Revision as of 21:05, 19 January 2024 by Admin (Created page with "'''Answer: B''' <math>\mathrm{EPV}(</math> premiums <math>)=\mathrm{EPV}(</math> benefits <math>)</math> <math>P\left(1+v p_{x}+v_{2}^{2} p_{x}\right)=P\left(v q_{x}+2 v^{2} p_{x} q_{x+1}\right)+10000\left(v^{3}{ }_{2} p_{x} q_{x+2}\right)</math> <math>P\left(1+\frac{0.9}{1.04}+\frac{0.9 \times 0.88}{1.04^{2}}\right)=P\left(\frac{0.1}{1.04}+\frac{2 \times 0.9 \times 0.12}{1.04^{2}}\right)+10000\left(\frac{0.9 \times 0.88 \times 0.15}{1.04^{3}}\right)</math> <math>2.5...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: B
[math]\mathrm{EPV}([/math] premiums [math])=\mathrm{EPV}([/math] benefits [math])[/math]
[math]P\left(1+v p_{x}+v_{2}^{2} p_{x}\right)=P\left(v q_{x}+2 v^{2} p_{x} q_{x+1}\right)+10000\left(v^{3}{ }_{2} p_{x} q_{x+2}\right)[/math]
[math]P\left(1+\frac{0.9}{1.04}+\frac{0.9 \times 0.88}{1.04^{2}}\right)=P\left(\frac{0.1}{1.04}+\frac{2 \times 0.9 \times 0.12}{1.04^{2}}\right)+10000\left(\frac{0.9 \times 0.88 \times 0.15}{1.04^{3}}\right)[/math]
[math]2.5976 P=0.29588 P+1056.13[/math]
[math]P=459[/math]
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