0.9\) in each year. Find \(A^{1}_{x:\overline{2}|}\) using the temporary annuity identity. (Answer: 0.1376)
  3. For a hypothetical three-year term, \(i=25\%\) and the survival probabilities at times 1, 2, and 3 are \(0.9\), \(0.8\), and \(0.7\). Find the unit end-of-year-of-death insurance value. (Answer: 0.1952)
  4. A hypothetical contract has discount rate \(d=0.05\), term insurance value \(A^{1}_{x:\overline{n}|}=0.30\), and temporary annuity-due value \(\ddot{a}_{x:\overline{n}|}=2.4\), all on the same basis. Find the pure endowment factor. (Answer: 0.58)
  5. A hypothetical three-year policy has \(v=0.9\) and conditional survival probability \(0.8\) in each year. Find its unit end-of-year-of-death term insurance value by the identity, then round to four decimals. (Answer: 0.4029)

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