Parametric Survival Model

Study guide: Exam FAM

A Parametric Survival Model specifies the distribution of the future lifetime by a formula with a few parameters, usually a law for the force of mortality $\mu_x$ , so that every survival probability and moment follows by integration. A life table is the nonparametric alternative, listing $\ell_x$ or $q_x$ at integer ages. The two are linked, because a law generates a table and a table can be fitted by a law.

${}_tp_x = \exp\left(-\int_0^t \mu_{x+s}\,ds\right)$

$\text{Makeham: } \mu_x = A + Bc^x$

$\text{Makeham: } {}_tp_x = \exp\left(-At - \frac{Bc^x\left(c^t - 1\right)}{\ln c}\right)$

$\text{Weibull: } {}_tp_x = \exp\left(-\frac{k\left[(x+t)^{n+1} - x^{n+1}\right]}{n+1}\right)$

$\text{de Moivre: } {}_tp_x = \frac{\omega - x - t}{\omega - x}$

The laws: Gompertz's law is Makeham's with $A = 0$ , a force that grows exponentially with age, and Makeham's constant $A$ adds an age-independent risk such as accidents. Weibull's law has $\mu_x = kx^n$ . The exponential model has a constant force $\mu$ , so ${}_tp_x = e^{-\mu t}$ at every age. De Moivre's law has $\mu_x = 1/(\omega - x)$ , so $T_x$ is uniform on $(0, \omega - x)$ .
Moments: Under a constant force, $\mathring{e}_x = 1/\mu$ and $\text{Var}(T_x) = 1/\mu^2$ , as for any exponential distribution. Under de Moivre's law, $\mathring{e}_x = (\omega - x)/2$ and $\text{Var}(T_x) = (\omega - x)^2/12$ . Gompertz and Makeham moments have no closed form, so they are found numerically or from a table.
From law to table: A law generates a table through $\ell_{x+t} = \ell_x\,{}_tp_x$ . The Standard Ultimate Life Table (SULT) used on FAM is Makeham's law with $A = 0.00022$ , $B = 2.7 \times 10^{-6}$ and $c = 1.124$ , tabulated from a radix at age 20. A table gives values only at integer ages and needs a fractional age assumption between them, while a law gives ${}_tp_x$ at every age directly.
Strengths and limits: A law is smooth, needs few parameters and extrapolates to the oldest ages, where data are thin. No simple law fits the whole human mortality curve, with its infant mortality and its accident hump in early adulthood, so tables are used where accuracy at every age matters.

The SULT as a Table Generated by Makeham's Law Under the SULT, find ${}_{10}p_{65}$ and ${}_{20}p_{65}$ from Makeham's law and use them to find ${}_{10|10}q_{65}$ . Check the answer against the table values $\ell_{65} = 94{,}579.7$ , $\ell_{75} = 85{,}203.5$ and $\ell_{85} = 61{,}184.9$ .
Answer Here $Bc^{65} = 0.00538485$ and $\ln c = 0.116894$ .
 $\begin{align*} \int_0^{10}\mu_{65+s}\,ds &= 0.00022(10) + \frac{0.00538485(1.124^{10} - 1)}{0.116894} \\ &= 0.0022 + 0.102201 \\ &= 0.104401 \\ {}_{10}p_{65} &= e^{-0.104401} \\ &= 0.90086 \\ \int_0^{20}\mu_{65+s}\,ds &= 0.0044 + \frac{0.00538485(1.124^{20} - 1)}{0.116894} \\ &= 0.0044 + 0.431142 \\ &= 0.435542 \\ {}_{20}p_{65} &= e^{-0.435542} \\ &= 0.64691 \end{align*}$ 
So ${}_{10|10}q_{65} = 0.90086 - 0.64691 = 0.25395$ . The table gives $(85{,}203.5 - 61{,}184.9)/94{,}579.7 = 0.25395$ , the same value, because the table was generated from the law.

Two Models with the Same Expectation of Life For a life aged 40, compare de Moivre's law with $\omega = 100$ and a constant force $\mu = 1/30$ . Find $\mathring{e}_{40}$ , ${}_{30}p_{40}$ and $\text{Var}(T_{40})$ under each.
Answer Under de Moivre's law $T_{40}$ is uniform on $(0, 60)$ , and under the constant force it is exponential with mean 30.
 $\begin{align*} \text{de Moivre: } \mathring{e}_{40} &= 60/2 \\ &= 30 \\ {}_{30}p_{40} &= 30/60 \\ &= 0.5 \\ \text{Var}(T_{40}) &= 60^2/12 \\ &= 300 \\ \text{Exponential: } \mathring{e}_{40} &= 1/\mu \\ &= 30 \\ {}_{30}p_{40} &= e^{-1} \\ &= 0.3679 \\ \text{Var}(T_{40}) &= 30^2 \\ &= 900 \end{align*}$ 
The means agree, but the exponential model kills more lives early and lets some live without limit, so its variance is three times as large. Its force is also flat, while de Moivre's force rises from $1/60$ at age 40 to $1/40$ at age 60.

Fitting Gompertz's Law to Two Forces Mortality follows Gompertz's law, $\mu_x = Bc^x$ , with $\mu_{50} = 0.003$ and $\mu_{60} = 0.009$ . Find $c$ and ${}_{10}p_{60}$ .
Answer Dividing the two forces gives $c^{10} = 3$ , so $c = 3^{0.1} = 1.116123$ and $\ln c = 0.109861$ . The Gompertz survival probability is ${}_tp_x = \exp\left(-\mu_x(c^t - 1)/\ln c\right)$ .
 $\begin{align*} {}_{10}p_{60} &= \exp\left(-\frac{0.009(3 - 1)}{0.109861}\right) \\ &= e^{-0.163843} \\ &= 0.84888 \end{align*}$ 
$B$ is never needed, because $Bc^{60}$ is the given force $\mu_{60}$ .

How it appears on the exam

FAM gives a law (constant force, de Moivre, Gompertz, Makeham or Weibull) with its parameters and asks for ${}_tp_x$ , ${}_tq_x$ , $\mu_x$ , $\mathring{e}_x$ or $\text{Var}(T_x)$ at a stated age.

Relationship to tables: A question can give a few table values and ask which law produced them, or ask for a parameter from two forces or two survival probabilities, as in the Gompertz example.

SULT questions: The SULT is itself Makeham's law, so a question can ask for a fractional-age probability directly from the law where a table alone would need UDD or a constant force.

Exam traps

Age in the Makeham term: The factor is $Bc^x$ at the current age $x$ , not $Bc^0$ or $Bc^t$ . Using $B$ alone treats the life as a newborn.

Dividing by $\ln c$ : The integral of $Bc^{x+s}$ over $(0, t)$ is $Bc^x(c^t - 1)/\ln c$ . Forgetting the $\ln c$ understates the integral by a factor of about 8.6 for the SULT.

De Moivre at age $x$ : $T_x$ is uniform on $(0, \omega - x)$ , not on $(0, \omega)$ . Using $\omega$ for a life aged $x$ overstates $\mathring{e}_x$ by $x/2$ .

Constant force and age: Under a constant force, ${}_tp_x$ does not depend on $x$ . The age given with a constant force plays no part in the answer, and adjusting for it is an error.