Recovering Survival Probabilities from Term Values

Why Insurance and Annuity Values Reveal Survival

A present value for a life-contingent contract combines two ingredients: the timing and amount of its payments, and the probability that the insured survives long enough for those payments to be made. When the interest assumption is known, carefully chosen combinations of insurance and annuity values can separate out the survival probability. This is useful when a problem gives contract values but asks for a probability that is not stated directly.

The calculations below use annual periods and benefits payable at the end of the year of death unless a benefit is explicitly payable at the end of the term. Let $i$ be the effective annual interest rate and $v=1/(1+i)$. Write ${}_kp_x$ for the probability that a life aged $x$ survives $k$ years, with ${}_0p_x=1$. The corresponding one-year conditional death probability is $q_{x+k}=1-p_{x+k}$.

Definition: An $n$-year discrete term insurance pays 1 at the end of the year of death if death occurs within $n$ years; it pays nothing if the insured survives the term. A pure endowment pays 1 at time $n$ if the insured survives $n$ years; it pays nothing otherwise. A temporary life annuity-due with $n$ possible payments pays 1 at times $0,1,\ldots,n-1$, provided the life is alive at each payment time. Denote its expected present value by $\ddot{a}_{x:\overline{n}|}$.

These definitions fix the payment timing. In particular, the term insurance benefit for a death in year $k+1$ is discounted $k+1$ years, while an annuity-due payment at time $k$ is discounted $k$ years. The survival probability associated with a pure endowment or the final payment of an annuity is ${}_np_x$, not the conditional one-year probability $p_{x+n-1}$.

Values That Directly Isolate Survival

The pure endowment value is the most direct route. Its benefit is paid only if the life survives the full term, and it is paid at a fixed time. Therefore its expected present value is the survival probability multiplied by the discount factor for that time.

{}_nE_x=v^n\,{}_np_x \qquad\Longrightarrow\qquad {}_np_x=\frac{{}_nE_x}{v^n}.

Here ${}_nE_x$ denotes the expected present value of a unit pure endowment. The result assumes the benefit is paid at time $n$ only on survival to that time. It does not apply unchanged to a benefit paid at the beginning of the term, or to a benefit payable on death.

Temporary annuity values can also isolate survival by comparing two terms. Increasing the number of possible annuity-due payments from $n$ to $n+1$ adds exactly one possible payment, at time $n$. That payment is made only if the life survives to time $n$.

\ddot{a}_{x:\overline{n+1}|} - \ddot{a}_{x:\overline{n}|} = v^n\,{}_np_x \qquad\Longrightarrow\qquad {}_np_x= \frac{\ddot{a}_{x:\overline{n+1}|}-\ddot{a}_{x:\overline{n}|}}{v^n}.

This difference requires values for the same life, interest rate, and payment convention. It is not a comparison of annuities beginning at different ages. The added payment occurs at time $n$, so discounting it by $v^{n+1}$ would use the wrong time.

Worked Example: Recovering Survival from a Pure Endowment

A hypothetical insurer reports a unit pure endowment value of $0.4096$ for a three-year term on a life aged $x$. The effective annual interest rate is $25\%$. Find the probability that the life survives all three years.

Method. A pure endowment is paid at the end of year 3 only if the life survives to that time, so divide its present value by the three-year discount factor. The common mistake to avoid is dividing by $v$, rather than $v^3$.

First calculate the discount factor:

v=\frac{1}{1.25}=0.8, \qquad v^3=0.8^3=0.512.

Then solve for the survival probability:

{}_3p_x=\frac{0.4096}{0.512}=0.8.

Check. Discounting the recovered probability gives $0.512(0.8)=0.4096$, exactly the stated pure endowment value. The result is a three-year survival probability, not a one-year survival probability.

Combining Term Insurance and Annuity Values

A term insurance value alone generally does not identify a survival probability: it combines probabilities of death in several different years. A temporary annuity value alone also combines survival probabilities at several payment times. Their relationship, however, allows the two values together to reveal the pure endowment component.

Theorem: For an $n$-year discrete term insurance with unit benefit payable at the end of the year of death, a unit pure endowment payable at time $n$, and an $n$-payment temporary life annuity-due, $$ A^1_{x:\overline{n}|}+{}_nE_x = 1-d\,\ddot{a}_{x:\overline{n}|}, \qquad d=1-v. $$ Consequently, $$ {}_np_x= \frac{1-d\,\ddot{a}_{x:\overline{n}|}-A^1_{x:\overline{n}|}}{v^n}. $$

Here $A^1_{x:\overline{n}|}$ denotes the expected present value of the unit term insurance just defined. The identity assumes annual payment periods and the stated end-of-year death benefit. For a benefit payable immediately on death, or for an annuity with a different payment timing, these values cannot be substituted without adjusting the relation.

The identity follows by collecting the contributions for each survival duration. The term insurance value and pure endowment value are

A^1_{x:\overline{n}|} = \sum_{k=0}^{n-1}v^{k+1}{}_kp_xq_{x+k}, \qquad {}_nE_x=v^n{}_np_x.

Since ${}_kp_xq_{x+k}={}_kp_x-{}_{k+1}p_x$, the insurance sum records the probability of leaving the in-force group in each year. On adding the pure endowment term, the final survival contribution cancels the final negative term in this difference. The resulting expression is

A^1_{x:\overline{n}|}+{}_nE_x = v-(1-v)\sum_{k=1}^{n-1}v^k{}_kp_x.

