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Exam FAM · Discrete Life Insurance Values · Tutorial 225 of 988

Solving for Interest From a Discrete Insurance APV

Form a discrete insurance APV equation, solve for its discount factor, and convert the admissible solution to an effective annual interest rate.

Beginner 9 min read

What You'll Learn

  • Express a term insurance APV using death-year probabilities and end-of-year benefit timing
  • Turn a stated APV into an equation in the discount factor
  • Solve linear, quadratic, and cubic discount equations
  • Convert a valid discount factor into an effective annual interest rate
  • Use monotonicity and model assumptions to check whether a solution is admissible
  • Apply the geometric-series formula to a whole-life insurance APV under constant mortality

Why Solve for the Interest Rate?

A discrete life insurance APV depends on both mortality and interest. If the mortality assumptions and benefit are known but the APV is given, the interest rate can be the unknown. The central task is to write the present value using the correct death-year probabilities and benefit timing, then solve for the discount factor before converting it to an annual rate.

For a benefit paid at the end of the year of death, a death in the first policy year is discounted for one year, not zero years. Keeping that timing straight is essential: it determines the powers of the discount factor in the equation being solved.

Definition: Let \(i\) be the effective annual interest rate and \(v=1/(1+i)\) the annual discount factor. If a level benefit \(B\) is paid at the end of the policy year of death, then the probability of payment at the end of year \(k+1\) is \({}_kp_x q_{x+k}\). For a term of \(n\) years, the APV is
$$ \operatorname{APV} = B\sum_{k=0}^{n-1} {}_kp_x q_{x+k}v^{k+1}. $$
Here \({}_0p_x=1\). The formula assumes the benefit is paid only if death occurs within the term, and is paid at the end of the year of death. It does not describe a benefit paid at the moment of death or a benefit payable on survival.

From an APV Equation to an Interest Rate

Each death-year probability in the term formula is fixed by the mortality assumptions. Divide the stated APV by the benefit amount to obtain an equation in \(v\). The exponents come from the number of years between issue and payment: the first death-year probability multiplies \(v\), the second multiplies \(v^2\), and so on.

When interest is assumed nonnegative, \(0<v\leq 1\). Once a valid value of \(v\) has been found, convert it to the effective annual rate using

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

If a polynomial equation has multiple real roots, do not accept a root merely because it solves the algebra. Check that it is positive and within the range allowed by the assumed interest-rate domain. Then substitute it back into the original APV equation.

Why a valid rate is usually unique: For nonnegative interest, \(v\) lies in \((0,1]\). Every term in a discrete insurance APV has a nonnegative coefficient and a positive power of \(v\). If at least one benefit-payment probability is positive, increasing \(v\) strictly increases the APV. Since \(v=1/(1+i)\) decreases as \(i\) increases, the APV strictly decreases as the interest rate increases. Thus there can be at most one admissible rate for a stated APV.

This monotonicity also gives a useful range check. For a finite term, the APV at \(i=0\) is \(B\) times the probability of death within the term; as \(i\) grows without bound, the APV approaches zero. Under the nonnegative-interest assumption, a stated APV must therefore be positive and no greater than the undiscounted expected benefit. A value outside that range signals inconsistent data or an incorrect model.

1
Identify the payment timing.
For an end-of-year-of-death benefit, use \(v^{k+1}\) for a death in policy year \(k+1\).
2
Write the payment probabilities.
Use \({}_kp_x q_{x+k}\), the probability of surviving \(k\) years and then dying in the following year.
3
Set up and solve the equation.
Equate the benefit times the discounted probability sum to the stated APV. Solve for \(v\), using a polynomial method or a numerical root method as needed.
4
Check and convert.
Keep only roots allowed by the interest assumption, verify the APV, and calculate \(i=1/v-1\).

Worked Term Insurance Problems

Worked Example: Solving a Two-Year Term APV

A hypothetical two-year term policy pays $10,000 at the end of the year of death. The probabilities of death in policy years 1 and 2 are 0.20 and 0.30, respectively. The stated APV is $3,520. Assuming a nonnegative effective annual interest rate, find the rate.

