Jan 17'24

Exercise

You are doing a mortality study of insureds between ages 70 and 90 . Two specific lives contributed this data to the study:

Life Age at Entry Age at Exit Cause of exit
1 70.0 90.0 End of study
2 70.0 Between 89.0 and 90.0 Death

You assume mortality follows Gompertz law [math]\mu_{x}=B \times c^{x}[/math] and plan to use maximum likelihood estimation.

[math]L[/math] is the likelihood function associated with these two lives.

[math]L^{*}[/math] denotes the value of [math]L[/math] if the Gompertz parameters are [math]B=0.000003[/math] and [math]c=1.1[/math].

Calculate [math]L^{*}[/math].

  • 0.0115
  • 0.0131
  • 0.0147
  • 0.0163
  • 0.0179

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

1 Answer
Jan 17'24

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].

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

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