The Finite-Term Telescoping Identity

Why Three Finite-Term Values Fit Together

A finite-term life contract can be viewed in three parts: a benefit paid at the end of the year of death, a stream of payments made while the insured is alive, and a benefit paid at the end of the term if the insured survives. These parts refer to different events and payment times. The finite-term telescoping identity shows that their actuarial present values are connected by a simple equation.

The connection is useful for calculating a missing value and for checking a valuation. Its accuracy depends on using consistent benefit amounts, timing, mortality assumptions, and interest. In particular, the adjustment to the annuity value uses the annual discount rate $d$, not the effective annual interest rate $i$.

Definition: Consider a life aged $x$ and an $n$-year term, with integer $n$. Let $A^1_{x:\overline{n}|}$ be the actuarial present value of a unit benefit paid at the end of the year of death if death occurs within $n$ years. Let $\ddot{a}_{x:\overline{n}|}$ be the actuarial present value of a unit annuity-due paying at times $0,1,\ldots,n-1$, provided the life is alive at each payment time. Let ${}_nE_x$ be the actuarial present value of a unit benefit paid at time $n$ if the life survives $n$ years. The finite-term identity is $$ A^1_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}+{}_nE_x=1, $$ where $i$ is the effective annual interest rate, $v=1/(1+i)$, and $d=1-v=i/(1+i)$ is the annual discount rate.

Assumptions and Timing

Use an annual discrete life-contingency model. Write ${}_kp_x$ for the probability that a life aged $x$ survives $k$ years, and ${}_kq_x=1-{}_kp_x$ for the probability of death within $k$ years. In particular, the probability of surviving $k$ years and then dying in the following year is ${}_kp_xq_{x+k}$, where $q_{x+k}$ is the one-year death probability at age $x+k$.

Assume a fixed effective annual interest rate $i$, with discount factor $v$ for each year. The term insurance benefit is paid at the end of the year of death; the annuity-due payments occur at times $0$ through $n-1$ if the life is then alive; and the pure endowment is paid at time $n$ only if the life survives to that time. Each benefit is one unit. These are expected present values, so the identity does not require a particular mortality table or a specific distribution for the future lifetime.

The death benefit timing matters. The formula for $A^1_{x:\overline{n}|}$ above is for payment at the end of the year of death, not immediately at death. Likewise, the annuity value is for payments in advance. Changing either timing changes the present-value expressions and the stated identity need not apply in this form.

Why the Identity Holds

The key is to compare discounted survival from one policy year to the next. For each $k=0,1,\ldots,n-1$, define the discounted probability of being alive at time $k$ as $v^k{}_kp_x$. The difference between this quantity and the corresponding quantity for time $k+1$ accounts for both discounting and death during the year.

Write the next-year survival probability.
To survive $k+1$ years, the life must survive $k$ years and then survive the following year, so ${}_{k+1}p_x={}_kp_xp_{x+k}$, where $p_{x+k}=1-q_{x+k}$.

Take the difference of consecutive discounted survival values.
Since $1-v=d$, the difference separates into a discount-rate term and a death-probability term.

Sum over all $n$ years.
The consecutive terms cancel in pairs, leaving the initial value 1 minus the discounted probability of survival to time $n$.

In detail, for each year $k$,

\begin{aligned} v^k{}_kp_x-v^{k+1}{}_{k+1}p_x &=v^k{}_kp_x(1-vp_{x+k})\\ &=v^k{}_kp_x\bigl((1-v)+vq_{x+k}\bigr)\\ &=d\,v^k{}_kp_x+v^{k+1}{}_kp_xq_{x+k}. \end{aligned}

The last term is the present value contribution from a unit death benefit in year $k+1$, weighted by the probability of surviving to year $k$ and dying in the next year. The first term is $d$ times the present value contribution from the annuity payment at time $k$. Summing the left side over $k=0$ to $n-1$ cancels each intermediate discounted survival value:

\sum_{k=0}^{n-1}\left(v^k{}_kp_x-v^{k+1}{}_{k+1}p_x\right) =1-v^n{}_np_x =1-{}_nE_x.

Summing the right side gives $d\ddot{a}_{x:\overline{n}|}+A^1_{x:\overline{n}|}$. Equating the two sums and rearranging proves the identity. This cancellation is the reason the relationship is called a telescoping identity.

