Revision as of 21:28, 19 January 2024 by Admin (Created page with "'''Answer: A''' The loss at issue is given by: <math>L_{0}=100 v^{K+1}+0.05 G+0.05 G \ddot{a}_{\overline{K+1}}-G \ddot{a}_{\overline{K+1}}</math> <math>=100 v^{K+1}+0.05 G-0.95 G\left(\frac{1-v^{K+1}}{d}\right)</math> <math>=\left(100+\frac{0.95 G}{d}\right) v^{K+1}+0.05 G-0.95 \frac{G}{d}</math> Thus, the variance is <math display="block"> \begin{aligned} \operatorname{Var}\left(L_{0}\right)=[100 & \left.+\frac{0.95(2.338)}{0.04 / 1.04}\right]^{2}\left({ }^{2} A_...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: A
The loss at issue is given by:
[math]L_{0}=100 v^{K+1}+0.05 G+0.05 G \ddot{a}_{\overline{K+1}}-G \ddot{a}_{\overline{K+1}}[/math]
[math]=100 v^{K+1}+0.05 G-0.95 G\left(\frac{1-v^{K+1}}{d}\right)[/math]
[math]=\left(100+\frac{0.95 G}{d}\right) v^{K+1}+0.05 G-0.95 \frac{G}{d}[/math]
Thus, the variance is
[[math]]
\begin{aligned}
\operatorname{Var}\left(L_{0}\right)=[100 & \left.+\frac{0.95(2.338)}{0.04 / 1.04}\right]^{2}\left({ }^{2} A_{x}-\left(A_{x}\right)^{2}\right) \\
& =\left[100+\frac{0.95(2.338)}{0.04 / 1.04}\right]^{2}\left(0.17-\left(1-\frac{0.04}{1.04}(16.50)\right)^{2}\right) \\
& =908.1414
\end{aligned}
[[/math]]
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