The temporary annuity-due value is $\ddot{a}_{x:\overline{n}|}=\sum_{k=0}^{n-1}v^k{}_kp_x$, whose time-zero term is 1. Thus $1-d\ddot{a}_{x:\overline{n}|}=v-(1-v)\sum_{k=1}^{n-1}v^k{}_kp_x$, because $d=1-v$. This proves the identity. Solving it first for ${}_nE_x$, then using ${}_nE_x=v^n{}_np_x$, gives the recovery formula.

Worked Example: Combining Term Insurance and an Annuity-Due Value

For a hypothetical two-year contract on a life aged $x$, the unit term insurance value is $0.1952$, and the two-payment temporary annuity-due value is $1.72$. The effective annual interest rate is $25\%$. Find ${}_2p_x$.

Method. Apply the identity using the discount rate $d=1-v$, then divide the implied pure endowment value by $v^2$. This avoids trying to interpret the term insurance value as if it were itself a survival probability.

Calculate the discount quantities:

v=\frac{1}{1.25}=0.8, \qquad d=1-0.8=0.2, \qquad v^2=0.64.

Use the term-insurance and annuity values to recover the pure endowment value:

{}_2E_x = 1-d\,\ddot{a}_{x:\overline{2}|}-A^1_{x:\overline{2}|} = 1-(0.2)(1.72)-0.1952 = 1-0.344-0.1952 = 0.4608.

Now remove the two-year discounting:

{}_2p_x=\frac{0.4608}{0.64}=0.72.

Check. The values imply $A^1+{}_2E_x=0.1952+0.4608=0.656$, while $1-d\ddot{a}=1-0.344=0.656$. The two sides agree, and discounting the recovered survival probability gives $0.64(0.72)=0.4608$.

Worked Example: Recovering Survival by Comparing Annuity Terms

For a hypothetical life aged $x$, a two-payment temporary annuity-due has value $1.72$, and a three-payment temporary annuity-due has value $2.1808$. The effective annual interest rate is $25\%$. Find the probability of surviving two years.

Method. Subtract the shorter annuity value from the longer one. The additional payment is at time 2, so the difference equals $v^2{}_2p_x$. The trap is to treat the difference as an undiscounted probability or to discount it for three years.

The difference in values is

$$ 2.1808-1.72=0.4608. $$

At $25\%$ interest, $v=0.8$, so $v^2=0.64$. Therefore,

{}_2p_x=\frac{0.4608}{0.64}=0.72.

Check. The extra payment has present value $0.64(0.72)=0.4608$, matching the difference between the annuity values. As a further check on the timing, an $n$-payment annuity-due has payments at times $0$ through $n-1$; the extra payment when moving from two to three payments is at time 2.

Reading the Values Carefully

A reliable calculation begins by identifying exactly which event triggers each payment and when it is made. The following distinctions prevent most errors in survival recovery:

Pure endowment: survival to time $n$ triggers a payment at time $n$, giving value $v^n{}_np_x$.
Temporary annuity-due difference: the additional payment is at time $n$, also giving $v^n{}_np_x$.
Term insurance: death in year $k+1$ triggers a payment at the end of that year, with contribution $v^{k+1}{}_kp_xq_{x+k}$. The term value aggregates contributions over several possible death years.
Conditional versus unconditional survival: $p_{x+k}$ is survival through the next year conditional on reaching age $x+k$; ${}_np_x$ is survival from age $x$ for the entire $n$-year period.

For example, if the first two conditional annual survival probabilities are known, then ${}_2p_x=p_xp_{x+1}$. A recovered two-year survival probability is not generally equal to $p_{x+1}$; it includes the probability of surviving the first year as well.

Key takeaway: First identify the payment time and survival event. A pure endowment value, an annuity-value difference, or a term-insurance-plus-annuity identity can then isolate $v^n{}_np_x$. Divide by the correct $v^n$ to obtain the survival probability.

These identities are useful for checking related contract values as well as solving for an unknown probability. A result should be between 0 and 1, and substituting it back into the relevant value equation should reproduce the supplied values. If either check fails, reconsider the term length, discount exponent, benefit timing, or whether the requested probability is conditional or unconditional.

Check Your Understanding

For each problem, use the payment timing stated and give the requested survival probability.

A unit pure endowment payable after two years has value $0.64$. The effective annual interest rate is $25\%$. Find ${}_2p_x$. (Answer: $1.0000$)
At an effective annual interest rate of $25\%$, the values of the three-payment and two-payment temporary annuities-due are $2.1808$ and $1.72$, respectively. Find ${}_2p_x$. (Answer: $0.7200$)
A two-year unit term insurance has value $0.15$, and the two-payment temporary annuity-due has value $1.75$. If $i=20\%$, find ${}_2p_x$. (Answer: $0.8040$)
For a two-year contract, the unit endowment insurance value is $0.656$, and the unit term insurance value is $0.1952$. At $i=25\%$, find the two-year survival probability. (Answer: $0.7200$)
At zero interest, a two-year unit term insurance has value $0.32$. The conditional probability of surviving the first year is $p_x=0.8$. Find the conditional probability of surviving the second year, $p_{x+1}$. (Answer: $0.8500$)

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