Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Mortality Models Practice Questions

In a life table:
- $l_x$ = expected number alive at exact age $x$
- $d_x$ = expected number of deaths between ages $x$ and x+1

Therefore: $d_x = l_x - l_{x+1}$

Equivalently: $d_x = l_x \cdot q_x$ (number alive times probability of death).

Choice (B) gives a negative number. Choice (C) gives $l_x \cdot p_x = l_{x+1}$, the survivors. Choices (D) and (E) are not meaningful in this context.

Statement (A) is FALSE. Select mortality rates are typically LOWER than the corresponding ultimate mortality rates at the same attained age. This is because underwriting at issue selects for healthier lives, so a recently underwritten life has lower mortality than someone of the same age who was underwritten years ago.

For example, $q_{[40]} < q_{[39]+1} < q_{40}$ (attained age 40, but the most recently selected life has the lowest mortality).

All other statements are true:
- (B) Selection does arise from underwriting.
- (C) The select period is the duration after which selection effects wear off.
- (D) [x]+s correctly denotes a life selected at age $x$, now $s$ years later.
- (E) After the select period, mortality follows the ultimate table.

For Gompertz, the survival probability is:
$${}_{10}p_x = \exp\left(-\frac{Bc^x(c^{10} - 1)}{\ln c}\right)$$

Taking logarithms of each:
$$\ln({}_{10}p_{50}) = -\frac{Bc^{50}(c^{10}-1)}{\ln c}$$
$$\ln({}_{10}p_{60}) = -\frac{Bc^{60}(c^{10}-1)}{\ln c}$$

Dividing:
$$\frac{\ln({}_{10}p_{60})}{\ln({}_{10}p_{50})} = \frac{c^{60}}{c^{50}} = c^{10}$$

$$\ln(0.920) = -0.08338$$
$$\ln(0.850) = -0.16252$$

$$c^{10} = \frac{-0.16252}{-0.08338} = 1.9492 \approx 1.949$$

The answer is 1.949.

Why other choices fail:
- Choice A (2.412): Incorrect formula or inverted ratio.
- Choice D (1.082): $0.920/0.850 = 1.082$. Ratio of probabilities, not logs.
- Choice C (1.649): $\sqrt{1.949} = 1.396$, not 1.649. Incorrect square root.
- Choice E (3.120): Sums logs: $-0.08338 + (-0.16252) = -0.2459$. Meaningless.