Hazard And Life Table Translation Workshop

Why Translation Matters

A mortality problem may describe the same risk in different ways: as a force of mortality at each age, as a probability of surviving an interval, or as columns in a life table. These quantities are connected, but they are not interchangeable. A force is an instantaneous rate; a death probability is a probability over a specified interval; and a life table records how a cohort’s survivors change from one age to the next.

The key to translating correctly is to keep track of both the age interval and the <strongassumption about how mortality behaves within it. When the force varies with age, survival depends on its integral over the interval. When the force is constant, the calculation simplifies to an exponential formula.

Definition: The force of mortality at age \(x+t\), written \(\mu_{x+t}\), is the instantaneous rate of death conditional on survival to that age. The probability of surviving \(t\) years from age \(x\) is \({}_tp_x\), and the probability of dying within those \(t\) years is \({}_tq_x=1-{}_tp_x\). In a life table, \(\ell_x\) is the number alive at exact age \(x\), \(d_x=\ell_x-\ell_{x+1}\) is the number dying between ages \(x\) and \(x+1\), and \(q_x=d_x/\ell_x\) is the one-year death probability.

From a Force to Survival

For a force that can vary with age, the survival probability from age \(x\) to age \(x+t\) is determined by the accumulated force over that interval. This relationship applies when the force function describes mortality throughout the interval; knowing only its value at the starting age is not enough if it changes later.

{}_tp_x=\exp\!\left(-\int_0^t \mu_{x+s}\,ds\right), \qquad {}_tq_x=1-\exp\!\left(-\int_0^t \mu_{x+s}\,ds\right).

If the force is constant at \(\mu\) throughout an interval of length \(h\), the integral is \(\mu h\). Thus the probability of surviving that interval is \(e^{-\mu h}\), and the probability of dying is \(1-e^{-\mu h}\). In particular, for one full year with constant force \(\mu_x\), \(p_x=e^{-\mu_x}\) and \(q_x=1-e^{-\mu_x}\). These formulas do not say that the force equals the death probability: the force is a rate, while \(q_x\) is a probability.

Derivation: Let \(S(t)={}_tp_x\) be survival from age \(x\) to age \(x+t\). By the definition of the force, over a short interval of length \(dt\), the conditional probability of death is approximately \(\mu_{x+t}\,dt\). Therefore \(S(t+dt)\) is approximately \(S(t)(1-\mu_{x+t}\,dt)\), giving \(dS(t)/dt=-\mu_{x+t}S(t)\). Dividing by \(S(t)\) and integrating from \(0\) to \(t\) gives \(\ln S(t)-\ln S(0)=-\int_0^t\mu_{x+s}\,ds\). Since \(S(0)=1\), exponentiating gives the survival formula.

For consecutive intervals, survival probabilities multiply. If a person must survive first from \(x\) to \(x+1\) and then from \(x+1\) to \(x+2\), then

{}_2p_x=p_xp_{x+1}, \qquad {}_2q_x=1-p_xp_{x+1}.

This multiplication also follows by adding the integrated force over adjacent intervals and then exponentiating. The forces do not generally add as probabilities; rather, their interval integrals add, and the corresponding survival probabilities multiply.

Worked Example: Building Life Table Entries from Piecewise-Constant Forces

A hypothetical cohort has \(\ell_{40}=100{,}000\). Assume the force is constant at \(0.020\) from age 40 to 41, at \(0.050\) from age 41 to 43, and at \(0.010\) from age 43 to 44. Find the annual \(q_x\) values and the life table survivor counts through age 44. Use full precision in intermediate calculations and round final counts to two decimal places and probabilities to four decimal places.

Method. For each one-year interval with constant force \(\mu\), calculate \(p=e^{-\mu}\) and \(q=1-e^{-\mu}\). Then multiply the current survivor count by \(p\) to obtain the next life table entry.

From age 40 to 41, \(p_{40}=e^{-0.020}=0.9801987\), so \(q_{40}=1-0.9801987=0.0198013\), or \(0.0198\) to four decimal places. Therefore \(\ell_{41}=100{,}000(0.9801987)=98{,}019.87\).

The force is \(0.050\) in each of the next two one-year intervals. For each year, \(p=e^{-0.050}=0.9512294\) and \(q=1-0.9512294=0.0487706\), or \(0.0488\). Thus \(\ell_{42}=98{,}019.87(0.9512294)=93{,}239.38\), and \(\ell_{43}=93{,}239.38(0.9512294)=88{,}692.04\), using unrounded counts in the multiplications.