Method. Form the term APV equation using the given unconditional death-year probabilities. Because the benefit is paid at the end of the year of death, the two discount factors are \(v\) and \(v^2\).

The equation is

$$ 10{,}000(0.20v+0.30v^2)=3{,}520. $$

Divide by $10,000:

$$ 0.30v^2+0.20v-0.352=0. $$

The discriminant is

$$ (0.20)^2-4(0.30)(-0.352) =0.0400+0.4224 =0.4624, \qquad \sqrt{0.4624}=0.68. $$

The quadratic formula gives

$$ v=\frac{-0.20\pm0.68}{0.60}. $$

The two roots are \(0.48/0.60=0.80\) and \(-0.88/0.60=-1.4667\). The negative root is inadmissible because \(v=1/(1+i)\) must be positive. Therefore \(v=0.80\), and

$$ i=\frac{1}{0.80}-1=0.25. $$

Answer and check. The effective annual interest rate is 25%. Substitution gives \(10{,}000[0.20(0.80)+0.30(0.80)^2]=10{,}000(0.160+0.192)=\$3{,}520\), as stated. The common mistake to avoid is assigning the first death-year probability a factor of \(v^0\); its payment occurs one year after issue.

Worked Example: Finding a Rate From a Three-Year Term APV

A hypothetical three-year policy pays $40,000 at the end of the year of death. The unconditional probabilities of death in years 1, 2, and 3 are 0.10, 0.20, and 0.40. The stated APV is $14,250. Find the nonnegative effective annual interest rate.

Method. Divide the APV by the benefit to obtain a cubic equation in \(v\). A numerical root search can locate a root; the domain and monotonicity checks then establish whether it is the admissible solution.

The discount equation is

$$ 0.10v+0.20v^2+0.40v^3 = \frac{14{,}250}{40{,}000} = 0.35625. $$

Define \(f(v)=0.10v+0.20v^2+0.40v^3-0.35625\). To bracket a root, evaluate \(f\) at 0.74 and 0.76:

$$ \begin{aligned} f(0.74) &=0.074+0.10952+0.1620896-0.35625\\ &=-0.0106404,\\ f(0.76) &=0.076+0.11552+0.1755904-0.35625\\ &=0.0108604. \end{aligned} $$

The root lies between 0.74 and 0.76. The midpoint is 0.75; testing it gives

$$ 0.10(0.75)+0.20(0.75)^2+0.40(0.75)^3 =0.075+0.1125+0.16875 =0.35625. $$

Thus \(v=0.75\) is a root. For \(v>0\), the derivative of the left side is \(0.10+0.40v+1.20v^2\), which is positive. The left side is strictly increasing, so there is no second positive root. The admissible root also satisfies \(v\leq1\), as required for nonnegative interest.

Convert to the annual rate:

$$ i=\frac{1}{0.75}-1=\frac{1}{3}=0.3333\ldots. $$

Answer and check. The rate is 33.33%, rounded to two decimal places. Multiplying the verified APV ratio by $40,000 gives \(0.35625(40{,}000)=\$14{,}250\). The common mistake to avoid is accepting an arbitrary real cubic root without checking its sign and the assumed range \(0<v\leq1\).

Whole-Life Insurance Under Constant Mortality

A whole-life insurance APV is an infinite sum over possible death years. Under a constant mortality assumption, let the probability of death in each year conditional on being alive at its start be \(q\), with \(0<q\leq1\). The probability of death in policy year \(k+1\) is then \((1-q)^kq\). For a level benefit \(B\) paid at the end of the year of death,

$$ \operatorname{APV} = B\sum_{k=0}^{\infty}(1-q)^kqv^{k+1}. $$

This is a geometric series with first term \(qv\) and common ratio \((1-q)v\). When \((1-q)v<1\), its sum is

$$ \frac{\operatorname{APV}}{B} = \frac{qv}{1-(1-q)v}. $$

For \(q>0\) and nonnegative interest, \(v\leq1\), so the series converges. This expression can be rearranged into an equation in \(v\), or used directly to solve for \(i\). The constant-mortality assumption matters: if mortality varies by duration, the death-year probabilities do not form this geometric sequence, and this particular shortcut does not apply.

