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(Created page with "'''Answer: D''' The contribution from Life 1 is <math>{ }_{14.5} p_{60}</math>. With Gompertz and the selected parameters, the contribution is <math display = "block">{ }_{14.5} p_{60}=\exp \left[-\frac{B}{\ln c} c^{60}\left(c^{14.5}-1\right)\right]=\exp \left[-\frac{0.000004}{\ln 1.12} 1.12^{60}\left(1.12^{14.5}-1\right)\right]=0.87619.</math> The contribution from Life 2 is <math>{ }_{14.5} p_{60} \times \mu_{74.5}</math>. The contribution is <math display = "block"...") |
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The contribution to the log-likelihood is <math>\ln (0.01426)=-4.25</math>. | The contribution to the log-likelihood is <math>\ln (0.01426)=-4.25</math>. | ||
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Latest revision as of 02:38, 18 January 2024
Answer: D
The contribution from Life 1 is [math]{ }_{14.5} p_{60}[/math]. With Gompertz and the selected parameters, the contribution is
[[math]]{ }_{14.5} p_{60}=\exp \left[-\frac{B}{\ln c} c^{60}\left(c^{14.5}-1\right)\right]=\exp \left[-\frac{0.000004}{\ln 1.12} 1.12^{60}\left(1.12^{14.5}-1\right)\right]=0.87619.[[/math]]
The contribution from Life 2 is [math]{ }_{14.5} p_{60} \times \mu_{74.5}[/math]. The contribution is
[[math]]
{ }_{14.5} p_{60} \times \mu_{74.5}=0.87619 \times 0.000004 \times 1.12^{74.5}=0.01627
[[/math]]
The contribution to the likelihood is [math]0.87619(0.01627)=0.01426[/math].
The contribution to the log-likelihood is [math]\ln (0.01426)=-4.25[/math].
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.