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Exam FAM · Insurance-Annuity Relationships · Tutorial 358 of 988

Solving for the Discount Rate from Related Values

Learn to isolate the discount rate from whole-life, term, and endowment present-value relationships and convert it to an effective annual interest rate.

Intermediate 9 min read

What You'll Learn

  • Apply the whole-life insurance and annuity-due identity to isolate the discount rate.
  • Convert an effective discount rate to an effective annual interest rate.
  • Use finite-term identities with the pure endowment value included.
  • Distinguish term insurance from endowment insurance when setting up the equation.
  • Check that supplied values use consistent timing and assumptions.

Why Related Present Values Reveal the Discount Rate

An insurance value and a life annuity value can be linked because their payments occur at complementary times. If both present values are supplied on the same mortality and interest basis, their relationship may determine the discount rate without requiring the underlying survival probabilities. The key is to identify the contract type and payment timing before rearranging an identity.

This tutorial uses discrete annual contracts: insurance benefits are paid at the end of the year of death, and life annuity-due payments are made at the beginning of each year while the insured is alive. All values in an identity must be calculated using the same interest and mortality assumptions. These relationships are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let \(i\) be the effective annual interest rate, \(v=1/(1+i)\) the annual discount factor, and \(d=1-v=i/(1+i)\) the effective annual discount rate. The whole-life insurance present value per unit benefit is \(A_x\), and the whole-life annuity-due present value per unit payment is \(\ddot{a}_x\). For a term of \(n\) years, \(A^1_{x:\overline{n}|}\) denotes the present value of unit term insurance, \({}_nE_x=v^n{}_np_x\) the unit pure endowment value, and \(\ddot{a}_{x:\overline{n}|}\) the temporary life annuity-due value.

The Whole-Life Identity and Rate Conversion

For whole-life insurance with a unit benefit payable at the end of the year of death, and a whole-life annuity-due with unit payments at times \(0,1,2,\ldots\) while alive, the present values satisfy:

$$ A_x+d\ddot{a}_x=1. $$

This identity is useful for rate recovery because the unknown rate appears in the discount rate \(d\), which can be isolated directly:

$$ d=\frac{1-A_x}{\ddot{a}_x}. $$

Once \(d\) is found, convert it to the effective annual interest rate. Since \(d=i/(1+i)\), solving for \(i\) gives:

$$ i=\frac{d}{1-d}. $$

The annuity value in the denominator must be positive. For a nonnegative effective annual interest rate, the recovered discount rate must satisfy \(0\le d<1\), and the corresponding interest rate is also nonnegative. A result outside this range is a signal to check the input values, the contract definitions, and whether all values share the same basis.

Why the Whole-Life Identity Holds

The identity follows from a pathwise relationship between the insurance payment and the annuity-due payments. Let \(K=\lfloor T_x\rfloor\) be the curtate future lifetime. Death during year \(K+1\) produces a unit insurance benefit at time \(K+1\), with present value \(v^{K+1}\). The annuity-due payments occur at times \(0,\ldots,K\), so their present value is \(\sum_{j=0}^{K}v^j\). By the finite geometric-series formula:

$$ d\sum_{j=0}^{K}v^j =(1-v)\frac{1-v^{K+1}}{1-v} =1-v^{K+1}. $$

Thus, for each such lifetime outcome, the insurance present value plus \(d\) times the annuity present value is \(v^{K+1}+(1-v^{K+1})=1\). Taking expectations gives \(A_x+d\ddot{a}_x=1\). The timing matters: the death benefit is paid at the end of the year of death, while the annuity payments are due at the beginning of each year the person is alive.

Worked Example: Recovering a Whole-Life Interest Rate

A hypothetical insurer’s valuation supplies \(A_x=0.42\) and \(\ddot{a}_x=9.5\) for unit whole-life insurance and a unit whole-life annuity-due on the same basis. Recover the effective annual interest rate. Assume the insurance pays at the end of the year of death and the annuity pays at the beginning of each year while alive.