Worked Example: Inferring Interest From a Whole-Life APV

A hypothetical whole-life policy pays $10,000 at the end of the year of death. Assume a constant annual conditional death probability \(q=0.20\), with death eventually occurring with probability 1. The stated APV is $4,000. Find the effective annual interest rate.

Method. Use the constant-mortality geometric-series formula and solve its discount equation. The constant mortality assumption means the probability of death in year \(k+1\) is \(0.20(0.80)^k\).

The APV divided by the benefit is \(4{,}000/10{,}000=0.40\), so

$$ 0.40=\frac{0.20v}{1-0.80v}. $$

The denominator is positive for \(0<v\leq1\), since \(1-0.80v\geq0.20\). Cross-multiplying gives

$$ 0.40(1-0.80v)=0.20v, \qquad 0.40-0.32v=0.20v, \qquad 0.40=0.52v. $$

Therefore \(v=0.40/0.52=10/13\), which is positive and no greater than 1. The rate is

$$ i=\frac{1}{10/13}-1=\frac{13}{10}-1=0.30. $$

Answer and check. The effective annual interest rate is 30%. Substituting \(v=10/13\) into the APV ratio gives \((0.20)(10/13)/[1-(0.80)(10/13)]=(2/13)/(5/13)=0.40\), or $4,000 for a $10,000 benefit. The common mistake to avoid is treating \(q\) as the probability of death in every year without conditioning on survival; the unconditional year-\(k+1\) probability is \((1-q)^kq\).

Common Mistake: Shifting the Payment by One Year

In the first worked example, using \(v^k\) with \(k=0,1\) for the first and second policy years would give

$$ 10{,}000\left[0.20(1)+0.30(0.80)\right] = 10{,}000(0.44) = \$4{,}400. $$

That calculation discounts the first-year death benefit for zero years and the second-year death benefit for only one year. Both payments occur at the end of their policy year, so the correct powers are \(v^1\) and \(v^2\):

$$ 10{,}000\left[0.20(0.80)+0.30(0.80)^2\right] = \$3{,}520. $$

The safest check is to state the actual payment time before writing an exponent: a benefit at the end of policy year \(j\) is discounted by \(v^j\). Also distinguish conditional mortality rates from unconditional death-year probabilities. If the problem gives \(q_x\) and \(q_{x+1}\), for example, the year-2 death probability is \({}_1p_xq_{x+1}\), not \(q_{x+1}\) alone.

Key takeaway: Build the APV with the correct death-year probabilities and end-of-year discount powers, solve for \(v\), and only then calculate \(i=1/v-1\). Check the root against the assumed interest domain and substitute it into the original APV equation.

Check Your Understanding

For each problem, assume the benefit is paid at the end of the year of death and interest is nonnegative. State the effective annual interest rate.

  1. A two-year term benefit of $20,000 has unconditional death-year probabilities 0.15 and 0.25. Its APV is $5,600. (Answer: 25%)
  2. A one-year term benefit of $50,000 has probability of death 0.08 during the year and an APV of $3,200. (Answer: 25%)
  3. A three-year term benefit has unconditional death-year probabilities 0.10, 0.20, and 0.40. Its APV is 0.35625 times the benefit amount. (Answer: 33.33%)
  4. Under constant mortality with annual conditional death probability 0.25, a whole-life policy’s APV is 0.50 times its benefit. (Answer: 25%)
  5. A two-year term benefit of $10,000 has unconditional death-year probabilities 0.30 and 0.40. Its APV is $4,960. (Answer: 25%)
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