From age 43 to 44, \(p_{43}=e^{-0.010}=0.9900498\), so \(q_{43}=0.0099502\), or \(0.0100\). The final survivor count is \(\ell_{44}=88{,}692.04(0.9900498)=87{,}809.54\).

The resulting annual survivor sequence is \(100{,}000.00,\ 98{,}019.87,\ 93{,}239.38,\ 88{,}692.04,\ 87{,}809.54\). As a check over all four years, the accumulated force is \(0.020+0.050+0.050+0.010=0.130\). Direct calculation gives \(100{,}000e^{-0.130}=87{,}809.54\), agreeing with the sequential calculation.

Common mistake avoided: The force \(0.050\) applies over each of two separate years, so its cumulative contribution is \(0.100\), not \(0.050\). The two-year survival over ages 41 to 43 is \(e^{-0.100}=0.9048374\), not \(1-2(0.0487706)\).

From Life Table Columns to Forces

A life table gives an interval death probability directly from adjacent survivor counts. For a one-year interval, \(q_x=d_x/\ell_x\), where \(d_x=\ell_x-\ell_{x+1}\). If the force is assumed constant over that year, the corresponding force is \(\mu_x=-\ln(p_x)=-\ln(1-q_x)\). Without the constant-force assumption, the table identifies the integrated force over the year, but does not determine the force at every age within it.

p_x=\frac{\ell_{x+1}}{\ell_x}, \qquad q_x=\frac{\ell_x-\ell_{x+1}}{\ell_x}, \qquad \int_0^1\mu_{x+s}\,ds=-\ln(p_x).

The last equality is valid whether or not the force is constant. It gives the total force accumulated over the year. Calling that value the force at each age in the year requires the additional assumption that the force is constant there.

Worked Example: Recovering Interval Forces and Multi-Year Survival

A hypothetical life table has \(\ell_{30}=1{,}000{,}000\), \(\ell_{31}=980{,}000\), \(\ell_{32}=940{,}800\), and \(\ell_{33}=912{,}576\). For each year, find \(d_x\), \(q_x\), and the equivalent constant force. Then find the probability of surviving from age 31 to age 33 and the two-year death probability. Assume a constant force within each individual year. Give probabilities to four decimal places and forces to four significant figures.

Method. Subtract adjacent survivor counts to get deaths, divide by the starting survivor count to get \(q_x\), and use \(\mu_x=-\ln(1-q_x)\) for the equivalent constant annual force.

For age 30 to 31, \(d_{30}=1{,}000{,}000-980{,}000=20{,}000\). Hence \(q_{30}=20{,}000/1{,}000{,}000=0.0200\), and \(\mu_{30}=-\ln(0.9800)=0.02020\).

For age 31 to 32, \(d_{31}=980{,}000-940{,}800=39{,}200\). Hence \(q_{31}=39{,}200/980{,}000=0.0400\), and \(\mu_{31}=-\ln(0.9600)=0.04082\).

For age 32 to 33, \(d_{32}=940{,}800-912{,}576=28{,}224\). Hence \(q_{32}=28{,}224/940{,}800=0.0300\), and \(\mu_{32}=-\ln(0.9700)=0.03046\).

From age 31 to age 33, the survival probability is the ratio of survivors, \(912{,}576/980{,}000=0.9312\). Equivalently, the annual survival probabilities multiply: \(0.9600(0.9700)=0.9312\). The two-year death probability is \(1-0.9312=0.0688\). The accumulated force is \(-\ln(0.9312)=0.07128\), which also equals \(\mu_{31}+\mu_{32}\) using unrounded values.

Check. The death probability over ages 31 to 33 is \(1-\ell_{33}/\ell_{31}=0.0688\), matching the result from multiplying annual survival probabilities. The annual forces are equivalent constant forces for their respective intervals; they do not establish that mortality was actually constant within those years.

When the Force Changes Within a Year

A force need not be constant over a year. In that case, use the integral of the specified age-dependent function, rather than the starting force or an unweighted sum of its values. The life table still records the resulting one-year survival probability; it does not, by itself, specify the shape of the force inside the interval.

Worked Example: Integrating an Age-Varying Force

For a hypothetical cohort, let the force from age 60 to 61 be \(\mu_{60+t}=0.010+0.020t\), for \(0\le t\le1\). From age 61 to 62, let it be \(\mu_{61+s}=0.030+0.010s\), for \(0\le s\le1\). If \(\ell_{60}=100{,}000\), find the one-year death probabilities and survivor counts at ages 61 and 62. Round probabilities to four decimal places and counts to two decimal places.

