excans:627db70ba1

Exercise

Jan 20'24

Answer

Answer: A

This first solution recognizes that the full preliminary term reserve at the end of year 10 for a 30 year endowment insurance on (40) is the same as the net premium policy value at the end of year 9 for a 29 year endowment insurance on (41). Then, using superscripts of FPT for full preliminary term reserve and NLP for net premium policy value to distinguish the symbols, we have

[[math]] \begin{aligned} 1000_{10} V^{F P T} & =1000_{9} V^{N L P}=1000\left(A_{50: 20}-P_{41: 29} \ddot{a}_{50: \overline{20}}\right) \\ & =1000[0.38844-0.01622(12.8428)]=180 \end{aligned} [[/math]]

or [math]=1000\left(1-\frac{\ddot{a}_{50: 201}}{\ddot{a}_{41: 29}}\right)=1000\left(1-\frac{12.8428}{15.6640}\right)=180[/math]

where

[[math]] \begin{aligned} \ddot{a}_{41: \overline{29}} & =\ddot{a}_{41}-{ }_{29} E_{41} \ddot{a}_{70} \\ & =18.3403-(0.2228726)(12.0083) \\ & =15.6640 \\ A_{41: \overline{29}} & =1-d(15.6640)=0.254095 \\ { }_{29} E_{41} & =v^{29}\left(\frac{l_{70}}{l_{41}}\right)=(0.242946)\left(\frac{91,082.4}{99,285.9}\right)=0.2228726 \\ P_{41: \overline{29}} & =\frac{0.254095}{15.6640}=0.01622 \end{aligned} [[/math]]

Alternatively, working from the definition of full preliminary term reserves as having [math]{ }_{1} V^{F P T}=0[/math] and the discussion of modified net premium reserves in the Notation and Terminology Study Note, let [math]\alpha[/math] be the valuation premium in year 1 and [math]\beta[/math] be the valuation premium thereafter. Then (with some of the values taken from above),

[math]\alpha=1000 v q_{40}=0.5019[/math]

APV (valuation premiums) [math]=[/math] APV (benefits)

[math]\alpha+{ }_{1} E_{40}\left(\ddot{a}_{41: 29)}\right) \beta=1000 A_{40: 30}[/math]

[math]0.5019+0.95188(15.6640) \beta=242.37[/math]

[math]\beta=\frac{242.37-0.5019}{14.9102}=16.22[/math]

Where

[[math]] \begin{aligned} { }_{1} E_{40}= & (1-0.000527) v=0.95188 \\ A_{40: 30} & =A_{40}+{ }_{20} E_{40}\left({ }_{10} E_{60}\right)\left(1-A_{70}\right) \\ & =0.12106+0.36663(0.57864)(1-0.42818)=0.24237 \\ { }_{10} V^{F P T} & =1000 A_{50: 20}-\beta \ddot{a}_{50: 20 \mid}=1000(0.38844)-16.22(12.8427)=180 \end{aligned} [[/math]]