Revision as of 21:04, 19 January 2024 by Admin (Created page with "'''Answer: C''' Let <math>\pi</math> be the annual premium, so that <math>\pi \ddot{a}_{50}=A_{50}+0.01 \ddot{a}_{50}+0.19</math> <math>\Rightarrow \pi=\frac{A_{50}+0.19}{\ddot{a}_{50}}+0.01=\frac{0.18931+0.19}{17.0245}+0.01=0.03228</math> Loss at issue: <math>L_{0}=v^{k+1}-(\pi-0.01) \ddot{a}_{\overline{k+1}}\left(1-v^{k+1}\right) / d+0.19</math> <math display="block"> \begin{aligned} \Rightarrow \operatorname{Var}\left[L_{0}\right] & =\left(1+\frac{(\pi-0.01)}{d}\...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Exercise


ABy Admin
Jan 19'24

Answer

Answer: C

Let [math]\pi[/math] be the annual premium, so that [math]\pi \ddot{a}_{50}=A_{50}+0.01 \ddot{a}_{50}+0.19[/math]

[math]\Rightarrow \pi=\frac{A_{50}+0.19}{\ddot{a}_{50}}+0.01=\frac{0.18931+0.19}{17.0245}+0.01=0.03228[/math]

Loss at issue: [math]L_{0}=v^{k+1}-(\pi-0.01) \ddot{a}_{\overline{k+1}}\left(1-v^{k+1}\right) / d+0.19[/math]

[[math]] \begin{aligned} \Rightarrow \operatorname{Var}\left[L_{0}\right] & =\left(1+\frac{(\pi-0.01)}{d}\right)^{2}\left({ }^{2} A_{50}-A_{50}^{2}\right) \\ & =(2.15467)\left(0.05108-0.18931^{2}\right) \\ & =(2.15467)(0.015242) \\ & =0.033 \end{aligned} [[/math]]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

00