Revision as of 12:55, 18 January 2024 by Admin (Created page with "'''Answer: E''' <math>E(Z)=\frac{i}{\delta}\left[v q_{x}+v^{2} p_{x} q_{x+1}\right]=\frac{0.04}{\ln (1.04)}\left[\frac{1}{1.04}(0.04)+\frac{1}{1.04^{2}}(0.96)(0.06)\right]=0.09353831</math> <math>E\left(Z^{2}\right)=\frac{(1+i)^{2}-1}{2 \delta}\left[v^{2} q_{x}+v^{4} p_{x} q_{x+1}\right]=\frac{\left(1.04^{2}-1\right)}{2 \ln (1.04)}\left[\frac{1}{1.04^{2}}(0.04)+\frac{1}{1.04^{4}}(0.96)(0.06)\right]=0.08969072</math> <math>\operatorname{Var}(Z)=E\left(Z^{2}\right)-[E(Z...")
Exercise
Jan 18'24
Answer
Answer: E
[math]E(Z)=\frac{i}{\delta}\left[v q_{x}+v^{2} p_{x} q_{x+1}\right]=\frac{0.04}{\ln (1.04)}\left[\frac{1}{1.04}(0.04)+\frac{1}{1.04^{2}}(0.96)(0.06)\right]=0.09353831[/math]
[math]E\left(Z^{2}\right)=\frac{(1+i)^{2}-1}{2 \delta}\left[v^{2} q_{x}+v^{4} p_{x} q_{x+1}\right]=\frac{\left(1.04^{2}-1\right)}{2 \ln (1.04)}\left[\frac{1}{1.04^{2}}(0.04)+\frac{1}{1.04^{4}}(0.96)(0.06)\right]=0.08969072[/math]
[math]\operatorname{Var}(Z)=E\left(Z^{2}\right)-[E(Z)]^{2}=0.08969072-(0.09353831)^{2}=0.0809413[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.