Actuarial Notes

Endowment Insurance

Study guide: Exam FAM

Endowment Insurance pays a sum insured on the death of the insured within a term of n years, or at the end of the term if the insured survives it, so the benefit is certain to be paid by time n. For a life aged x with the death benefit paid at the end of the year of death, its present value random variable is Z = v^{\min(K_x+1,\,n)} and its expected present value (EPV) is A_{x:\overline{n}|}. It combines a term insurance with a pure endowment and behaves mainly as a savings contract.

Z = v^{\min(K_x+1,\,n)}

A_{x:\overline{n}|} = A^{1}_{x:\overline{n}|} + {}_nE_x

\text{Var}(Z) = {}^2A_{x:\overline{n}|} - \left(A_{x:\overline{n}|}\right)^2

\ddot{a}_{x:\overline{n}|} = \frac{1 - A_{x:\overline{n}|}}{d}

  • Symbols: K_x is the curtate future lifetime, and {}^2A_{x:\overline{n}|} is the EPV at v^2, which equals {}^2A^{1}_{x:\overline{n}|} + v^{2n}\,{}_np_x. When the death benefit is paid at the moment of death, \bar{Z} = v^{\min(T_x,\,n)} and \bar{a}_{x:\overline{n}|} = (1 - \bar{A}_{x:\overline{n}|})/\delta.
  • Two parts: Z is the sum of the term insurance and pure endowment present values, Z_1 + Z_2. They can never both be positive, so E[Z_1 Z_2] = 0 and \text{Cov}(Z_1, Z_2) = -A^{1}_{x:\overline{n}|}\,{}_nE_x. The second moments therefore add, but the variances do not.
  • The annuity identity: The present value of an n-year temporary life annuity-due is Y = \ddot{a}_{\overline{\min(K_x+1,\,n)}|} = (1 - Z)/d. Taking expectations gives the fourth block, and \text{Var}(Y) = \text{Var}(Z)/d^2. It follows that the net annual premium per unit is A_{x:\overline{n}|}/\ddot{a}_{x:\overline{n}|} = d\,A_{x:\overline{n}|}/(1 - A_{x:\overline{n}|}).
  • A savings contract: Because the benefit is paid by time n at the latest, A_{x:\overline{n}|} \ge v^n, and for a young life most of the value is the maturity payment. Its value depends mainly on the interest rate, and under UDD only the term part is converted to continuous, \bar{A}_{x:\overline{n}|} = \frac{i}{\delta}A^{1}_{x:\overline{n}|} + {}_nE_x.

Twenty-Year Endowment Insurance on the SULT On the SULT at 5\%, A_{45} = 0.15161, A_{65} = 0.35477, {}_{20}E_{45} = 0.35994 and \ddot{a}_{45} = 17.8162. For a fully discrete 20-year endowment insurance of $100,000 on (45), find A_{45:\overline{20}|}, \ddot{a}_{45:\overline{20}|} and the net annual premium. Compare the premium with that for a fully discrete whole life insurance of the same amount.

Answer The term part is the whole life value less a whole life insurance deferred 20 years.

\begin{align*}
A^{1}_{45:\overline{20}|} &= 0.15161 - 0.35994(0.35477) \\
&= 0.023914 \\
A_{45:\overline{20}|} &= 0.023914 + 0.35994 \\
&= 0.383854 \\
\ddot{a}_{45:\overline{20}|} &= \frac{1 - 0.383854}{0.05/1.05} \\
&= 12.9391 \\
P &= \frac{38{,}385.4}{12.9391} \\
&= 2{,}966.62
\end{align*}

The whole life premium is 15{,}161/17.8162 = 850.97. The endowment costs three and a half times as much, because it pays $100,000 by age 65 for certain, and 94\% of its EPV is the pure endowment.

Paying the Death Benefit at the Moment of Death For the policy in the first example, assume UDD and find \bar{A}_{45:\overline{20}|}, with the death benefit paid at the moment of death. Then find \bar{a}_{45:\overline{20}|} with \delta = \ln 1.05 = 0.048790.

Answer Only the death benefit is paid within the year, so only the term part is multiplied by i/\delta = 1.024797.

\begin{align*}
\bar{A}_{45:\overline{20}|} &= 1.024797(0.023914) + 0.35994 \\
&= 0.024507 + 0.35994 \\
&= 0.384447 \\
\bar{a}_{45:\overline{20}|} &= \frac{1 - 0.384447}{0.048790} \\
&= 12.6164
\end{align*}

Paying claims sooner adds less than 0.2\% to the value, because the death benefit is a small part of it. The continuous annuity is smaller than \ddot{a}_{45:\overline{20}|} = 12.9391, because each payment is spread over its year instead of being paid at the start.

How it appears on the exam

  • FAM asks candidates to identify the present value random variable of an endowment insurance paid annually, 1/m-thly or continuously, and to find its EPV and variance, either from second moments or from the term and pure endowment parts with their covariance.
  • Relationships: FAM's objective on relationships between insurance, endowment and annuity values is built on \ddot{a}_{x:\overline{n}|} = (1 - A_{x:\overline{n}|})/d, which questions use in both directions.
  • Premiums and losses: Endowment insurance is one of the policies in FAM's premium section. Its future loss random variable is a linear function of Z, so its variance has the same shortcut as whole life.

Exam traps

  • Adding the variances of the parts: The variance of Z is not \text{Var}(Z_1) + \text{Var}(Z_2), because the covariance -A^{1}_{x:\overline{n}|}\,{}_nE_x is negative. Leaving it out overstates the variance.
  • The annuity from the wrong insurance: The identity \ddot{a}_{x:\overline{n}|} = (1 - A)/d needs the endowment insurance value. Using the term value A^{1}_{x:\overline{n}|} in the first example gives 20.50, which is impossible for a 20-year annuity-due at 5\%.
  • UDD on the whole value: Multiplying all of A_{x:\overline{n}|} by i/\delta treats the maturity payment as if it were paid early, and in the second example it overstates \bar{A} by 0.0089.
  • Death in the last year: A death in year n is paid at time n, the same time as the maturity benefit. So Z = v^n with probability {}_{n-1}p_x, not {}_np_x.