exercise:0e23d503a4

AAdmin

Jan 15'24

Exercise

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

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AAdmin

Jan 15'24

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]