Using the Identity in Valuations

The identity provides a direct check on three values calculated on the same basis. If two values are known, it can also be rearranged to obtain the third. For example, when $i$ is known, the term insurance value can be found from $1-d\ddot{a}_{x:\overline{n}|}-{}_nE_x$. When $d$ is the unknown, it can be isolated as $(1-A^1_{x:\overline{n}|}-{}_nE_x)/\ddot{a}_{x:\overline{n}|}$, provided the annuity value is nonzero.

These rearrangements do not change the assumptions. They are valid only when the insurance, annuity, and endowment factors use the same mortality basis, interest basis, term, and unit benefit. The examples show both a direct valuation and uses of the identity as a calculation and checking tool.

Worked Example: Calculate and Check All Three Values

A hypothetical life aged $x$ has annual survival probabilities $p_x=0.90$, $p_{x+1}=0.80$, and $p_{x+2}=0.75$. The effective annual interest rate is $i=5\%$. For a three-year term, calculate the unit term insurance, temporary annuity-due, and pure endowment values, then check the identity. Assume death benefits are paid at the end of the year of death.

Method. First calculate the survival probabilities to each year, then sum the discounted expected payments using their respective payment times. This avoids assigning the year-$k+1$ death probability to the wrong survival period.

The survival probabilities are ${}_0p_x=1$, ${}_1p_x=0.90$, ${}_2p_x=0.90(0.80)=0.72$, and ${}_3p_x=0.72(0.75)=0.54$. The corresponding death probabilities in years 1, 2, and 3, including survival to the start of each year, are $0.10$, $0.90(0.20)=0.18$, and $0.72(0.25)=0.18$.

The interest quantities are $v=1/1.05=0.952380952$ and $d=1-v=0.047619048$. The term insurance value is

\begin{aligned} A^1_{x:\overline{3}|} &=0.10v+0.18v^2+0.18v^3\\ &=0.10(0.952380952)+0.18(0.907029478)+0.18(0.863837598)\\ &=0.095238095+0.163265306+0.155490768\\ &=0.413994169. \end{aligned}

The annuity-due payments are made at times 0, 1, and 2, conditional on being alive at each payment time:

\begin{aligned} \ddot{a}_{x:\overline{3}|} &=1+0.90v+0.72v^2\\ &=1+0.857142857+0.653061224\\ &=2.510204081. \end{aligned}

The pure endowment is paid at time 3 only if the life survives to that time:

{}_3E_x=0.54v^3=0.54(0.863837598)=0.466472303.

Now $d\ddot{a}_{x:\overline{3}|}=0.047619048(2.510204081)=0.119533527$. Therefore the sum is $0.413994169+0.119533527+0.466472303=0.999999999$, which equals 1 apart from rounding.

Check. The identity holds to the displayed precision. The tiny difference from 1 is due to rounding intermediate values, not a discrepancy in the valuation.

Worked Example: Use the Identity to Recover a Term Value

For a hypothetical two-year policy, the effective annual interest rate is $4\%$, with $p_x=0.75$ and $p_{x+1}=0.80$. Calculate the term insurance value using the identity and verify it by valuing the death benefit directly.

Method. Calculate the temporary annuity-due and pure endowment first, then use the identity to obtain the term insurance value. A direct death-benefit calculation provides an independent check. The mortality assumption is annual conditional survival as specified; the death benefit is paid at the end of the year of death.

The probability of surviving two years is ${}_2p_x=0.75(0.80)=0.60$. The interest quantities are $v=1/1.04=0.961538462$ and $d=1-v=0.038461538$. Thus,

\ddot{a}_{x:\overline{2}|}=1+0.75v =1+0.721153846 =1.721153846, \qquad {}_2E_x=0.60v^2 =0.60(0.924556213) =0.554733728.

Applying the identity gives

\begin{aligned} A^1_{x:\overline{2}|} &=1-d\ddot{a}_{x:\overline{2}|}-{}_2E_x\\ &=1-(0.038461538)(1.721153846)-0.554733728\\ &=1-0.066198225-0.554733728\\ &=0.379068047. \end{aligned}

For the direct check, the probability of death in year 1 is $1-0.75=0.25$. The probability of surviving year 1 and dying in year 2 is $0.75(1-0.80)=0.15$. Hence

A^1_{x:\overline{2}|} =0.25v+0.15v^2 =0.25(0.961538462)+0.15(0.924556213) =0.240384616+0.138683432 =0.379068048.

The two results agree to the displayed precision. The difference in the final decimal is rounding. The direct calculation confirms that the second-year death probability must include survival through the first year.

