You are given:

i) [math]\mu_{x+t}=\beta t^{2}, t \geq 0[/math]

ii) [math]l_{x}=1000[/math]

iii) [math]l_{x+10}=400[/math]

Calculate [math]1000 \beta[/math].

• 2.75
• 2.80
• 2.85
• 2.90
• 2.95

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

Answer: A

[math]{ }_{10} p_{x}=\frac{l_{x+10}}{l_{x}}=e^{-\int_{0}^{10} \mu_{x+1} \cdot d t}=\gt\frac{400}{1000}=e^{-\int_{0}^{10} \beta t^{2} \cdot d t}=\gt0.4=e^{-\beta t^{3} / 3 b^{10}}[/math]

[math]==\gt0.4=e^{-\beta \cdot 100^{3} / 3}==\gt\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==\gt\beta=-\ln (0.4)(.003)=0.0027489[/math]

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

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

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