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Jan 15'24

You are given the survival function:

[[math]]S_{0}(x)=\left(1-\frac{x}{60}\right)^{\frac{1}{3}}, \quad 0 \leq x \leq 60.[[/math]]

Calculate [math]1000 \mu_{35}[/math].

  • 5.6
  • 6.7
  • 13.3
  • 16.7
  • 20.1

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Jan 15'24

You are given the following survival function of a newborn:

[[math]] S_{0}(x)= \begin{cases}1-\frac{x}{250}, & 0 \leq x\lt40 \\ 1-\left(\frac{x}{100}\right)^{2}, & 40 \leq x \leq 100\end{cases} [[/math]]


Calculate the probability that (30) dies within the next 20 years.

  • 0.13
  • 0.15
  • 0.17
  • 0.19
  • 0.21

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Dec 04'25

You are given the survival function:

[[math]]S_{0}(x)=\left(1-\frac{x}{60}\right)^{k}, \quad 0 \leq x \leq 60[[/math]]

where [math]k[/math] is an unknown constant.

You are also given that the force of mortality at age 30 is [math]\mu_{30} = \frac{1}{12}[/math].

Calculate [math]\mathbf{1000 \mu_{45}}[/math].

  • 22.22
  • 55.56
  • 83.33
  • 133.33
  • 166.67
Dec 04'25

You are given the following survival function of a newborn:

[[math]] S_{0}(x)= \begin{cases}1-\frac{x}{125}, & 0 \leq x\lt25 \\ 0.8 \cdot \exp\left(-\frac{x-25}{50}\right), & 25 \leq x\lt75 \\ \left(1-\frac{x}{100}\right)^{2}, & 75 \leq x \leq 100\end{cases} [[/math]]

Calculate the probability that a life aged [math]\mathbf{20}[/math] survives for at least [math]\mathbf{60}[/math] more years.

  • 0.035
  • 0.048
  • 0.10
  • 0.317
  • 0.95
Jan 15'24

You are given that mortality follows Makeham's Law with the following parameters:

[[math]] \begin{array}{ll} \text { i) } & A=0.004 \\ \text { ii) } & B=0.00003 \\ \text { iii) } & c=1.1 \end{array} [[/math]]


Let [math]L_{15}[/math] be the random variable representing the number of lives alive at the end of 15 years if there are 10,000 lives age 50 at time 0 .

Calculate [math]\operatorname{Var}\left[L_{15}\right][/math].

  • 1,317
  • 1,328
  • 1,339
  • 1,350
  • 1,361

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Dec 04'25

You are given:
(i) The survival function is [math]S_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}[/math], for [math]0 \leq t \leq \omega[/math]
(ii) The derivative of the survival function evaluated at age 65 is [math]\frac{d}{dt} S_{0}(t)|_{t=65} = -\frac{1}{120} \left(\frac{5}{14}\right)^{\frac{3}{4}}[/math]

Calculate [math]\mathring{e}_{106}[/math], the complete expectation of life at age 106.

  • 4.8
  • 12.8
  • 16
  • 27.2
  • 34
Dec 04'25

You are given that mortality follows Gompertz Law with parameters:

  • [math]B = 0.0002[/math]
  • [math]c = 1.08[/math]

A life is currently aged [math]x=30[/math].

Calculate the time [math]t[/math] at which the probability density function of the future lifetime, [math]f_{30}(t)[/math], reaches its maximum value.

  • 43.95
  • 47.35
  • 52.75
  • 77.35
  • 85.05
Dec 05'25

You are given that mortality follows Makeham's Law with the following parameters:

[[math]] \begin{array}{ll} \text { i) } & A=0.001 \\ \text { ii) } & B=0.00011 \\ \text { iii) } & c=1.07 \end{array} [[/math]]

A group of 1,000 lives, all aged 50 at time 0, is observed.

Let [math]L_{20}[/math] be the random variable representing the number of lives alive at the end of 20 years.

Given that 900 lives were observed to be alive at the end of 10 years, calculate the expected number of lives still alive at the end of 20 years.

[[math]]\text{Calculate } E\left[L_{20} \mid L_{10} = 900\right][[/math]]

  • 783
  • 810
  • 855
  • 878
  • 900
Dec 06'25

The probability that the system is still operational at time [math]t[/math] is:

[[math]]S(t) = 1 - \left(\frac{t}{\tau}\right)^\alpha[[/math]]

for [math]0 \leq t \leq \tau[/math], where [math]\tau \gt 0[/math] is the maximum lifetime and [math]\alpha \gt 0[/math] is a shape parameter.

Derive the formula for the [math]n[/math]-th moment of the lifetime as a function of the parameters [math]\tau[/math], [math]\alpha[/math], and [math]n[/math].

  • [math]\frac{\tau^n}{n+\alpha}[/math]
  • [math]\frac{n \tau^n}{n+\alpha}[/math]
  • [math]\frac{\alpha \tau^n}{n+\alpha}[/math]
  • [math]\tau^n \left(1 - \frac{n}{n+\alpha}\right)[/math]
  • [math]\frac{\alpha \tau^\alpha}{n+\alpha}[/math]
Dec 06'25

For a new system, the operational lifetime is governed by the following consistent data:

i) The expected future lifetime at age 50 is 35: [math]\mathbf{e_{50} = 35}[/math]

ii) The one-year mortality rate at age 50 is 0.005: [math]\mathbf{q_{50} = 0.005}[/math]

iii) The one-year survival probability at age 51 is 0.992: [math]\mathbf{p_{51} = 0.992}[/math]

Calculate the expected future lifetime at age 52, [math]\mathbf{e_{52}}[/math].

  • 33.34
  • 33.45
  • 34.17
  • 34.28
  • 34.45