An insurance company sells 15 -year pure endowments of 10,000 to 500 lives, each age [math]x[/math], with independent future lifetimes. The single premium for each pure endowment is determined by the equivalence principle.

(i) You are given:

(ii) [math]\quad i=0.03[/math]

(iii) [math]\quad \mu_{x}(t)=0.02 t, \quad t \geq 0[/math]

(iv) [math]{ }_{0} L[/math] is the aggregate loss at issue random variable for these pure endowments.

Using the normal approximation without continuity correction, calculate [math]\operatorname{Pr}\left({ }_{0} L\gt50,000\right)[/math].

• 0.08
• 0.13
• 0.18
• 0.23
• 0.28

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

Answer: B

The probability that the endowment payment will be made for a given contract is:

[[math]] \begin{aligned} { }_{15} p_{x} & =\exp \left(-\int_{0}^{15} 0.02 t d t\right) \\ & =\exp \left(-\left.0.01 t^{2}\right|_{0} ^{15}\right) \\ & =\exp \left(-0.01(15)^{2}\right) \\ & =0.1054 \end{aligned} [[/math]]

Because the premium is set by the equivalence principle, we have [math]E\left[{ }_{0} L\right]=0[/math]. Further,

[[math]] \begin{aligned} \operatorname{Var}\left({ }_{0} L\right) & =500\left[\left(10,000 v^{15}\right)^{2}\left({ }_{15} p_{x}\right)\left(1-{ }_{15} p_{x}\right)\right] \\ & =1,942,329,000 \end{aligned} [[/math]]

Then, using the normal approximation, the approximate probability that the aggregate losses exceed 50,000 is

[math]P\left({ }_{0} L\gt50,000\right)=P\left(Z\gt\frac{50,000-0}{\sqrt{1,942,329,000}}\right)=P(Z\gt1.13)=0.13[/math]

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