Fractional Age Assumption

Study guides: Exam FAM, Exam ALTAM

A Fractional Age Assumption fixes how deaths, or other decrements, are spread within each year of age. With it, a life table that gives only $\ell_x$ or $q_x$ at integer ages also yields ${}_sp_x$ , ${}_sq_x$ and the force of mortality $\mu_{x+s}$ for $0 < s < 1$ . The two in AMLCR Section 3.3 are uniform distribution of deaths (UDD) and constant force of mortality.

$\text{UDD:}\quad {}_sq_x = s\,q_x$

$\text{UDD:}\quad \mu_{x+s} = \frac{q_x}{1 - s\,q_x}$

$\text{Constant force:}\quad {}_sp_x = \left(p_x\right)^s$

$\text{Constant force:}\quad \mu_{x+s} = -\ln p_x$

What UDD means. $\ell_{x+s}$ is linear in $s$ , so the density ${}_sp_x\,\mu_{x+s} = q_x$ is flat over the year. Equivalently, $T_x = K_x + R$ , where $R$ is uniform on $(0,1)$ and independent of the curtate lifetime $K_x$ . This gives $\mathring{e}_x = e_x + \tfrac12$ . Under a constant force, $\ell_{x+s}$ is exponential in $s$ and $\mu$ is flat within the year.
Within a year, not from age $x$ . For $0 \le t < t+s \le 1$ , UDD gives ${}_sq_{x+t} = s\,q_x/(1 - t\,q_x)$ , while a constant force gives ${}_sp_{x+t} = (p_x)^s$ . Across an integer age, split the period and multiply the pieces.
Insurance values (FAM). Under UDD, $\bar{A}_x = \tfrac{i}{\delta}A_x$ and $A^{(m)}_x = \tfrac{i}{i^{(m)}}A_x$ . The claims acceleration approximation $\bar{A}_x \approx (1+i)^{1/2}A_x$ assumes instead that deaths are paid, on average, half a year early.
Multiple decrements (ALTAM). UDD in the multiple decrement table, or constant forces, gives $p'^{(j)}_x = \left(p^{(\tau)}_x\right)^{q^{(j)}_x/q^{(\tau)}_x}$ . UDD in each single decrement table instead gives, for two decrements, $q^{(1)}_x = q'^{(1)}_x\left(1 - \tfrac12 q'^{(2)}_x\right)$ .

Survival Across a Birthday Given $q_{60} = 0.02$ and $q_{61} = 0.025$ , find ${}_{0.75}q_{60.5}$ under UDD and under a constant force of mortality.
Answer Split at age 61: ${}_{0.75}p_{60.5} = {}_{0.5}p_{60.5}\cdot{}_{0.25}p_{61}$ .
 $\begin{align*} \text{UDD:}\ {}_{0.75}p_{60.5} &= \frac{p_{60}}{1 - 0.5\,q_{60}}\left(1 - 0.25\,q_{61}\right) \\ &= \frac{0.98}{0.99}(0.99375) \\ &= 0.983712 \\ \text{Constant force:}\ {}_{0.75}p_{60.5} &= (0.98)^{0.5}(0.975)^{0.25} \\ &= 0.983703 \end{align*}$ 
So ${}_{0.75}q_{60.5} = 0.016288$ under UDD and $0.016297$ under a constant force. When $q$ is small the two assumptions agree to about five decimals.

From Annual to Continuous Insurance Values On the SULT at $i = 5\%$ , $A_{50} = 0.18931$ . Find $\bar{A}_{50}$ under UDD and by claims acceleration, and find $A^{(12)}_{50}$ under UDD.
Answer
 $\begin{align*} \bar{A}_{50} &= \frac{i}{\delta}A_{50} \\ &= \frac{0.05}{\ln 1.05}(0.18931) \\ &= 1.024797(0.18931) \\ &= 0.19400 \\ \bar{A}_{50} &\approx (1.05)^{1/2}(0.18931) \\ &= 0.19399 \\ A^{(12)}_{50} &= \frac{0.05}{0.048889}(0.18931) \\ &= 0.19361 \end{align*}$ 
Paying at the moment of death raises the value by about $2.5\%$ . Monthly payment raises it by a little less.

Deconstructing a Multiple Decrement Table A two-decrement model has $q^{(1)}_x = 0.03$ (death) and $q^{(2)}_x = 0.12$ (withdrawal). Assuming UDD in the multiple decrement table, find the associated single decrement rates $q'^{(1)}_x$ and $q'^{(2)}_x$ .
Answer $q^{(\tau)}_x = 0.15$ , so $p^{(\tau)}_x = 0.85$ .
 $\begin{align*} p'^{(1)}_x &= (0.85)^{0.03/0.15} \\ &= 0.968019 \\ p'^{(2)}_x &= (0.85)^{0.12/0.15} \\ &= 0.878082 \end{align*}$ 
So $q'^{(1)}_x = 0.03198$ and $q'^{(2)}_x = 0.12192$ , and $p'^{(1)}_x\,p'^{(2)}_x = 0.85$ . Each independent rate is higher than its dependent rate. With withdrawal removed, more lives remain exposed to death, and the reverse holds for withdrawal.

How it appears on the exam

FAM gives a life table at integer ages and asks for a probability, $\mu_{x+s}$ , or the complete expectation $\mathring{e}_x$ or ${}_n\mathring{e}_x$ at a fractional age or duration. The question states which assumption to use.

FAM converts annual insurance values to continuous or $m$ -thly ones with $i/\delta$ or $i/i^{(m)}$ under UDD, or with claims acceleration. These are then used in a premium or variance calculation.

ALTAM builds a multiple decrement table from single decrement rates, or recovers them from it, under UDD in the multiple decrement table or in the single decrement tables. The table then feeds a premium, policy value or pension valuation.

Exam traps

${}_sq_{x+t} \neq s\,q_x$ . Under UDD the denominator is $1 - t\,q_x$ . Dropping it understates the probability slightly, which matters when the answer choices are close.

Mixing the assumptions. $(p_x)^s$ is the constant-force answer, while $1 - s\,q_x$ is UDD. A question that says UDD and is worked with exponents lands on the neighbouring choice.

Annuities do not convert like insurance. $\bar{a}_x = (1 - \bar{A}_x)/\delta$ must be built from the converted $\bar{A}_x$ . Multiplying $\ddot{a}_x$ by $i/\delta$ is wrong.

Two kinds of multiple decrement UDD. UDD in the multiple decrement table, or constant forces, uses the power formula. UDD in each single decrement table uses $q'^{(1)}\left(1 - \tfrac12 q'^{(2)}\right)$ . They give different tables.