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Exercise


Jan 17'24

Answer

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.

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