Deriving Constant-Force Survival Interpolation

Why Fractional-Age Survival Needs an Assumption

A life table may give the probability that a person aged $x$ survives to age $x+1$, while a calculation asks about survival for only part of that year. The annual probability alone does not determine how deaths are distributed within the year. A fractional-age calculation therefore requires an assumption about mortality between integer ages.

Under the constant-force assumption within the year of age $x$, the force of mortality is constant from age $x$ up to, but not including, age $x+1$. This assumption gives an exponential form for survival during that interval. The key result is that survival for a fraction $t$ of the year is the annual survival probability raised to the power $t$.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, and let ${}_tp_x=P(T_x>t)$ be the probability of surviving at least $t$ years from age $x$. Write $p_x={}_1p_x=1-{}_1q_x$ for the one-year survival probability. Under constant force within the age-$x$ year, for $0\leq t\leq1$, $$ {}_tp_x=(p_x)^t. $$ Here $t$ is measured in years, and the assumption concerns only the interval from age $x$ to age $x+1$.

Deriving the Fractional-Age Formula

The force of mortality describes the instantaneous rate of mortality for a life that has survived to a given age. For an age-$x$ year with constant force, write that force as $\mu_x$. The standard survival relation over an interval is the exponential of the negative accumulated force. Since the force does not vary within this particular year, its accumulated value over $t$ years is $\mu_x t$.

Write survival over $t$ years under the constant force.
For $0\leq t\leq1$, constant force gives ${}_tp_x=\exp(-\mu_x t)$.

Match the model to the full-year probability.
At $t=1$, the formula must reproduce the known annual survival probability: $p_x={}_1p_x=\exp(-\mu_x)$.

Eliminate the force parameter.
Raise the annual relation to the power $t$: $p_x^t=(\exp(-\mu_x))^t=\exp(-\mu_x t)={}_tp_x$.

This derivation explains why the exponent is the fraction of a year: the annual probability represents one full unit of exposure to the constant force, and a $t$-year interval accumulates $t$ times that force. The result is also a useful consistency check. At $t=0$, it gives survival probability $1$; at $t=1$, it returns $p_x$.

{}_tp_x=(p_x)^t=(1-{}_1q_x)^t,\qquad 0\leq t\leq1,\quad 0<p_x\leq1

The condition $p_x>0$ is needed for a finite constant force to reproduce the annual probability. If $p_x=1$, the force is zero and the formula gives survival probability $1$ throughout the year. A one-year survival probability of zero cannot be generated by a finite constant force; it would require an infinite force in this model. In applications, use the formula only for a fractional duration within the same age year. Extending it across an integer age boundary requires specifying the force for each year involved.

Using the Formula Carefully

Before calculating, identify the start age, the duration in years, and the annual survival probability for that age year. If the problem gives an annual death probability, first take its complement. If it gives life-table counts at ages $x$ and $x+1$, the annual survival probability is the ratio of the count at age $x+1$ to the count at age $x$, provided both counts refer to the same life table.

The formula calculates a probability, not a number of survivors. If a problem asks for an expected number of survivors from an initial group, multiply the group size by the probability after calculating it. Do not confuse the fractional survival probability with an interpolated life-table count unless the problem explicitly asks for that count.

Worked Example: Survival for Half a Year

A hypothetical mortality model has one-year survival probability $p_x=0.81$ for lives aged $x$. Under constant force within this age year, find the probability that a life aged $x$ survives six months. Report the probability to four decimal places.

Model and inputs. The mortality assumption is constant force from age $x$ to $x+1$. The duration is $t=6/12=0.5$ years, and the annual survival parameter is $p_x=0.81$. There is no interest rate or financial discounting in this probability calculation.

Method. Apply the fractional-age survival formula:

{}_{0.5}p_x=(p_x)^{0.5}=(0.81)^{0.5}=0.90.

Thus the probability of surviving six months is $0.9000$ to four decimal places. As a check, the square of the half-year survival probability is $0.90^2=0.81$, which recovers the specified annual survival probability. The common mistake avoided here is to treat half a year as half of the annual death probability; that is not the constant-force rule.

Worked Example: Survival for Nine Months

For a hypothetical age-$y$ group, the annual death probability is $0.36$. Assuming constant force within the age-$y$ year, calculate the probability of surviving nine months. Give the answer to four decimal places.

Model and inputs. The force is constant within the single age-$y$ year. The annual death probability is ${}_1q_y=0.36$, so the annual survival probability is $p_y=1-0.36=0.64$. Nine months is $t=9/12=0.75$ years. Interest is not applicable.

