Exercise


Jan 18'24

Answer

Answer: D

[math]A_{[50]: 3]}^{1}=v q_{[50]}+v^{2} p_{[50]} q_{[50]+1}+v^{3} p_{[50]} p_{[50]+1} q_{52}[/math]

where: [math]v=\frac{1}{1.04}[/math]

[math]q_{[50]}=0.7(0.045)=0.0315[/math]

[math]p_{[50]}=1-q_{[50]}=0.9685[/math]

[math]q_{[50]+1}=0.8(0.050)=0.040[/math]

[math]p_{[50]+1}=1-q_{[50]+1}=0.960[/math]

[math]q_{52}=0.055[/math]

So: [math]A_{[50]: 3]}^{1}=0.1116[/math]

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

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