Exercise
For fully discrete whole life insurance policies of 10,000 issued on 600 lives with independent future lifetimes, each age 62 , you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) [math]\quad i=0.05[/math]
(iii) Expenses of [math]5 \%[/math] of the first year gross premium are incurred at issue
(iv) Expenses of 5 per policy are incurred at the beginning of each policy year
(v) The gross premium is [math]103 \%[/math] of the net premium.
(vi) [math]{ }_{0} L[/math] is the aggregate present value of future loss at issue random variable
Calculate [math]\operatorname{Pr}\left({ }_{0} L\lt40,000\right)[/math], using the normal approximation.
- 0.75
- 0.79
- 0.83
- 0.87
- 0.91
Answer: B
Net Premium [math]=10,000 A_{62} / \ddot{a}_{62}=10,000(0.31495) / 14.3861=218.93[/math]
[math]G=218.93(1.03)=225.50[/math]
Let [math]{ }_{0} L^{*}[/math] be the present value of future loss at issue for one policy.
[math]E\left({ }_{0} L^{*}\right)=14,630.50 A_{62}-4619.22=14,630.50(0.31495)-4619.22=-11.34[/math]
[math]\operatorname{Var}\left({ }_{0} L^{*}\right)=(14,630.50)^{2}\left({ }^{2} A_{62}-A_{62}^{2}\right)=(14,630.50)^{2}\left(0.12506-0.31495^{2}\right)=5,536,763[/math]
Let [math]{ }_{0} L[/math] be the aggregate loss for 600 such policies.
[math]\operatorname{Var}\left({ }_{0} L\right)=600 \operatorname{Var}\left({ }_{0} L^{*}\right)=600(5,536,763)=3,322,057,800[/math]
[math]\operatorname{StdDev}\left({ }_{0} L\right)=3,322,057,800^{0.5}=57,637[/math]
[math]\operatorname{Pr}\left({ }_{0} L\lt40,000\right)=\Phi\left(\frac{40,000+6804}{57,637}\right)=\Phi(0.81)=0.7910[/math]