Decomposing Whole-Life Annuities into Temporary Terms

Why Split a Whole-Life Annuity?

A whole-life annuity has payments extending for as long as the life survives. For valuation, it is often helpful to separate its cash flows into two time ranges: an initial block of payments and everything after that block. This gives a direct way to check an annuity value, isolate the value of its later payments, or compare values at different durations.

The split depends on payment timing. An annuity-due makes a payment at time 0, while an annuity-immediate has its first possible payment at time 1. Therefore, when the split is made after $n$ years, the temporary part contains different payment dates for the two annuity types. The tail begins at time $n$ for a due annuity and at time $n+1$ for an immediate annuity.

Definition: For a life aged $x$, let ${}_kp_x$ be the probability that the life survives at least $k$ years, with ${}_0p_x=1$. Let $v=1/(1+i)$, where $i$ is the effective annual interest rate. A discrete life annuity pays only at the stated integer times and only if the life is then alive. Its expected present value weights each payment by both its discount factor and its probability of being made.

Whole-Life and Temporary Values

For level payments of 1 per year, a whole-life annuity-due pays at times $0,1,2,\ldots$, provided the life is alive at each payment time. A whole-life annuity-immediate pays at times $1,2,3,\ldots$, provided the life is alive at each payment time. The corresponding expected present values are

\ddot{a}_x=\sum_{k=0}^{\infty}v^k\,{}_kp_x, \qquad a_x=\sum_{k=1}^{\infty}v^k\,{}_kp_x.

The temporary annuity-due lasting $n$ years contains payments at times $0$ through $n-1$. The temporary annuity-immediate lasting $n$ years contains payments at times $1$ through $n$. In each case, a payment is made only if the life is alive at its payment time:

\ddot{a}_{x:\overline{n}|}=\sum_{k=0}^{n-1}v^k\,{}_kp_x, \qquad a_{x:\overline{n}|}=\sum_{k=1}^{n}v^k\,{}_kp_x.

Here $n$ is a positive integer. These sums describe a finite initial block of a whole-life annuity; they do not assume that payments in that block are guaranteed.

The Decomposition

Theorem: For integer $n\geq 1$, a whole-life annuity-due can be divided into an $n$-year temporary annuity-due and the value of the payments from time $n$ onward: $$ \ddot{a}_x=\ddot{a}_{x:\overline{n}|}+v^n\,{}_np_x\,\ddot{a}_{x+n}. $$ A whole-life annuity-immediate can be divided into an $n$-year temporary annuity-immediate and the value of payments from time $n+1$ onward: $$ a_x=a_{x:\overline{n}|}+v^n\,{}_np_x\,a_{x+n}. $$

The term $v^n{}_np_x$ discounts the value at time $n$ back to time 0 and accounts for the probability that the life survives to time $n$. The annuity value at age $x+n$ is conditional on being alive at that time. For the due annuity, its payment at time 0 relative to age $x+n$ is at calendar time $n$; for the immediate annuity, its first payment relative to age $x+n$ is one year later, at calendar time $n+1$.

Choose the split time.
Fix an integer duration $n$. Put the due payments at times $0,\ldots,n-1$, or the immediate payments at times $1,\ldots,n$, in the temporary part.

Locate the first tail payment.
It is at time $n$ for the due annuity and at time $n+1$ for the immediate annuity. The tail is valued as an annuity beginning at age $x+n$.

Move the tail value to time 0.
Multiply its value at time $n$ by both $v^n$ and ${}_np_x$, then add the temporary value.

To justify the due formula, split its infinite sum after the term at $k=n-1$. In the remaining sum, write $k=n+j$. The multiplication rule for conditional survival probabilities gives ${}_{n+j}p_x={}_np_x\,{}_jp_{x+n}$. Thus:

\begin{aligned} \ddot{a}_x &=\sum_{k=0}^{n-1}v^k\,{}_kp_x+\sum_{j=0}^{\infty}v^{n+j}\,{}_{n+j}p_x\\ &=\ddot{a}_{x:\overline{n}|} +v^n\,{}_np_x\sum_{j=0}^{\infty}v^j\,{}_jp_{x+n}\\ &=\ddot{a}_{x:\overline{n}|}+v^n\,{}_np_x\,\ddot{a}_{x+n}. \end{aligned}

For the immediate formula, split the sum after $k=n$; the remaining terms have $k=n+j$ for $j=1,2,\ldots$. The same survival factorization leaves $a_{x+n}$, not $\ddot{a}_{x+n}$, because there is no payment at $j=0$. These decompositions require the stated discrete payment timing and conditional survival probabilities. They do not turn the tail into payments that are certain to occur.

