Revision as of 22:38, 17 January 2024 by Admin (Created page with "'''Answer: A''' The contribution from Life 1 is <math>{ }_{20} p_{70}</math>. With Gompertz and the selected parameters, the contribution is <math>{ }_{20} p_{70}=\exp \left[-\frac{B}{\ln c} c^{70}\left(c^{20}-1\right)\right]=\exp \left[-\frac{0.000003}{\ln 1.1} 1.1^{70}\left(1.1^{20}-1\right)\right]=0.86730</math>. The contribution from Life 2 is <math>{ }_{19} p_{70}-{ }_{20} p_{70}</math>. The contribution is <math display = "block"> \begin{aligned} & { }_{19} p_{...")
Exercise
Jan 17'24
Answer
Answer: A
The contribution from Life 1 is [math]{ }_{20} p_{70}[/math]. With Gompertz and the selected parameters, the contribution is [math]{ }_{20} p_{70}=\exp \left[-\frac{B}{\ln c} c^{70}\left(c^{20}-1\right)\right]=\exp \left[-\frac{0.000003}{\ln 1.1} 1.1^{70}\left(1.1^{20}-1\right)\right]=0.86730[/math].
The contribution from Life 2 is [math]{ }_{19} p_{70}-{ }_{20} p_{70}[/math]. The contribution is
[[math]]
\begin{aligned}
& { }_{19} p_{70}-{ }_{20} p_{70}=\exp \left[-\frac{B}{\ln c} c^{70}\left(c^{19}-1\right)\right]-0.86730=\exp \left[-\frac{0.000003}{\ln 1.1} 1.1^{70}\left(1.1^{19}-1\right)\right]-0.86730 \\
& =0.88058-0.86730=0.01328
\end{aligned}
[[/math]]
The contribution to the likelihood is [math]0.86730(0.01328)=0.01152[/math].