Revision as of 01:13, 18 November 2023 by Admin (Created page with "At a force of interest <math>\delta_t=\frac{0.5}{5+0.5 t}, 0 \leq t \leq 6</math> an investment of 1000 at time <math>t=2</math> will accumulate to <math>X</math> at time <math>t=6</math>. At an annual nominal rate of discount of <math>8 \%</math> convertible quarterly, an investment of <math>Y</math> will accumulate to <math>X</math> at the end of two years. <ul class="mw-excansopts"><li>1124</li><li>1129</li><li>1134</li><li>1138</li><li>1143</li></ul> {{soacopyright...")
ABy Admin
Nov 18'23
Exercise
At a force of interest [math]\delta_t=\frac{0.5}{5+0.5 t}, 0 \leq t \leq 6[/math] an investment of 1000 at time [math]t=2[/math] will accumulate to [math]X[/math] at time [math]t=6[/math]. At an annual nominal rate of discount of [math]8 \%[/math] convertible quarterly, an investment of [math]Y[/math] will accumulate to [math]X[/math] at the end of two years.
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Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
ABy Admin
Nov 18'23
Solution: C
[[math]]\begin{aligned} & X=1000 \exp \left(\int_2^6 \frac{0.5}{5+0.5 t} d t\right) \\ & =1000 \exp \left[\left.\ln (5+0.5 t)\right|_2 ^6\right] \\ & =1000 \exp (\ln 8-\ln 6)=1000\left(\frac{8}{6}\right)=1333.33 \\ & 1333.33=Y\left[1-\frac{0.08}{4}\right]^{-4(2)} \\ & Y=1134.35\end{aligned}[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.