1 Survival Models

1.1 Suplementary Concepts

Let $X$ be the random variable of age-at-death of a newborn. Then $F(x)=\mathrm{Pr}[X \leq x]$ is the cumulative distribution function and $s(x)=1-F(x)$ is the survival function.
A valid survival function $s(x)$ must satisfy five properties:

$s(0)=1$ (normalization condition)
$s(x)$ is non-increasing in $x$ (i.e., survival cannot increase with age)
$\lim_{x\to\infty} s(x)=0$ (i.e., nobody lives forever)
$s(x) \geq 0,\forall x \geq 0$
$s(x)$ must be right-continuous

The conditional probability that a newborn dies between ages $x$ and $z$ given survival to age $x$ is given by

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{Pr}[x < X \leq z] = \frac{F(z) - F(x)}{1 - F(x)} = \frac{s(x) - s(z)}{s(x)} \quad \\ \\ \end{gathered} } \]

The force of mortality (also known as the hazard rate) is defined as

\[ \boxed{ \begin{gathered} \\ \quad \mu_x = \frac{f(x)}{1 - F(x)} = \frac{-s'(x)}{s(x)} = -\frac{d}{dx}\log{s(x)} \quad \\ \\ \end{gathered} } \]

Substituting $x$ by $y$ and reordering, $-\mu_ydy=d\log{s(y)}$. And after integration, we get

\[ \boxed{ \begin{gathered} \\ \quad -\int_x^{x+n}\mu_ydy=\log{\frac{s(x+n)}{s(x)}} \quad \\ \\ \end{gathered} } \]

Finally, through exponentiation of both sides of the equation above, we get

\[ \boxed{ \begin{gathered} \\ \quad \frac{s(x+n)}{s(x)}=\exp{\left( -\int_x^{x+n}\mu_ydy \right) } \quad \\ \\ \end{gathered} } \]

And analogously,

\[ \boxed{ \begin{gathered} \\ \quad s(x) = \exp{\left( -\int_0^{x}\mu_sds \right) } \quad \\ \\ \end{gathered} } \]

The following table relates $F(x)$, $f(x)$, $s(x)$ and $\mu_x$

	$F(x)$	$f(x)$	$s(x)$	$\mu_x$
$F(x)$	$-$	$F'(x)$	$1-F(x)$	$\frac{F'(x)}{1-F(x)}$
$f(x)$	$\int_0^\infty f(r)dr$	$-$	$1-\int_0^\infty f(r)dr$	$\frac{f(x)}{\int_x^\infty f(r)dr}$
$s(x)$	$1-s(x)$	$-s'(x)$	$-$	$\frac{-s'(x)}{s(x)}$
$\mu_x$	$1-\exp{\left( -\int_0^x \mu_rdr\right)}$	$\mu_x \cdot\exp{\left( -\int_0^x \mu_rdr\right)}$	$\exp{\left( -\int_0^x \mu_rdr\right)}$	$-$

The future lifetime of a person aged $x$ is defined by the random variable $T(x)=X-x$, where $X$ is the random variable of age-at-death. Thus, the probability that a person aged $x$ dies before $t$ years is given by

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{Pr}[\,T(x) \leq t\,] = F_x(t) = 1 - s_x(t) = {_t}q_x \quad \\ \\ \end{gathered} } \]

where the symbol $_tq_x$ conforms with the International Actuarial Notation.

Similarly, the probability that $(x)$ survives to age $x+t$ is given by

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{Pr}[\,T(x) > t\,] = s_x(t) = \frac{s_0(x+t)}{s_o(x)} = {_t}p_x \quad \\ \\ \end{gathered} } \]

where the symbol $_tp_x$ conforms with the International Actuarial Notation.

The random variable $K(x)$ denotes the number of complete years of future lifetime at age $x$. A practical way to think of $K$ is the number of future birthday parties that (hopefully!) $(x)$ will celebrate.

