The Continuous Term Insurance-Annuity Identity

Why the Three Values Belong Together

Over a fixed term, a life aged $x$ has two mutually exclusive outcomes: death before the term ends, or survival to its end. A continuous term insurance pays on the first outcome; a term pure endowment pays on the second. A continuous temporary life annuity accumulates value while the person is alive during the term. The identity in this tutorial links the present values of these three benefits.

The central point is that the annuity value by itself is not the complement of the term insurance value. The annuity’s stream of payments must first be multiplied by the force of interest. With that adjustment, the three present values add to one for unit benefits and a unit annuity payment rate.

Definition: Fix a term of $n$ years and a life aged $x$. Let $\bar{A}^{1}_{x:\overline{n}|}$ be the expected present value of a unit benefit paid at the instant of death if death occurs within $n$ years. Let $\bar{a}_{x:\overline{n}|}$ be the expected present value of a continuous annuity paying at rate one per year while the life is alive during those $n$ years. Let ${}_{n}E_x$ be the expected present value of a unit benefit paid at time $n$ if the life survives to time $n$.

Use a constant force of interest $\delta$, so that a payment at time $t$ has discount factor $e^{-\delta t}$. Write ${}_tp_x$ for the probability that a life aged $x$ survives at least $t$ years, and $\mu_{x+t}$ for the force of mortality at time $t$, conditional on survival to that time. Assume a continuous lifetime distribution, with a suitably integrable force of mortality and no probability mass at the term boundary.

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

The insurance integral uses the density of death at time $t$, which is ${}_tp_x\mu_{x+t}$ under this continuous model. The annuity integral weights each instant’s payment rate by the probability the life is then alive. The pure endowment includes both survival to the end and discounting back to time zero.

Deriving the Identity

The identity follows by considering the discounted probability of being alive at time $t$. Define $g(t)=e^{-\delta t}{}_tp_x$. The survival probability satisfies $\frac{d}{dt}{}_tp_x=-{}_tp_x\mu_{x+t}$ wherever the derivative exists. Applying the product rule gives the change in $g(t)$.

g'(t)=-(\delta+\mu_{x+t})e^{-\delta t}{}_tp_x.

Integrate this equation from zero to $n$. Since $g(0)=1$ and $g(n)={}_{n}E_x$, the resulting difference is the combined value of the death benefit and the interest-adjusted annuity payments.

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

Theorem: For a continuous lifetime model and a constant force of interest $\delta$, the unit-benefit values over the same term satisfy $$ \bar{A}^{1}_{x:\overline{n}|}+\delta\bar{a}_{x:\overline{n}|}+{}_{n}E_x=1. $$

The argument also explains the economic structure of the result. At each instant while the life is alive, the discounted survival value declines through two mechanisms: interest, represented by $\delta$, and mortality, represented by $\mu_{x+t}$. The annuity term accounts for the interest part of that decline; the insurance term accounts for the mortality part. The discounted survival value remaining at time $n$ is the pure endowment.

Worked Calculations

With a constant force of mortality $\mu$, survival is ${}_tp_x=e^{-\mu t}$. For $\delta+\mu>0$, the three values can be evaluated directly:

\bar{A}^{1}_{x:\overline{n}|} =\frac{\mu}{\delta+\mu}\left(1-e^{-(\delta+\mu)n}\right),\qquad \bar{a}_{x:\overline{n}|} =\frac{1-e^{-(\delta+\mu)n}}{\delta+\mu},\qquad {}_{n}E_x=e^{-(\delta+\mu)n}.

These formulas are specific to constant forces over the term. When mortality varies with time, use the defining integrals or divide the term into intervals on which the forces are constant.

Worked Example: All Three Components Over Two Years

A hypothetical policy covers a life aged $x$ for $n=2$ years. The force of interest is $\delta=\ln 2$ per year, corresponding to effective annual interest $i=e^\delta-1=1$, or 100%. The force of mortality is constant at $\mu=\ln 2$ per year, so ${}_tp_x=e^{-(\ln 2)t}$. Calculate the term insurance, temporary annuity, and pure endowment values, then verify the identity.

Method. Apply the constant-force formulas. The combined force is $\delta+\mu=2\ln 2$, and over two years the exponential endpoint factor is $e^{-(2\ln 2)(2)}=e^{-4\ln 2}=1/16$.

For insurance, the mortality share of the combined force is $\mu/(\delta+\mu)=1/2$. Therefore,

\bar{A}^{1}_{x:\overline{2}|} =\frac{1}{2}\left(1-\frac{1}{16}\right) =\frac{1}{2}\cdot\frac{15}{16} =\frac{15}{32}.

For the annuity, first calculate its value and then apply the required interest adjustment:

\bar{a}_{x:\overline{2}|} =\frac{15/16}{2\ln 2} =\frac{15}{32\ln 2}, \qquad \delta\bar{a}_{x:\overline{2}|} =(\ln 2)\frac{15}{32\ln 2} =\frac{15}{32}.

The pure endowment value is ${}_{2}E_x=1/16$. Adding the three contributions gives $15/32+15/32+1/16=30/32+2/32=1$. The calculation checks exactly. The common mistake this method avoids is adding the unadjusted annuity value $15/(32\ln 2)$ to the other components.

Worked Example: Different Constant Forces

For a hypothetical one-year contract, let the force of interest be $\delta=\ln 2$ per year, so $i=100\%$ effective annually. Assume constant mortality force $\mu=\ln 4$ per year, giving ${}_tp_x=e^{-(\ln 4)t}$. Find each component and verify the identity.

