Life Table

Study guides: Exam FAM, Exam MAS-I

A Life Table is the tabular form of a Survival Model: starting from a radix $\ell_0$ of newborns, it records $\ell_x$ , the expected number still alive at age $x$ , and $d_x = \ell_x - \ell_{x+1}$ , the number dying between ages $x$ and $x+1$ . Every survival and mortality probability the exam asks for can be read off these two columns.

$_{t}p_x = \frac{\ell_{x+t}}{\ell_x}, \qquad _{t}q_x = 1 - {}_{t}p_x = \frac{\ell_x - \ell_{x+t}}{\ell_x}$

$e_x = \sum_{t=1}^{\infty} {}_{t}p_x = \frac{\ell_{x+1} + \ell_{x+2} + \cdots}{\ell_x}$

$_{t}p_x$ is the probability a life aged $x$ survives $t$ more years; $_{t|}q_x = {}_{t}p_x \cdot q_{x+t}$ is the probability of dying in the year after surviving $t$ years (a deferred death probability)
Survival probabilities multiply across periods: $_{s+t}p_x = {}_{s}p_x \cdot {}_{t}p_{x+s}$ , which is why $\ell_x$ alone is enough to answer any question
The table is the discrete counterpart of the continuous survival function: $\ell_x = \ell_0\, S(x)$ , and the Hazard Rate (force of mortality) is $\mu_x = -\frac{d}{dx}\ln \ell_x$
The curtate expectation $e_x$ counts complete years survived; if deaths are uniformly distributed within each year of age, the complete expectation is exactly $\mathring{e}_x = e_x + \tfrac12$ (under a constant force it is only an approximation, close when death rates are small)
Fractional-age assumptions fill in between integer ages — UDD gives $_{s}q_x = s\,q_x$ , a constant force gives $_{s}p_x = (p_x)^s$
Life-table probabilities feed straight into Whole Life Insurance and Life Annuity values, and into Joint Life statuses when two independent lives are combined by multiplying their survival probabilities

Reading Probabilities off a Table A life table gives $\ell_{60} = 8{,}000$ , $\ell_{61} = 7{,}840$ , $\ell_{62} = 7{,}650$ , $\ell_{63} = 7{,}430$ . Find $q_{60}$ , $_{2}p_{60}$ , and $_{2|}q_{60}$ .

Answer $q_{60} = \frac{\ell_{60} - \ell_{61}}{\ell_{60}} = \frac{160}{8{,}000} = 0.02$ $_{2}p_{60} = \frac{\ell_{62}}{\ell_{60}} = \frac{7{,}650}{8{,}000} = 0.95625$ $_{2|}q_{60} = \frac{\ell_{62} - \ell_{63}}{\ell_{60}} = \frac{220}{8{,}000} = 0.0275$

Curtate Expectation from a Terminating Table A table ends at age $63$ with $\ell_{63} = 0$ and has $\ell_{60} = 8{,}000$ , $\ell_{61} = 7{,}840$ , $\ell_{62} = 7{,}650$ . Find $e_{60}$ .

Answer $e_{60} = \frac{\ell_{61} + \ell_{62} + \ell_{63}}{\ell_{60}} = \frac{7{,}840 + 7{,}650 + 0}{8{,}000} = 1.936 \text{ years}$ A life aged $60$ completes on average $1.936$ further whole years; if deaths are uniform within each year, the complete expectation of life is $1.936 + 0.5 = 2.436$ years.