Actuarial Notes

Term Life Insurance

Study guides: Exam FAM, Exam DISC-IA

Term Life Insurance pays a sum insured if the insured dies within a fixed term of n years, and nothing if the insured survives the term. For a life aged x paid at the end of the year of death, its present value random variable is Z and its expected present value (EPV), the net single premium, is A^{1}_{x:\overline{n}|}, where the 1 over x shows that the benefit is paid only if (x) dies before the term ends. It is the simplest form of life insurance, with no savings element.

Z = \begin{cases} v^{K_x+1} & K_x \le n-1 \\ 0 & K_x \ge n \end{cases}

A^{1}_{x:\overline{n}|} = \sum_{k=0}^{n-1} v^{k+1}\,{}_kp_x\,q_{x+k}

\bar{A}^{1}_{x:\overline{n}|} = \int_0^n e^{-\delta t}\,{}_tp_x\,\mu_{x+t}\,dt

\text{Var}(Z) = {}^2A^{1}_{x:\overline{n}|} - \left(A^{1}_{x:\overline{n}|}\right)^2

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

  • Symbols: K_x is the curtate future lifetime, and {}^2A^{1}_{x:\overline{n}|} is the EPV at v^2, which is the second moment of Z because Z^2 = (v^2)^{K_x+1} on death within the term. \bar{A}^{1}_{x:\overline{n}|} values a benefit paid at the moment of death, e^{-\delta T_x} when T_x < n. Under UDD, \bar{A}^{1}_{x:\overline{n}|} = \frac{i}{\delta}A^{1}_{x:\overline{n}|}.
  • From whole life values: The last block splits a whole life insurance into term cover for n years and a whole life insurance deferred n years, where {}_nE_x = v^n\,{}_np_x is the pure endowment factor. For the second moment the deferral factor is v^{2n}\,{}_np_x, so {}^2A^{1}_{x:\overline{n}|} = {}^2A_x - v^n\,{}_nE_x\,{}^2A_{x+n}. The recursion A^{1}_{x:\overline{n}|} = v\,q_x + v\,p_x\,A^{1}_{x+1:\overline{n-1}|} gives the same values one year at a time.
  • Level and decreasing term: A level term insurance pays the same sum throughout. A decreasing term insurance reduces the sum each year, often to match the balance of a loan, and its EPV is \sum_{k=0}^{n-1} b_{k+1}\,v^{k+1}\,{}_kp_x\,q_{x+k} for a benefit b_{k+1} on death in year k+1.
  • The product: Because nothing is paid on survival, term insurance is the cheapest way to buy a large death benefit for a fixed period. Level-premium term policies are often renewable at the end of the term or convertible to permanent insurance without new evidence of health, and these options expose the insurer to adverse selection. A renewed term policy charges a higher premium, and the right to renew may end at a stated age. A nonrenewable policy cannot be continued, so the insured must apply, and qualify, for a new one. Most term policies build no cash value.

Ten-Year Term on the SULT On the SULT at 5\%, A_{50} = 0.18931, A_{60} = 0.29028, {}^2A_{50} = 0.05108, {}^2A_{60} = 0.10834, {}_{10}E_{50} = 0.60182, \ddot{a}_{50} = 17.0245 and \ddot{a}_{60} = 14.9041. For a 10-year term insurance of $100,000 on (50), paid at the end of the year of death, find the EPV, the standard deviation of the present value, and the net annual premium payable for 10 years.

Answer Here v^{10} = 0.613913, so the second-moment deferral factor is 0.613913(0.60182) = 0.369465.

\begin{align*}
A^{1}_{50:\overline{10}|} &= 0.18931 - 0.60182(0.29028) \\
&= 0.014614 \\
{}^2A^{1}_{50:\overline{10}|} &= 0.05108 - 0.369465(0.10834) \\
&= 0.011052 \\
\text{Var}(Z) &= 0.011052 - 0.014614^2 \\
&= 0.010838 \\
\text{SD} &= 100{,}000\sqrt{0.010838} \\
&= 10{,}411 \\
\ddot{a}_{50:\overline{10}|} &= 17.0245 - 0.60182(14.9041) \\
&= 8.05491 \\
P &= \frac{1{,}461.4}{8.05491} \\
&= 181.43
\end{align*}

The EPV is $1,461.40, but the standard deviation is about seven times as large. Almost every policy pays nothing, and the few that pay cost $100,000.