Method. Use the whole-life identity to find the effective discount rate, then convert it to the effective annual interest rate.

Step 1: Solve for \(d\).

$$ d=\frac{1-A_x}{\ddot{a}_x} =\frac{1-0.42}{9.5} =\frac{0.58}{9.5} =0.0610526316\ldots $$

Step 2: Convert \(d\) to \(i\).

$$ i=\frac{d}{1-d} =\frac{0.0610526316\ldots}{1-0.0610526316\ldots} =0.0650. $$

The effective annual interest rate is 6.50%. As a check, \(d=i/(1+i)=0.065/1.065=0.0610329\ldots\) would not equal the intermediate value above, so the displayed conversion requires careful arithmetic: using the exact fraction \(d=0.58/9.5=58/950=29/475\), the correct conversion is

$$ i=\frac{29/475}{1-29/475} =\frac{29}{446} =0.0650224\ldots. $$

Rounded to four decimal places, \(i=0.0650\), or 6.50%. The identity check is \(0.42+(29/475)(9.5)=0.42+0.58=1.0000\). The method also avoids treating \(d\) itself as the interest rate.

Finite-Term Contracts: Include the Pure Endowment

The finite-term relationship requires attention to what happens if the insured survives the term. A unit term insurance pays only on death during the \(n\)-year term. A pure endowment pays a unit amount at time \(n\) if the insured survives to that time. Together, these two benefits make an endowment insurance. The finite-term identities are:

$$ A^1_{x:\overline{n}|}+{}_nE_x+d\ddot{a}_{x:\overline{n}|}=1, \qquad A_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}=1, $$

where \(A_{x:\overline{n}|}=A^1_{x:\overline{n}|}+{}_nE_x\) is the present value of unit endowment insurance. Therefore, if the term insurance value, pure endowment value, and temporary annuity-due value are supplied on the same basis:

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

Alternatively, if the endowment insurance value is supplied directly, use \(d=(1-A_{x:\overline{n}|})/\ddot{a}_{x:\overline{n}|}\). If only term insurance and temporary annuity values are given, with no pure endowment value or equivalent information, the displayed identity does not determine \(d\) by itself: the survival benefit at time \(n\) has not been accounted for.

Justification: If death occurs during year \(k+1\), for \(0\le k<n\), the unit term benefit has present value \(v^{k+1}\), and the temporary annuity-due payments occur at times \(0,\ldots,k\). Their combined value after multiplying the annuity by \(d\) is \(v^{k+1}+d\sum_{j=0}^{k}v^j=1\). If the insured survives to time \(n\), the pure endowment value is \(v^n\), while the annuity-due payments occur at times \(0,\ldots,n-1\); then \(v^n+d\sum_{j=0}^{n-1}v^j=1\). Taking expectations yields the finite-term identity.

Worked Example: Using a Term Value and a Pure Endowment

A hypothetical four-year contract has \(A^1_{x:\overline{4}|}=0.24\), \({}_4E_x=0.46\), and \(\ddot{a}_{x:\overline{4}|}=6.0\). The term benefit is payable at the end of the year of death, and the pure endowment is payable at time 4 if the insured survives. Recover the effective annual interest rate.

Method. Apply the finite-term identity with both the term insurance and pure endowment present values included.

Step 1: Find the portion of the identity left for the annuity term.

$$ 1-A^1_{x:\overline{4}|}-{}_4E_x =1-0.24-0.46 =0.30. $$

Step 2: Solve for the discount rate.

$$ d=\frac{0.30}{6.0}=0.05. $$

Step 3: Convert to the effective annual interest rate.

$$ i=\frac{0.05}{1-0.05} =\frac{0.05}{0.95} =0.0526315789\ldots. $$

The effective annual interest rate is 5.26%, rounded to two decimal places. Check: \(0.24+0.46+0.05(6.0)=0.24+0.46+0.30=1.00\). The calculation uses the pure endowment value rather than silently treating term insurance as endowment insurance.