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{Pr}[K(x) = k] = \mathrm{Pr}[k \leq T(x) < k+1] = \frac{s_0(x+k)-s_0(x+k+1)}{s_0(x)} = {_{k\vert}}q_x \quad \\ \\ \end{gathered} } \]

where the symbol ${_{k\vert}}q_x$ conforms with the International Actuarial Notation.

Life expectancy at age $x$ is given by

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{E}[\,T(x)\,] = \int_0^\infty t\cdot f_x(t)\,dt = \int_0^\infty t\cdot {_t}p_x \,\mu_{x+t}\,dt = \int_0^\infty {_t}p_x\,dt \quad \\ \\ \end{gathered} } \]

Analogously, the expected value of $K$ is given by

\[ \boxed{ \begin{gathered} \\ \quad \mathrm{E}[\,K(x)\,] = \sum_0^\infty k\cdot \mathrm{Pr}[\,K(x)=k\,] = \sum_0^\infty k\cdot {_{k\vert}}q_x = \sum_0^\infty {_k}p_x \quad \\ \\ \end{gathered} } \]

Let $l_x$ be the number of survivors of exactly age $x$ from a group of $l_0$ newborns. Thus,

$d_x=l_x-l_{x+1}$ is the number of deaths between ages $x$ and $x+1$;
$q_x=\frac{d_x}{l_x}$ is the probability of dying between ages $x$ and $x+1$;
$p_x=1-q_x=\frac{l_{x+1}}{l_x}$ is the probability of surviving to age $x+1$;

If $l_0$ and $q_x$ values are available, a mortality table can be constructed as follows:

$l_1=l_0(1-q_0)=l_0-d_0$
$l_2=l_1(1-q_1)=l_1-d_1=l_0-(d_0+d_1)$$</li> <li id="ind50">$$</li> <li id="ind50">$l_x=l_{x-1},(1-q_{x-1})=l_{x-1}-d_{x-1}=l_0-_{k=0}^{x-1}d_k$

The average number of years lived between ages $x$ and $x+1$ by an initial group of $l_0$ newborns is given by

\[ \boxed{ \begin{gathered} \\ \quad L_x = \int_0^1 t\cdot {_t}p_x \,\mu_{x+t}\,dt\, +\, l_{x+1} = \int_0^1 l_{x+t}\,dt \quad \\ \\ \end{gathered} } \]

The central rate of mortality is given by

\[ \boxed{ \begin{gathered} \\ \quad m_x = \frac{\int_0^1 l_{x+t}\,\mu_{x+t} \, dt}{\int_0^1 l_{x+t}\,dt} = \frac{d_x}{L_x} \quad \\ \\ \end{gathered} } \]

1.2 Exercises

Let $s(x)= \large \frac{20,000\,-\,100x\,-\, x^2}{20,000}$.

verify that $s(x)$ satisfies the conditions of a survival function
determine $\omega$
calculate the probability that $(0)$ survives to age $35$
calculate the probability that $(15)$ dies before age $42$
calculate the probability that $(20)$ dies between ages $50$ and $70$

Let $s(x)=1-0.01x$, $0\leq x\leq 100$.

Calculate the probability that

$(0)$ reaches age $25$
$(0)$ dies between ages $15$ and $42$
$(15)$ dies before age $42$
$(36)$ survives to age $64$

Let $\varphi(x)= \large{-\frac{d}{dx}}\normalsize s(x)$.

Show that

$\int_{0}^{\omega}\varphi(x),dx=1$
$\int_{x}^{x+n}\varphi(y),dy$ is the probability that a newborn dies between ages $x$ and $x+n$
$s(x)=1-\int_{0}^{x}\varphi (y),dy$
If $\varphi(x)=\frac {2}{75}x^{-{\frac{1}{3}}}$, find $s(x)$ and $\omega$

Let $s(x)=exp{(\frac{-x^3}{12})}$, $x \geq0$.

