Revision as of 01:58, 19 January 2024 by Admin (Created page with "For a fully continuous whole life insurance policy of 100,000 on (35), you are given: (i) The density function of the future lifetime of a newborn: <math display="block"> f(t)= \begin{cases}0.01 e^{-0.01 t}, & 0 \leq t<70 \\ g(t), & t \geq 70\end{cases} </math> (ii) <math>\delta=0.05</math> (iii) <math>\quad \bar{A}_{70}=0.51791</math> Calculate the annual net premium rate for this policy. <ul class="mw-excansopts"><li> 1000<li> 1110</li><li> 1220<li> 1330<li> 1...")
ABy Admin
Jan 19'24
Exercise
For a fully continuous whole life insurance policy of 100,000 on (35), you are given:
(i) The density function of the future lifetime of a newborn:
[[math]]
f(t)= \begin{cases}0.01 e^{-0.01 t}, & 0 \leq t\lt70 \\ g(t), & t \geq 70\end{cases}
[[/math]]
(ii) [math]\delta=0.05[/math]
(iii) [math]\quad \bar{A}_{70}=0.51791[/math]
Calculate the annual net premium rate for this policy.
- 1000
- 1110
- 1220
- 1330
- 1440
ABy Admin
Jan 19'24
Answer: D
[math]\bar{A}_{35}=\left(1-e^{-35(\mu+\delta)}\right) \times\left(\frac{\mu}{\mu+\delta}\right)+e^{-35(\mu+\delta)} \bar{A}_{70}=0.063421+0.146257=0.209679[/math]
[math]\bar{a}_{35}=\frac{1-\bar{A}_{35}}{\delta}=\frac{1-0.209679}{0.05}=15.80642[/math]
[math]\bar{P}_{35}=\frac{\bar{A}_{35}}{\bar{a}_{35}}=\frac{0.209679}{15.80642}=0.0132654[/math]
The annual net premium for this policy is therefore [math]100,000 \times 0.0132654=1,326.54[/math]