For a fully discrete 3 -year endowment insurance of 1000 on [math](x)[/math], you are given:

(i) [math]\quad \mu_{x+t}=0.06[/math], for [math]0 \leq t \leq 3[/math]

(ii) [math]\delta=0.06[/math]

(iii) The annual premium is 315.80

(iv) [math]\quad L_{0}[/math] is the present value random variable for the loss at issue for this insurance Calculate [math]\operatorname{Pr}\left[L_{0}\gt0\right][/math].

• 0.03
• 0.06
• 0.08
• 0.11
• 0.15

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

Answer: D

The policy is fully discrete, so all cash flows occur at the start or end of a year.

Die Year 1[math] \implies L_{0}=1000 v-315.80=625.96[/math]

Die Year 2[math]\implies L_{0}=1000 v^{2}-315.80(1+v)=273.71[/math]

Survive Year 2[math]\implies L_{0}=1000 v^{3}-315.80\left(1+v+v^{2}\right)=-58.03[/math]

There is a loss if death occurs in year 1 or year 2, otherwise the policy was profitable.

[math]\operatorname{Pr}([/math] death in year 1 or 2[math])=1-e^{-2 \mu}=0.113[/math]

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