Method. The combined force is $\ln 2+\ln 4=\ln 8=3\ln 2$. The endpoint factor is $e^{-\ln 8}=1/8$, so the common difference from one is $1-1/8=7/8$.

The mortality share of the combined force is $\ln 4/\ln 8=2/3$. Thus the insurance value is

\bar{A}^{1}_{x:\overline{1}|} =\frac{2}{3}\cdot\frac{7}{8} =\frac{7}{12}.

The annuity value and its interest-adjusted contribution are

\bar{a}_{x:\overline{1}|} =\frac{7/8}{3\ln 2} =\frac{7}{24\ln 2}, \qquad \delta\bar{a}_{x:\overline{1}|} =(\ln 2)\frac{7}{24\ln 2} =\frac{7}{24}.

The pure endowment is ${}_{1}E_x=1/8$. Verification gives $7/12+7/24+1/8=14/24+7/24+3/24=1$. The common mistake avoided here is using $\mu/(\delta+\mu)$ as the annuity factor: the annuity is obtained from the interest force, not the mortality force.

Worked Example: Mortality Force Changes During the Term

Consider a hypothetical two-year contract with $\delta=\ln 2$ per year, hence $i=100\%$ effective annually. The force of mortality is zero in the first year and $\ln 4$ per year in the second year: $\mu_{x+t}=0$ for $0\le t<1$ and $\mu_{x+t}=\ln 4$ for $1\le t\le2$. Calculate the three values.

Method. Use the integrals, splitting them at $t=1$. Survival is one through the first year and is $e^{-(\ln 4)(t-1)}$ during the second year. At $t=2$, survival is $1/4$, while the discount factor is $e^{-2\ln 2}=1/4$. Hence ${}_{2}E_x=1/16$.

There is no death benefit in the first year because the mortality force is zero. In the second year, set $s=t-1$. The discounted survival factor becomes $\frac12 e^{-(\ln 8)s}$. Therefore,

\bar{A}^{1}_{x:\overline{2}|} =(\ln 4)\int_0^1\frac12e^{-(\ln 8)s}\,ds =(2\ln 2)\frac12\frac{1-1/8}{3\ln 2} =\frac{7}{24}.

For the annuity, the first-year integral is $\int_0^1 e^{-(\ln 2)t}\,dt=1/(2\ln 2)$. The second-year integral is $\frac12(1-1/8)/(3\ln 2)=7/(48\ln 2)$. Thus

\bar{a}_{x:\overline{2}|} =\frac{1}{2\ln 2}+\frac{7}{48\ln 2} =\frac{31}{48\ln 2}, \qquad \delta\bar{a}_{x:\overline{2}|}=\frac{31}{48}.

The check is $7/24+31/48+1/16=14/48+31/48+3/48=1$. The common mistake avoided is applying the second-year mortality force to the first year as well. The integral for insurance only includes death risk where the force of mortality is positive.

Interpretation, Scope, and Common Traps

The identity is useful as a consistency check and as a way to find one value when the other two are known. Its terms must refer to the same life, term, interest basis, and payment timing. The insurance benefit is paid at the instant of death within the term; the annuity is paid continuously only while alive within the term; and the pure endowment is paid only at the term’s end, conditional on survival. Changing one of these conditions changes the values and may invalidate the identity as stated.

Forgetting the force of interest. The correct relation uses $\delta\bar{a}_{x:\overline{n}|}$, not $\bar{a}_{x:\overline{n}|}$ alone.
Mixing benefit timing. A benefit payable at the end of the year of death is not the continuous term insurance defined here.
Omitting the survival condition. The pure endowment is discounted and conditional on surviving the term; it is not just the discount factor.
Using inconsistent terms. An annuity extending beyond $n$ cannot be paired with insurance that ends at $n$ in this identity.

The constant-force formulas are convenient computational shortcuts, not assumptions required by the identity itself. The derivation applies with a time-varying mortality force, as long as the continuous survival model and regularity conditions hold. If the benefit amounts differ from one, each expected present value must reflect its own benefit amount before a complement calculation can be used.

Key takeaway: Over a common term, the discounted survival value falls through interest and mortality. The continuous annuity accounts for the interest component, continuous term insurance accounts for the mortality component, and the term pure endowment is the discounted survival value left at the endpoint. This identity is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the appropriate term values and use the same timing conventions throughout.

A hypothetical one-year contract has $\delta=\ln 2$ and constant $\mu=\ln 2$, both per year. Find the continuous term insurance value, the interest-adjusted temporary annuity value, and the pure endowment value. (Answer: $3/8,\ 3/8,\ 1/4$, respectively.)
A two-year contract has $\delta=\ln 2$ and constant $\mu=\ln 4$, both per year. Find the pure endowment value and the interest-adjusted annuity value, then verify the insurance value. (Answer: pure endowment $1/64$, interest-adjusted annuity $21/64$, insurance $21/32$.)
For a one-year term, $\delta=0.05$, the continuous term insurance value is $0.42$, and the temporary annuity value is $8$. Find the term pure endowment value. (Answer: $0.18$.)
For a two-year term, $\delta=0.04$, the temporary annuity value is $6.5$, and the pure endowment value is $0.30$. Find the continuous term insurance value. (Answer: $0.44$.)
A hypothetical two-year contract has $\delta=\ln 2$, zero mortality force in year one, and constant mortality force $\ln 4$ in year two. Find the pure endowment and the interest-adjusted annuity values, then use the identity to recover the insurance value. (Answer: pure endowment $1/16$, interest-adjusted annuity $31/48$, insurance $7/24$.)

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