Revision as of 21:41, 19 January 2024 by Admin (Created page with "'''Answer: C''' Let <math>P</math> be the annual net premium at <math>x+1</math>. Also, let <math>A_{y}^{*}</math> be the expected present value for the special insurance described in the problem issued to <math>(y)</math>. <math>P \ddot{a}_{x+1}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v_{k \mid}^{k+1} q_{x+1}=1000 A_{x+1}^{*}</math> We are given <math>110 \ddot{a}_{x}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v^{k+1}{ }_{k} q_{x}=1000 A_{x}^{*}</math> Which implies that <...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: C
Let [math]P[/math] be the annual net premium at [math]x+1[/math]. Also, let [math]A_{y}^{*}[/math] be the expected present value for the special insurance described in the problem issued to [math](y)[/math].
[math]P \ddot{a}_{x+1}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v_{k \mid}^{k+1} q_{x+1}=1000 A_{x+1}^{*}[/math]
We are given
[math]110 \ddot{a}_{x}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v^{k+1}{ }_{k} q_{x}=1000 A_{x}^{*}[/math]
Which implies that
[math]110\left(1+v p_{x} \ddot{a}_{x+1}\right)=1000\left(1.03 v q_{x}+1.03 v p_{x} A_{x+1}^{*}\right)[/math]
Solving for [math]A_{x+1}^{*}[/math], we get
[[math]]
A_{x+1}^{*}=\frac{\frac{110}{1000}[1+v(0.95)(7)]-1.03 v(0.05)}{1.03 v(0.95)}=0.8141032
[[/math]]
Thus, we have
[math]P=\frac{1000(0.8141032)}{7}=116.3005[/math]