For a fully discrete whole life insurance of 10,000 on (45), you are given:

(i) [math]\quad i=0.05[/math]

(ii) [math]\quad{ }_{0} L[/math] denotes the loss at issue random variable based on the net premium

(iii) If [math]K_{45}=10[/math], then [math]{ }_{0} L=4450[/math]

(iv) [math]\quad \ddot{a}_{55}=13.4205[/math]

Calculate [math]{ }_{10} V[/math], the net premium policy value at the end of year 10 for this insurance.

• 1010
• 1460
• 1820
• 2140
• 2300

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

Answer: B

[math]L=10,000 v^{K_{45}+1}-P \ddot{a}_{\overline{K_{45}+1 \mid}}=10,000 v^{11}-P \ddot{a}_{\overline{11}}[/math]

[math]4450=10,000(0.58468)-8.7217 P[/math]

[math]P=(5,846.8-4,450) / 8.7217=160.15[/math]

[math]A_{55}=1-d \ddot{a}_{55}=1-(0.05 / 1.05)(13.4205)=0.36093[/math]

[math]{ }_{10} V=10,000 A_{55}-P \ddot{a}_{55}=(10,000)(0.36093)-(160.15)(13.4205)=1,460[/math]

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