Mixed Hazard And Life Table Problems

Combining Mortality Information Across Age Intervals

A survival problem may give mortality in more than one form: a force of mortality over one interval, annual death probabilities over another, and life-table entries at selected ages. The central task is to keep each assumption attached to the correct age interval, then combine the corresponding survival probabilities. A force is not itself a probability, and an annual death probability is not itself a force; confusing them can change the answer even when every age is indexed correctly.

This tutorial develops a consistent method for combining cumulative hazard, conditional survival, annual mortality rates, and life-table entries at integer ages. The examples use hypothetical mortality assumptions. No fractional-age interpolation is needed: each interval begins and ends at an integer age.

Definition: For a life aged $x$, let ${}_tp_x$ be the probability of surviving $t$ years, conditional on being alive at age $x$, and let ${}_tq_x=1-{}_tp_x$. The force of mortality at age $x+t$ is $\mu_{x+t}$. If the force is specified over the interval from age $x$ to age $x+t$, the cumulative hazard over that interval is $$ H_x(t)=\int_0^t \mu_{x+s}\,ds, $$ and the conditional survival probability is ${}_tp_x=e^{-H_x(t)}$. For a life table, $\ell_x$ is the number of lives surviving to exact age $x$; consequently, ${}_tp_x=\ell_{x+t}/\ell_x$ when the entries refer to the same mortality basis.

One Interval, One Survival Factor

Every interval contributes a survival factor. If the force is given, integrate it over the interval and exponentiate the negative cumulative hazard. If an annual death probability $q_x$ is given directly, the one-year survival factor is $1-q_x$. If life-table entries are given, their ratio gives survival between the corresponding ages.

For successive intervals, multiply the survival factors. Equivalently, add the cumulative hazards for intervals governed by forces, while using the stated annual survival factors for intervals governed by annual death probabilities. The information supplied for an interval determines which conversion is appropriate.

Key result: If an age interval has a specified annual death probability $q_x$, its survival factor is $1-q_x$. If it instead has a specified force $\mu_{x+s}$, its survival factor over one year is $$ {}_1p_x=\exp\left(-\int_0^1\mu_{x+s}\,ds\right). $$ These expressions agree when the assumptions agree. In particular, under a constant force $\mu$ throughout the year, $q_x=1-e^{-\mu}$, so $\mu=-\ln(1-q_x)$. Do not use this constant-force conversion unless constant force over that year is assumed.

The distinction also clarifies conditional survival. If a person is known to have survived to age $x$, the probability of surviving to age $x+n$ is the product of the survival factors for the $n$ intervening intervals. Life-table entries provide a direct check: that same probability must equal $\ell_{x+n}/\ell_x$, if the entries and interval assumptions describe the same mortality basis.

Why the Interval Factors Multiply

The multiplication rule follows from conditioning on survival to each interval boundary. For example, survival from age $x$ to age $x+2$ requires survival from $x$ to $x+1$, followed by survival from $x+1$ to $x+2$. Thus

{}_2p_x =P(T>x+2\mid T>x) ={}_1p_x\,{}_1p_{x+1},

where $T$ denotes age at death. For intervals governed by forces, substituting ${}_1p_x=\exp(-\int_0^1\mu_{x+s}\,ds)$ gives

{}_2p_x =\exp\left(-\int_0^1\mu_{x+s}\,ds\right) \exp\left(-\int_0^1\mu_{x+1+s}\,ds\right) =\exp\left(-\int_0^2\mu_{x+s}\,ds\right).

The last equality holds because the two integrals cover adjacent, nonoverlapping age intervals. It is the reason cumulative hazards add across those intervals. If an interval instead has a specified annual $q$, use its survival factor $1-q$; do not insert a force-based factor unless a force assumption has been supplied or is explicitly derived under an appropriate model.

Worked Problems Across Multiple Ages

Worked Example: A Force Interval Between Annual Mortality Rates

A hypothetical life table has $\ell_{40}=10{,}000$. From age 40 to 41, the force of mortality is constant at $0.10$ per year. From age 41 to 42, the annual death probability is $q_{41}=0.20$. From age 42 to 43, the force is constant at $0.05$ per year. Find ${}_3p_{40}$ and $\ell_{43}$, and check the result using the intervening table entries.

Method. Find each one-year survival factor using the assumption specified for that interval, then multiply the three factors. A constant force $\mu$ over one year gives survival $e^{-\mu}$; a stated annual death probability $q$ gives survival $1-q$.

For ages 40 to 41, the survival factor is $e^{-0.10}=0.9048374180$. Therefore, $\ell_{41}=10{,}000(0.9048374180)=9{,}048.374180$.

For ages 41 to 42, the survival factor is $1-0.20=0.8000000000$. Therefore, $\ell_{42}=9{,}048.374180(0.8000000000)=7{,}238.699344$.

For ages 42 to 43, the survival factor is $e^{-0.05}=0.9512294245$. Therefore, $\ell_{43}=7{,}238.699344(0.9512294245)=6{,}885.663811$.

The combined conditional survival probability is

{}_3p_{40} =e^{-0.10}(1-0.20)e^{-0.05} =0.8e^{-0.15} =0.6885663811\ldots.

The table-ratio check gives $\ell_{43}/\ell_{40}=6{,}885.663811/10{,}000=0.6885663811\ldots$, agreeing with the product. Thus ${}_3p_{40}=0.6886$ to four decimal places and $\ell_{43}=6{,}885.66$ to the nearest cent. The common mistake is to treat $0.10$ or $0.05$ as a death probability. They are force values; their survival factors are exponentials.