Worked Example: Scale the Identity to a Dollar Benefit

A hypothetical two-year contract pays a benefit of $40,000 at the end of the year of death. It also has an annuity-due benefit of $40,000 per scheduled payment and a $40,000 pure endowment benefit. Assume a constant annual survival probability of $0.625$ in each year and an effective annual interest rate of $25\%$. Calculate the present values of all three benefits and check the scaled identity.

Method. Calculate the unit factors first, then multiply each by $40,000. The model uses independent annual survival probabilities conditional on being alive at the start of each year; equivalently, the one-year survival probability at each duration is $0.625$. The stated interest rate gives $v=1/1.25=0.8$ and $d=1-v=0.2$.

The survival probabilities are ${}_1p_x=0.625$ and ${}_2p_x=0.625^2=0.390625$. The one-year death probability is $0.375$. The unit term insurance value is

A^1_{x:\overline{2}|} =0.375(0.8)+(0.625)(0.375)(0.8^2) =0.300000+0.150000 =0.450000.

The unit temporary annuity-due value is $1+0.625(0.8)=1.5$, and the unit pure endowment value is $0.390625(0.8^2)=0.25$. The identity check is

$$ 0.450000+0.2(1.500000)+0.250000 =0.450000+0.300000+0.250000 =1.000000. $$

Multiplying each factor by $40,000 gives a term insurance value of $18,000.00, an annuity-due value of $60,000.00, and a pure endowment value of $10,000.00. The discount-rate adjustment to the annuity is $0.2($60,000.00)=$12,000.00$, so $18,000.00 + $12,000.00 + $10,000.00 = $40,000.00. The dollar amounts reconcile exactly.

Common Mistakes and Scope

A frequent error is to multiply the temporary annuity-due value by $i$ instead of $d$. The identity uses $d=1-v=i/(1+i)$, because the year-by-year telescoping difference contains $1-v$. In the last example, using $i=0.25$ would give $0.45+0.25(1.5)+0.25=1.075$, which is not 1. The correction is to use $d=0.20$, giving $0.45+0.20(1.5)+0.25=1$.

Other errors usually come from inconsistent timing or indexing. The annuity has a payment at time 0, while the pure endowment is paid at time $n$. The term insurance death probability for year $k+1$ is ${}_kp_xq_{x+k}$, and its payment is discounted $k+1$ years. Omitting survival to the start of the year, or discounting that benefit only $k$ years, breaks the telescoping calculation.

The identity is a relationship among specified expected present values, not a statement that every combination of an insurance and an annuity has these values. Before applying it, confirm that the term is the same for all three factors, each benefit amount is the same unit (or that the values have been scaled consistently), and the benefit and payment timing match the definitions. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: With end-of-year death benefits, an annuity-due, a pure endowment, and a common annual interest and mortality basis, the finite-term values satisfy $A^1_{x:\overline{n}|}+d\ddot{a}_{x:\overline{n}|}+{}_nE_x=1$. The discount-rate factor is $d=1-v$, and the identity follows by summing consecutive differences in discounted survival probabilities.

Check Your Understanding

For each problem, use the stated payment timing and keep the mortality and interest assumptions consistent. Answers are rounded only where necessary.

A hypothetical two-year contract has $i=5\%$, $p_x=0.90$, and $p_{x+1}=0.80$. Calculate $A^1_{x:\overline{2}|}$, $\ddot{a}_{x:\overline{2}|}$, and ${}_2E_x$, then verify the identity. (Answer: $A^1_{x:\overline{2}|}=0.258503$, $\ddot{a}_{x:\overline{2}|}=1.857143$, ${}_2E_x=0.653061$, and $A^1+d\ddot{a}+{}_2E=1.000000$.)
For a two-year term, $i=4\%$, $p_x=0.75$, and $p_{x+1}=0.80$. Calculate the term insurance value both from the identity and by summing the two possible year-of-death contributions. (Answer: $A^1_{x:\overline{2}|}=0.379068$ by either method.)
A valuation has $i=6\%$, $A^1_{x:\overline{n}|}=0.28$, and ${}_nE_x=0.52$, all on a common basis with a unit benefit. Find $\ddot{a}_{x:\overline{n}|}$. (Answer: $d=0.06/1.06$, so $\ddot{a}_{x:\overline{n}|}=(1-0.28-0.52)/d=3.5333$.)
On a common finite-term basis, $A^1_{x:\overline{n}|}=0.41$, $\ddot{a}_{x:\overline{n}|}=2.4$, and ${}_nE_x=0.46$. Use the identity to find the implied effective annual interest rate. (Answer: $d=(1-0.41-0.46)/2.4=0.0541667$, so $i=d/(1-d)=0.05727$, or $5.727\%$.)

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