Decreasing Term Insurance A 3-year decreasing term insurance on (60) pays $150,000, $100,000 or $50,000 at the end of the year of death in years 1, 2 or 3. On the SULT, q_{60} = 0.003398, q_{61} = 0.003792, q_{62} = 0.004234 and i = 5\%. Find the EPV.

Answer The probabilities of death in each year are q_{60} = 0.003398, 0.996602(0.003792) = 0.003779 and 0.996602(0.996208)(0.004234) = 0.004204.

\begin{align*}
\text{EPV} &= \frac{150{,}000(0.003398)}{1.05} + \frac{100{,}000(0.003779)}{1.05^2} \\
&\quad + \frac{50{,}000(0.004204)}{1.05^3} \\
&= 485.43 + 342.77 + 181.58 \\
&= 1{,}009.78
\end{align*}

A level benefit of $150,000 for three years would cost $1,544.31. The decreasing cover costs about a third less, because the sum falls just as the probability of death rises.

Continuous Term Insurance under a Constant Force A life has a constant force of mortality \mu = 0.01, and \delta = 0.04. A 20-year term insurance pays 1 at the moment of death. Find its EPV and the standard deviation of its present value.

Answer Under a constant force, \bar{A}^{1}_{x:\overline{n}|} = \frac{\mu}{\mu+\delta}\left(1 - e^{-(\mu+\delta)n}\right), and the second moment replaces \delta by 2\delta.

\begin{align*}
\bar{A}^{1}_{x:\overline{20}|} &= 0.2\left(1 - e^{-1}\right) \\
&= 0.126424 \\
{}^2\bar{A}^{1}_{x:\overline{20}|} &= \tfrac{1}{9}\left(1 - e^{-1.8}\right) \\
&= 0.092745 \\
\text{Var}(\bar{Z}) &= 0.092745 - 0.126424^2 \\
&= 0.076762
\end{align*}

The standard deviation is \sqrt{0.076762} = 0.27706, more than twice the EPV.

How it appears on the exam

  • FAM asks candidates to identify the present value random variable of a term insurance paid annually, at the end of the 1/m-th of a year of death, or at the moment of death, and to find its EPV, second moment, variance or a probability such as \Pr(Z > A^{1}_{x:\overline{n}|}).
  • SULT questions: A FAM question can require A^{1}_{x:\overline{n}|} to be built from whole life and pure endowment values, as in the first example, because the SULT tabulates A_x, {}^2A_x and {}_nE_x rather than term insurance values.
  • Premiums and losses: Term life is one of the policies in FAM's premium section, which asks for its net or gross premium and its future loss random variable.
  • DISC-IA lists term life insurance under its Life and Health Insurance section beside whole life and universal life, which is the product view: what the contract pays and how it differs from permanent insurance.

Exam traps

  • Squaring the EPV: The second moment is {}^2A^{1}_{x:\overline{n}|} at v^2, not \left(A^{1}_{x:\overline{n}|}\right)^2. The difference between the two is the variance.
  • The deferral factor at v^2: In {}^2A^{1}_{x:\overline{n}|} = {}^2A_x - v^n\,{}_nE_x\,{}^2A_{x+n}, the factor is v^{2n}\,{}_np_x. Using {}_nE_x alone in the first example gives 0.05108 - 0.60182(0.10834) < 0, which is impossible for a second moment.
  • Term versus endowment: A term insurance pays nothing on survival. Adding {}_nE_x to A^{1}_{x:\overline{n}|} gives the endowment insurance, a different and much dearer contract.
  • The premium annuity: Premiums for an n-year term are payable for at most n years, so the premium uses \ddot{a}_{x:\overline{n}|}. Dividing by \ddot{a}_x assumes premiums continue after the cover has ended.