Revision as of 22:07, 17 November 2023 by Admin (Created page with "Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of interest <math display = "block"> \delta_{t}={\frac{t^{2}}{100}},\,t>0. </math> The amount of interest earned from time 3 to time 6 is also X Calculate X. <ul class="mw-excansopts"><li>385</li><li>485</li><li>585</li><li>685</li><li>785</li></ul> {{soacopyright | 2023 }}")
ABy Admin
Nov 17'23
Exercise
Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of interest
[[math]]
\delta_{t}={\frac{t^{2}}{100}},\,t\gt0.
[[/math]]
The amount of interest earned from time 3 to time 6 is also X
Calculate X.
- 385
- 485
- 585
- 685
- 785
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
ABy Admin
Nov 17'23
Solution: E
The accumulation function is
[[math]]
a(t)=\exp\biggl[\int_{0}^{t}(s^{2}/100)d s\biggr]=\exp(t^{3}/300).
[[/math]]
The accumulated value of 100 at time 3 is [math]100 \exp(3^3 / 300) = 109.41743.[/math]
The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus
[[math]]
\begin{array}{l}{{\left(109.41743+X\right)[a(6)/a(3)-1]=X}}\\ {{\left(109.41743+X\right)(2.0544332/1..0941743-1)=X}}\\ {{\left(109.41743+X\right)(3.877613=X}}\\ {{X=784.61.}}\end{array}
[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.