Worked Example: Decomposing a Whole-Life Annuity-Due

A hypothetical annuity pays 1 annually in advance while its recipient is alive. The effective annual interest rate is $i=25\%$. Assume the conditional probability of surviving each year is $p=0.8$ at every age. Split the value after two years and check the result. Method: use the geometric survival model. This avoids treating the time-2 tail as though it were already payable at time 1.

First, $v=1/(1.25)=0.8$, so the discount-survival product for each year is $vp=0.8(0.8)=0.64$. Under this constant annual survival assumption, ${}_kp_x=0.8^k$, and the whole-life annuity-due is a geometric series:

\ddot{a}_x=\sum_{k=0}^{\infty}(0.64)^k =\frac{1}{1-0.64} =\frac{25}{9} =2.7778.

The two-year temporary annuity-due pays at times 0 and 1, so its value is $1+v\,{}_1p_x=1+0.64=1.6400$. The probability of surviving two years is ${}_2p_x=0.8^2=0.64$, and the two-year discount factor is $v^2=0.64$. Therefore the time-0 factor for the tail is $v^2{}_2p_x=0.64(0.64)=0.4096$. At age $x+2$, the same assumptions give $\ddot{a}_{x+2}=25/9$. The tail value is

0.4096\left(\frac{25}{9}\right)=\frac{10.24}{9}=1.1378.

Adding the two parts gives $1.6400+1.1378=2.7778$, matching the whole-life value. The check uses unrounded values: $1.64+10.24/9=25/9$. A common wrong calculation uses $v\,{}_2p_x$ for the tail factor, producing $0.512(25/9)=1.4222$ and a total of $3.0622$. That calculation discounts only one year even though the tail value is measured at time 2; the correct factor is $v^2{}_2p_x$.

Worked Example: Checking Due and Immediate Values

For a hypothetical life annuity, let the effective annual interest rate be $i=2/3$, and assume survival probability $p=0.8$ each year at every age. Find the whole-life due and immediate values, then split each after two years. Method: calculate the common product $vp$ first, then respect the different payment dates in the temporary terms. This avoids confusing the due and immediate tails.

Here $v=1/(1+2/3)=0.6$, and $vp=0.6(0.8)=0.48=12/25$. The whole-life values are

\ddot{a}_x=\frac{1}{1-0.48}=\frac{25}{13}=1.9231, \qquad a_x=\frac{0.48}{1-0.48}=\frac{12}{13}=0.9231.

For the due annuity, the two temporary payments occur at times 0 and 1, giving $1+0.48=1.4800$. The tail factor is $v^2{}_2p_x=(0.48)^2=0.2304$, and the due tail value at age $x+2$ is $25/13$. Thus the tail is $0.2304(25/13)=144/325=0.4431$. The sum is $1.4800+0.4430769=1.9230769$, which checks against $25/13$.

For the immediate annuity, the temporary payments occur at times 1 and 2, giving $0.48+0.48^2=0.7104$. Its tail begins at time 3, so it uses the same factor $0.2304$, but the annuity value at age $x+2$ is immediate: $12/13$. The tail is $0.2304(12/13)=1728/8125=0.2127$. The sum is $0.7104+0.2126769=0.9230769$, matching $12/13$. The due tail is $0.4431$, not $0.2127$, because the due annuity includes a payment at the start of its period while the immediate annuity does not.

Worked Example: Checking a Split with Age-Varying Survival

A hypothetical annuity-due pays 1 annually in advance. The effective annual interest rate is $i=1/9$, so $v=0.9$. Conditional survival probabilities are $p_x=0.9$, $p_{x+1}=0.8$, and $p_{x+2}=0.7$; at every age from $x+3$ onward, the annual conditional survival probability is $0.5$. Split the whole-life value after three years. Method: calculate the first three survival probabilities by multiplying the age-specific probabilities, then use the constant-survival tail. This avoids incorrectly assuming one survival probability applies at all ages.