Verify that $s(x)$ satisfies the conditions of a survival function.

Let $s(x)=e^{-2x}$, $x \geq0$.

Compute

The expected value of $X$
The median of $X$
The mode of $X$
The standard deviation of $X$

Let $s(x)=(1+x)^{-4}$, $x \geq0$.

Calculate the expected number of years of future lifetime for a person aged $41$.

Let $s(x)=1-0.008,x$, $0\leq x\leq 125$

Calculate the number of years people aged $70$ are expected to live before reaching $90$.

Let $f(x)=(20\sqrt {100-x})^{-1})$, $0\leq x\leq100$.

Calculate the probability that a person aged $69.75$ dies within the next decade.

Let $\mu_{x}=(4+x)^{-1}$, $x \geq0$.

Calculate the probability that the age at death of a newborn exceeds five.

Complete the following table:

$s(x)$	$F(x)$	$f(x)$	$\mu (x)$
-	-	-	$(110-x)^{-1}$, $0\leq x\leq 100$
$e^{-x}$, $x \geq 0$	-	-	-
-	$1-(1+x)^{-1}$, $x \geq0$	-	-

Assuming a constant force of mortality, the future lifetime of a person aged $x$ is two years.

Calculate the $75$th percentile of the distribution of $T$.

Let $\mu_{x}=3,(110-x)^{-1}+(150-x)^{-1}$, $0 \leq x\leq 110$.

Calculate $_{70}p_{10}$.

Let $z(w)=\exp(-\int_{x}^{x+n} \mu_{y} dy)$.

State in terms of $z(w)$ the probability that $(x)$ dies between ages $n$ and $m$, $x<n<m$.

Let $s(x)=exp(-ex)$, $x>0$.

Find the average age at death of those currently aged $10e$.

Assume the following cumulative distribution function for $T$:

\[ F(t) = \left{ \begin{array}{ll} t \cdot (100-x)^{-1} & 0 \leq t < 100-x \ 0 & \text{otherwise} \ \end{array} \right. \]

Compute:

$\mathrm{E[T]}$
$\mathrm{Var[T]}$
$\text{median}[T]$

Let $\mu_{x}=1, \forall{x}$.

Calculate $E[T-K]$.

Let $s(x)=1-0.005x-0.0005x^{2}$, $0 \leq x \leq 100$.

Construct the $l_{x}$, $d_{x}$, $q_{x}$ and $p_{x}$ columns of the corresponding life table for ages $0, 1, 2$, assuming a radix of $100,000$.

From the following values of $q_{x}$, construct the $l_{x}$ and $d_{x}$ columns, assuming a radix of $1,000,000$.

$x$	$q_{x}$
$0$	$0.011$
$1$	$0.005$
$2$	$0.003$

From the following values of $l_{x}$,

$x$	$l_{x}$
$30$	$94\,401$
$35$	$93\,589$
$55$	$82\,463$
$64$	$67\,970$
$65$	$65\,834$
$66$	$63\,603$

calculate the probability that $(35)$

lives at least $30$ years
dies within the next $30$ years
dies in the thirtieth year from the current age
dies between ages $55$ and $65$

Let $l_{x}=1000-2x$, $0 \leq x \leq 500$.

Calculate the expected number of complete future years of life at age $497$.

Suppose $\int_{10}^{\infty}{*t}p*{50}dt=15$ and $\int_{0}^{\infty}{*t}p*{60}dt=20$.

Calculate $*{10}p*{50}$.

Let $s(x)=e^{-0.2x}-e^{-0.4x}$, $x>5\log 2$.

Calculate the mode of the distribution of the number of deaths in a group of people currently aged $x=5\log 2$.

Let $*{t}p*{x}=1-0.6t+0.12t^{2}-0.008t^{3}$, $0 \leq x \leq 5$.

Calculate $\mu_{x+3}$.

Let $l_{x}=100-15x-6x^{2}+x^{3}$, $0 \leq x \leq 5$.

