Exercise


Nov 20'23

Answer

Solution: A

The present value function and its derivatives are

[[math]] \begin{array}{l c r}{{P(i)=X+Y(1+i)^{-3}-500(1+i)^{-1}-1000(1+i)^{-4}}}\\ {{P^{\prime}(i)=-3Y(1+i)^{-4}+500(1+i)^{-5}-20,000(1+i)^{-5}}}\\ {{P^{\prime\prime}(i)=12Y(1+i)^{-5}-1000(1+i)^{-5}-20,000(1+i)^{-6}.}}\end{array} [[/math]]

The equations to solve for matching present values and duration (at i = 0.10) and their solution are

[[math]] \begin{array}{l c r}{{P(0.1)=X+0.7513Y-1137.56=0}}\\ {{P^{\prime}(0.1)=-2.0490Y+2896.91=0}}\\ {{Y=2896.91/2.0490=1413.82}}\\ {{X=1137.56-0.7513(1413.82)=75.36.}}\end{array} [[/math]]

The second derivative is

[[math]] P^{\prime\prime}(0.1)=12(1413.82)(1.1)^{-5}-1000(1.1)^{-5}-20,000(1.1)^{-6}=-1506.34. [[/math]]

Redington immunization requires a positive value for the second derivative, so the condition is not satisfied.

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

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