Method. Raise annual survival to the duration in years:

{}_{0.75}p_y=(0.64)^{0.75}.

Since $0.64=(0.8)^2$,

(0.64)^{0.75}=((0.8)^2)^{3/4}=(0.8)^{3/2} =0.8\sqrt{0.8} =0.8(0.894427\ldots) =0.715541\ldots.

The probability of surviving nine months is $0.7155$ to four decimal places. It is less than the six-month survival probability $(0.64)^{0.5}=0.8$, as expected because nine months is the longer exposure period. The common mistake avoided is using $0.75$ as a multiplier on the annual death probability rather than raising the annual survival probability to $0.75$.

Worked Example: Extracting Annual Survival From Life-Table Counts

A hypothetical life table has $\ell_x=1{,}000$ and $\ell_{x+1}=810$. Assuming constant force within the age-$x$ year, find the probability that a life aged $x$ survives $0.4$ years. Report the result to four decimal places.

Model and inputs. The life-table counts refer to the same table, so the annual survival probability is their ratio. The force is constant within this age year, the requested duration is $t=0.4$ years, and no interest rate is involved.

Step 1: Find annual survival.

p_x=\frac{\ell_{x+1}}{\ell_x}=\frac{810}{1{,}000}=0.81.

Step 2: Apply constant-force interpolation.

{}_{0.4}p_x=(0.81)^{0.4} =\exp\bigl(0.4\ln(0.81)\bigr).

Using $\ln(0.81)=-0.2107210$, the exponent is $0.4(-0.2107210)=-0.0842884$. Therefore,

{}_{0.4}p_x=\exp(-0.0842884)=0.9192

to four decimal places. The answer is a survival probability, not a fractional-age life-table count. The common mistake avoided is reversing the life-table ratio: $810/1{,}000$ is the proportion surviving from age $x$ to $x+1$, whereas its complement $0.19$ is the annual death probability.

Common Exam Traps and Boundary Checks

The most frequent error is to scale the annual death probability linearly. For example, with $p_x=0.81$, the annual death probability is $0.19$. A tempting but incorrect constant-force calculation for half a year is

1-\frac{1}{2}(0.19)=0.905.

That calculation assumes that half the annual deaths occur in the first half of the year, which is a different within-year model. Under constant force, the correct half-year survival probability is $(0.81)^{0.5}=0.90$. The two values differ because they distribute mortality within the year differently; do not substitute a linear death-probability interpolation when the stated assumption is constant force.

Several quick checks help detect errors before reporting a result:

Duration: convert months or days to years using the time convention in the problem. For example, six months is $0.5$ years, not $6$.
Probability input: the power applies to annual survival $p_x$, not annual death probability ${}_1q_x$.
Range: for $0<p_x<1$ and $0<t<1$, $p_x<{}_tp_x<1$. A shorter exposure has a higher probability of survival than the full-year probability.
Endpoints: $t=0$ must give $1$, and $t=1$ must give $p_x$.
Age interval: use the annual probability for the age year in which the fractional duration lies. Do not carry the same annual probability across a birthday unless the model specifies it.

The formula matters because mortality data are often supplied at integer ages while actuarial questions require a probability at a fractional age. Constant-force interpolation provides a coherent way to make that calculation from the annual survival probability, while making its within-year assumption explicit. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: Under constant force within the age-$x$ year, the probability of surviving $t$ years is the annual survival probability raised to $t$: ${}_tp_x=(p_x)^t$, with $t$ expressed in years and $0\leq t\leq1$.

Check Your Understanding

For each problem, assume constant force within the stated age year. Probabilities should be reported to four decimal places when rounding is needed.

A hypothetical age-$r$ group has annual survival probability $0.49$. Find the probability of surviving six months. (Answer: $0.7000$)
A hypothetical age-$s$ group has annual death probability $0.36$. Find the probability of surviving three months. (Answer: $0.8944$)
For a hypothetical age-$w$ group, $\ell_w=2{,}000$ and $\ell_{w+1}=1{,}458$. Find the probability that a life aged $w$ survives one-third of a year. (Answer: $0.9000$)
A hypothetical age-$z$ group has annual survival probability $0.64$. Calculate the probability of surviving nine months, then verify that it is below the probability of surviving six months. (Answer: $0.7155$; the six-month probability is $0.8000$)
A hypothetical annual survival probability is $0.95$. Find survival over $0.2$ years and explain why multiplying the annual death probability by $0.2$ is not the constant-force calculation. (Answer: $0.9898$; constant force gives $0.95^{0.2}$, not $1-0.2(0.05)$)

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