Revision as of 20:00, 20 November 2023 by Admin (Created page with "'''Solution: A''' The present value function and its derivatives are <math display = "block"> \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 display = "block"> \begin{array}{l c r}{{...")
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.