Method. Integrate the force over each year, exponentiate the negative integral to obtain survival, and then apply that survival probability to the life table count.

For ages 60 to 61, the integrated force is \(\int_0^1(0.010+0.020t)\,dt=0.010+0.010=0.020\). Therefore \(p_{60}=e^{-0.020}=0.9801987\) and \(q_{60}=1-0.9801987=0.0198\). The survivor count is \(\ell_{61}=100{,}000(0.9801987)=98{,}019.87\).

For ages 61 to 62, the integrated force is \(\int_0^1(0.030+0.010s)\,ds=0.030+0.005=0.035\). Therefore \(p_{61}=e^{-0.035}=0.9656054\) and \(q_{61}=1-0.9656054=0.0344\). Applying this to the unrounded survivor count gives \(\ell_{62}=100{,}000e^{-0.020}e^{-0.035}=100{,}000e^{-0.055}=94{,}648.51\).

As a check, the two-year survival is \(0.9801987(0.9656054)=e^{-0.055}=0.9464851\). The two-year death probability is therefore \(1-0.9464851=0.0535\). Using only the starting forces \(0.010\) and \(0.030\) as if each held for a full year would give the wrong accumulated force because each force increases with age.

Translation Checks and Common Traps

When solving a multi-interval problem, organize the calculation around interval boundaries. For each interval, identify its length, the force assumption, and whether the requested quantity is a probability or a life table count. The following checks catch many errors:

Check the interval length. A constant force \(\mu\) over \(h\) years gives survival \(e^{-\mu h}\), not \(e^{-\mu}\) unless \(h=1\).
Keep force and probability distinct. Under a constant force, \(q=1-e^{-\mu}\); in general \(q\ne\mu\).
Multiply survival probabilities across years. For consecutive intervals, \( {}_tp_x\) is the product of the interval survival probabilities, while the total death probability is one minus that product.
Use the right life table denominator. The one-year death probability is \(d_x/\ell_x\), not \(d_x/\ell_{x+1}\).
Do not infer a within-year pattern from one annual probability. The table determines the integrated force over the year. A constant force is one possible assumption that gives an equivalent force value.

For example, if two successive annual death probabilities are \(0.10\) and \(0.20\), adding them gives \(0.30\), which is not the two-year death probability. The correct calculation is \(1-(1-0.10)(1-0.20)=1-(0.90)(0.80)=0.28\). The incorrect addition treats the second probability as though it applied to the original starting group rather than to those who survived the first year.

Key takeaway: Integrate the force over each interval to obtain survival, convert survival to death probability with \(q=1-p\), and use \(\ell_{x+1}=\ell_xp_x\) to move through a life table. Across consecutive intervals, multiply survival probabilities and preserve the age boundaries of every rate.

Check Your Understanding

For each problem, identify the interval assumptions before translating between forces, probabilities, and life table entries.

A constant force of \(0.030\) applies from age 45 to 46, followed by a constant force of \(0.020\) from age 46 to 48. Find survival from age 45 to 48 and the corresponding death probability. (Answer: survival \(e^{-0.070}=0.9324\); death probability \(0.0676\).)
A life table has \(\ell_{25}=80{,}000\), \(\ell_{26}=77{,}600\), and \(\ell_{27}=73{,}720\). Find \(q_{25}\), \(q_{26}\), and the probability of surviving from age 25 to 27. (Answer: \(q_{25}=0.0300\), \(q_{26}=0.0500\), and two-year survival \(=73{,}720/80{,}000=0.9215\).)
A constant force of \(0.025\) applies for two years. Find the two-year survival probability and death probability. Explain why using \(1-2(1-e^{-0.025})\) is not the correct survival calculation. (Answer: survival \(e^{-0.050}=0.9512\); death probability \(0.0488\).)
Over one year, a force is \(\mu_{70+t}=0.020+0.010t\) for \(0\le t\le1\). If \(\ell_{70}=50{,}000\), find \(q_{70}\) and \(\ell_{71}\). (Answer: integrated force \(0.025\), \(q_{70}=1-e^{-0.025}=0.0247\), and \(\ell_{71}=50{,}000e^{-0.025}=48{,}765.50\).)
A hypothetical table starts with \(\ell_{70}=120{,}000\), with \(q_{70}=0.025\), \(q_{71}=0.040\), and \(q_{72}=0.050\). Find \(\ell_{73}\) and the three annual death counts. (Answer: \(\ell_{73}=120{,}000(0.975)(0.960)(0.950)=106{,}704\); the death counts are \(3{,}000\), \(4{,}680\), and \(5{,}616\).)

What You'll Learn