Exercise


ABy Admin
Nov 18'23

Answer

Solution: D

[[math]] \overline{{{a}}}_{\overline{{{20}}|}}=1.5\overline{{{a}}}_{\overline{{{10}}|}},\;\;\;\frac{1-e^{-20\delta}}{\delta}=1.5\frac{1-e^{-10\delta}}{\delta},\;\;\;e^{-20\delta}-1.5e^{-10\delta}+0.5=0.[[/math]]

Let [math]X = e^{-10\delta}[/math]. We then have the quadratic equation

[[math]] X^2 -1.5X + 0.5 = 0 [[/math]]

with solution [math]X = 0.5[/math] for [math]\delta = \ln0.5\,/\,(-10)=0.069315. [/math]

Then, the accumulated value of a 7-year continuous annuity of 1 is

[[math]] \overline{{{s}}}_{\overline{{{7}}}|}=\frac{e^{7(0.069315)}-1}{0.069315}=9.01\,. [[/math]]

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

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