Exercise


ABy Admin
Nov 19'23

Answer

Solution: C

[[math]]\begin{aligned} & 100(1+0.05)^2 \exp \left(\int_2^5 \frac{1}{1+t}\right)=100(1.1025) \exp \left[\left.\ln (1+t)\right|_2 ^5\right] \\ & =110.25\left(\frac{6}{3}\right)=220.50 \\ & \frac{100}{(1-d)^5}=220.50 \\ & (1-d)^5=\frac{100}{220.50}=0.453515 \\ & 1-d=0.853726 \\ & d=0.1463\end{aligned}[[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

00