The survival probabilities needed for the temporary terms are ${}_0p_x=1$, ${}_1p_x=0.9$, and ${}_2p_x=0.9(0.8)=0.72$. The three-year temporary due value is therefore

\ddot{a}_{x:\overline{3}|} =1+0.9(0.9)+(0.9)^2(0.72) =1+0.81+0.5832 =2.3932.

Survival to time 3 is ${}_3p_x=0.9(0.8)(0.7)=0.504$, and $v^3=0.9^3=0.729$. At age $x+3$, the annual survival probability is constantly $0.5$, so $v p_{x+3}=0.9(0.5)=0.45$ and

\ddot{a}_{x+3}=\frac{1}{1-0.45}=\frac{20}{11}=1.8182.

The tail factor is $v^3{}_3p_x=0.729(0.504)=0.367416$. The tail value at time 0 is $0.367416(20/11)=0.6680291$. Adding the pieces gives $2.3932+0.6680291=3.0612291$, or $3.0612$ to four decimal places. As a direct check, the time-3 tail begins with probability-weighted discount $0.367416$; its subsequent payments have relative values $1,0.45,0.45^2,\ldots$, whose sum is $20/11$. This reproduces the same tail.

Reading and Checking the Split

The decomposition is useful because the temporary part and tail can be checked separately. The temporary terms are determined by the payment dates and survival to those dates. The tail requires three details to agree: the time at which its value is measured, the probability of reaching that time, and the annuity type used from that time onward.

Whole-life annuity	Temporary payments for duration $n$	First tail payment	Tail value at time 0
Annuity-due	Times $0,\ldots,n-1$	Time $n$	$v^n{}_np_x\ddot{a}_{x+n}$
Annuity-immediate	Times $1,\ldots,n$	Time $n+1$	$v^n{}_np_x a_{x+n}$

A particularly useful check is to add the temporary value and tail value and compare with the original whole-life value. If they disagree, inspect the payment boundary before recalculating: a due annuity’s temporary block includes time 0, while an immediate annuity’s block ends with the payment at time $n$. Also confirm that ${}_np_x$ is survival to time $n$, not survival to time $n-1$, and that the factor discounts $n$ full years.

Key takeaway: Split the payment sum at the chosen duration, then value the remaining payments conditional on survival to that split time. The factor $v^n{}_np_x$ moves that conditional tail value back to time 0; the annuity-due or annuity-immediate symbol determines the first payment in the tail.

Check Your Understanding

For each problem, use the stated annual survival and interest assumptions. Round final values to four decimal places where needed.

A hypothetical annuity-due has $i=1/3$ and constant annual survival probability $p=0.75$. Split its value after two years. Give the whole-life value, the temporary value, and the tail value. (Answer: whole-life $16/7=2.2857$; temporary $1.5625$; tail $81/112=0.7232$.)
A hypothetical annuity-immediate has $i=1/9$ and constant annual survival probability $p=0.9$. Split its value after three years. Give the whole-life, temporary, and tail values. (Answer: whole-life $81/19=4.2632$; temporary $1.9975$; tail $2.2656$.)
A hypothetical annuity-due has $i=0.25$. Its conditional survival probabilities are $p_x=0.8$, $p_{x+1}=0.7$, and $0.6$ at every age from $x+2$ onward. Split after two years. (Answer: whole-life $2.3292$; temporary $1.6400$; tail $0.6892$.)
A hypothetical annuity-immediate has $i=0.25$ and constant annual survival probability $p=0.85$. Split after one year. Identify the first tail payment date and calculate the temporary and tail values. (Answer: first tail payment at time 2; temporary $0.6800$; tail $1.4450$; whole-life $2.1250$.)
A hypothetical annuity-due has $i=0.2$. Its whole-life value is $4.1$, its three-year temporary due value is $2.3$, and ${}_3p_x=0.72$. Use the decomposition to find $\ddot{a}_{x+3}$. (Answer: $4.32$, since $v^3{}_3p_x=(5/6)^3(0.72)=5/12$, and $(4.1-2.3)/(5/12)=4.32$.)

Whole-life annuity	Temporary payments for duration \(n\)	First tail payment	Tail value at time 0
Annuity-due	Times \(0,\ldots,n-1\)	Time \(n\)	\(v^n{}_np_x\ddot{a}_{x+n}\)
Annuity-immediate	Times \(1,\ldots,n\)	Time \(n+1\)	\(v^n{}_np_x a_{x+n}\)

What You'll Learn