Worked Example: Starting With Endowment Insurance

A hypothetical eight-year endowment insurance has present value \(A_{x:\overline{8}|}=0.68\), and the corresponding eight-year temporary life annuity-due has present value \(\ddot{a}_{x:\overline{8}|}=8.0\). Both are for unit benefits or payments on the same mortality basis. Find the effective annual interest rate.

Method. Since the insurance pays on death within the term or survival to the end of the term, use the endowment insurance identity directly.

Step 1: Recover the discount rate.

$$ d=\frac{1-A_{x:\overline{8}|}}{\ddot{a}_{x:\overline{8}|}} =\frac{1-0.68}{8.0} =\frac{0.32}{8.0} =0.04. $$

Step 2: Recover the interest rate.

$$ i=\frac{0.04}{1-0.04} =\frac{0.04}{0.96} =0.0416666\ldots. $$

The effective annual interest rate is 4.17%, rounded to two decimal places. The check is \(0.68+0.04(8.0)=1.00\). No separate pure endowment value is needed here because it is already included in the endowment insurance value.

Common Traps and Consistency Checks

A rate-recovery calculation can be algebraically simple and still use the wrong identity. Before solving, confirm whether the insurance is whole life, term, or endowment; whether benefits are paid at the end of the year of death; and whether the annuity is due. Also verify that values are for compatible benefit and payment amounts. If the insurance benefit is not unit-sized, first ensure that the value used in the identity is expressed per unit benefit or is scaled consistently.

  • Omitting the pure endowment. Term insurance alone does not cover survival to the term’s end. For term insurance, include \({}_nE_x\) in the identity.
  • Confusing \(d\) with \(i\). The identity first gives \(d\). Convert using \(i=d/(1-d)\), not \(i=d\).
  • Mixing payment timings. The identities here use end-of-year death benefits and annuity-due payments. A different timing convention requires a different relationship.
  • Combining values from different bases. Values generated using different mortality assumptions, terms, benefit amounts, or payment frequencies need not satisfy the same identity.

For example, suppose term insurance has value \(0.24\), its pure endowment value is \(0.46\), and the temporary annuity-due value is \(6.0\). The incorrect calculation omits the survival benefit:

$$ d_{\text{wrong}}=\frac{1-0.24}{6.0}=0.1267. $$

This incorrectly implies \(i=0.1267/(1-0.1267)\approx 0.1450\), or about 14.50%. The correction includes the pure endowment:

$$ d_{\text{correct}}=\frac{1-0.24-0.46}{6.0}=0.05, \qquad i_{\text{correct}}=\frac{0.05}{0.95}=0.05263\ldots. $$

Thus the correct effective annual rate is 5.26%. The large difference arises from omitting a contract benefit, not from a rounding issue.

Key takeaway: Choose the identity from the contract’s actual benefits and payment times. Recover \(d\) from the related present values, convert it using \(i=d/(1-d)\), and substitute the result back into the identity as a final check.

Check Your Understanding

For each problem, use the stated discrete benefit and payment timing. Round interest rates to three decimal places where needed.

  1. A whole-life insurance value is \(A_x=0.55\), and the matching whole-life annuity-due value is \(\ddot{a}_x=9.0\). Find the effective annual interest rate. (Answer: 5.263%)
  2. A six-year endowment insurance has value \(0.64\), and its temporary annuity-due value is \(6.0\). Find the effective annual interest rate. (Answer: 6.383%)
  3. A five-year term insurance value is \(0.21\), the pure endowment value is \(0.49\), and the temporary annuity-due value is \(6.0\). Find the effective annual interest rate. (Answer: 5.263%)
  4. A whole-life insurance value is \(0.35\), and the matching whole-life annuity-due value is \(12.5\). Recover \(d\), then find \(i\). (Answer: \(d=0.052\), \(i=5.485\%\))
  5. A term insurance value is \(0.30\), its pure endowment value is \(0.50\), and the temporary annuity-due value is \(5.0\). Find the effective annual interest rate and state whether it is nonnegative. (Answer: \(d=0.040\), \(i=4.167\%\), nonnegative)
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