From a group of $100$ newborns, calculate the maximum number of deaths in any one-year interval.

Let $l_{1}=10,000$ and $d_{x}=x(x+1)$, $x=1, 2, 3, \dots , 30$.

Calculate $*{30}q*{1}$.

Let $L{x}=(40-x)^{2}$, $0 \leq x \leq 40$.

Calculate the central death rate at age $35$.

Let $s(x)=1-\frac{x}{110}$, $0 \leq x < 110$.

Calculate

the force of mortality at age $25$
the life expectancy at age $25$

The mortality pattern of $50,000$ newborns is such that $l_{x}\mu_{x}$ is constant, $0 \leq x \leq \omega$.

If $\frac{d^{6}}{dx^{6}}=\frac{45}{8}$ for $x=58$, calculate the expected number of survivors at age $42$.

Let $p_{x}=0.944$, $p_{x+1}=0.938$, $*{3}p*{x+1}=0.811$ and $q_{x+3}=0.0.074$.

Calculate the probability that $(x)$ dies between ages $x$ and $x+1$.

Let $s(x)=(1+x)^{-0.8}$, $x \geq 0$.

Calculate the $69$th percentile of the distribution of the random variable $T$ for future lifetime, at age $37$.

1.3 Code snippets

Definite integral:

The integral $\int_0^{25} t^2e^{-0.05t}dt$ can be evaluated as follows:

f <- function(t) {t^2 * exp(-0.05 * t)}

integrate(f, 0, 25)$value

[1] 2104.517

Calculation of present value of cash flows:

Consider the following sequence of annual cash flows of $2\,000, 3\,000, ..., 21\,000$.

To compute the present value of these cash flows at $5\%$ effective, we use:

(cf <- seq(2, 21))

 [1]  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21

(dsct <- 1.05 ^ -(1:20))

 [1] 0.9523810 0.9070295 0.8638376 0.8227025 0.7835262 0.7462154 0.7106813
 [8] 0.6768394 0.6446089 0.6139133 0.5846793 0.5568374 0.5303214 0.5050680
[15] 0.4810171 0.4581115 0.4362967 0.4155207 0.3957340 0.3768895

(sum(cf * dsct))

[1] 123.4128

	\(F(x)\)	\(f(x)\)	\(s(x)\)	\(\mu_x\)
\(F(x)\)	\(-\)	\(F'(x)\)	\(1-F(x)\)	\(\frac{F'(x)}{1-F(x)}\)
\(f(x)\)	\(\int_0^\infty f(r)dr\)	\(-\)	\(1-\int_0^\infty f(r)dr\)	\(\frac{f(x)}{\int_x^\infty f(r)dr}\)
\(s(x)\)	\(1-s(x)\)	\(-s'(x)\)	\(-\)	\(\frac{-s'(x)}{s(x)}\)
\(\mu_x\)	\(1-\exp{\left( -\int_0^x \mu_rdr\right)}\)	\(\mu_x \cdot\exp{\left( -\int_0^x \mu_rdr\right)}\)	\(\exp{\left( -\int_0^x \mu_rdr\right)}\)	\(-\)

\(s(x)\)	\(F(x)\)	\(f(x)\)	\(\mu (x)\)
-	-	-	\((110-x)^{-1}\), \(0\leq x\leq 100\)
\(e^{-x}\), \(x \geq 0\)	-	-	-
-	\(1-(1+x)^{-1}\), \(x \geq0\)	-	-

\(x\)	\(q_{x}\)
\(0\)	\(0.011\)
\(1\)	\(0.005\)
\(2\)	\(0.003\)

\(x\)	\(l_{x}\)
\(30\)	\(94\,401\)
\(35\)	\(93\,589\)
\(55\)	\(82\,463\)
\(64\)	\(67\,970\)
\(65\)	\(65\,834\)
\(66\)	\(63\,603\)