Worked Example: A Changing Force and Its Annual Table Entries

For a hypothetical cohort, $\ell_{40}=5{,}000$. At age $40+s$, for $0\le s\le 2$, the force is $\mu_{40+s}=0.02+0.01s$ per year. Find the cumulative hazard and survival from age 40 to 42, then calculate $\ell_{41}$, $\ell_{42}$, $q_{40}$, and $q_{41}$.

Method. Integrate the given force over each annual interval. The variable $s$ measures time since age 40, so its value resets only if the force expression is rewritten for a new interval; here the given expression uses the same age-40 origin throughout.

Over the full two-year interval,

H_{40}(2)=\int_0^2(0.02+0.01s)\,ds =0.02(2)+0.005(2^2) =0.06.

Hence ${}_2p_{40}=e^{-0.06}=0.9417645336$, and $\ell_{42}=5{,}000(0.9417645336)=4{,}708.822668$.

For the first year, the cumulative hazard is $\int_0^1(0.02+0.01s)\,ds=0.02+0.005=0.025$. Thus ${}_1p_{40}=e^{-0.025}=0.9753099120$, giving $\ell_{41}=5{,}000(0.9753099120)=4{,}876.549560$ and $q_{40}=1-e^{-0.025}=0.0246900880$.

For the second year, use $s$ from 1 to 2 in the original force expression. Its cumulative hazard is $\int_1^2(0.02+0.01s)\,ds=0.035$. Therefore, ${}_1p_{41}=e^{-0.035}=0.9656054163$ and $q_{41}=1-e^{-0.035}=0.0343945837$. The table calculation checks: $4{,}876.549560(0.9656054163)=4{,}708.822668$, matching the two-year result. The common mistake is integrating the first-year force expression again from $s=0$ to $s=1$. That would incorrectly assign the first year’s hazard to the second year.

Worked Example: Recovering a Life-Table Entry and a Conditional Probability

A hypothetical table has $\ell_{50}=12{,}000$, and the annual death probability from age 50 to 51 is $q_{50}=0.03$. From age 51 to 52, the force is constant at $0.04$ per year. Find $\ell_{51}$, $\ell_{52}$, $q_{51}$, and the probability of survival from age 50 to 52.

Method. Convert each interval’s assumption to a survival factor, update the table entry, and then use the ratio of entries as a check.

For ages 50 to 51, survival is $1-q_{50}=0.97$, so $\ell_{51}=12{,}000(0.97)=11{,}640$.

For ages 51 to 52, constant force gives survival $e^{-0.04}=0.9607894392$. Thus $\ell_{52}=11{,}640(0.9607894392)=11{,}183.589072$. The annual death probability on this interval is $q_{51}=1-e^{-0.04}=0.0392105608$.

The two-year conditional survival probability is

{}_2p_{50} =(1-q_{50})e^{-0.04} =(0.97)(0.9607894392) =0.9319657560.

The table-ratio check is $\ell_{52}/\ell_{50}=11{,}183.589072/12{,}000=0.9319657560$, as required. The common mistake is to apply $q_{50}$ to the second year as well. The problem specifies a force for ages 51 to 52, so the second survival factor is $e^{-0.04}$, not $1-0.03$.

Checks That Keep Ages and Assumptions Aligned

A good calculation has both a probability check and an age-index check. The probability check confirms that every factor is a survival probability between zero and one. The age-index check confirms that the factor from age $x$ to $x+1$ is applied to $\ell_x$, producing $\ell_{x+1}$, and not to a later or earlier entry.

For a sequence of intervals, these checks can be summarized as follows:

Mark the boundaries.
Write the attained ages at the start and end of every interval. Identify which mortality assumption applies between each pair.

Find each interval’s survival factor.
Use $1-q_x$ for a given annual death probability, or $e^{-\int\mu\,ds}$ for a force specified over the interval.

Combine and check.
Multiply the interval factors. When life-table entries are available, compare the result with the final entry divided by the initial entry.

If the two routes disagree, revisit the interval boundaries and assumptions before rounding. Frequent causes include using a force as if it were a probability, applying one year’s $q_x$ to a different age, and applying a force over the wrong integration limits. Only round after the combined probability or life-table entry has been calculated.

Key takeaway: A multi-interval survival calculation is a product of interval-specific survival factors. A force must be integrated before it becomes a survival probability; an annual $q_x$ already specifies the interval’s death probability. Life-table ratios provide a direct consistency check. These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical mortality model, calculate the requested quantity. Keep each interval attached to its stated age range.

A constant force of $0.08$ applies from age 30 to 32. Find ${}_2p_{30}$. (Answer: $0.8521$, to four decimal places.)
Annual death probabilities are $q_{55}=0.12$ and $q_{56}=0.05$. Find ${}_2p_{55}$. (Answer: $0.8360$, to four decimal places.)
A table has $\ell_{60}=8{,}000$ and $q_{60}=0.025$. A constant force of $0.03$ applies from age 61 to 62. Find $\ell_{62}$. (Answer: $7{,}569.48$, to the nearest cent.)
From age 40 to 41, the force is $\mu_{40+s}=0.01+0.02s$, for $0\le s\le1$. From age 41 to 42, $q_{41}=0.04$. Find ${}_2p_{40}$. (Answer: $e^{-0.02}(0.96)=0.9410$, to four decimal places.)
A hypothetical table has $\ell_{45}=10{,}000$ and $q_{45}=0.10$. A constant force of $0.02$ applies from age 46 to 47, followed by $q_{47}=0.05$ from age 47 to 48. Find $\ell_{48}$. (Answer: $8{,}380.70$, to the nearest cent.)

What You'll Learn