Revision as of 18:35, 17 January 2024 by Admin (Created page with "You are doing a mortality study of insureds between ages 60 and 90 . Two specific lives contributed this data to the study: {| class="table table-bordered" ! Life !! Age at Entry !! Age at Exit !! Cause of exit |- | 1 || 60.0 || 74.5 || Policy lapsed |- | 2 || 60.0 || 74.5 || 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 log-likelihood function (using natural...")
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Jan 17'24

Exercise

You are doing a mortality study of insureds between ages 60 and 90 . Two specific lives contributed this data to the study:

Life Age at Entry Age at Exit Cause of exit
1 60.0 74.5 Policy lapsed
2 60.0 74.5 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 log-likelihood function (using natural logs) associated with these two lives.

[math]L^{*}[/math] denotes the value of [math]L[/math] if the Gompertz parameters are [math]B=0.000004[/math] and [math]c=1.12[/math].

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

  • -4,67
  • -4.53
  • -4.39
  • -4.25
  • -4.11
1 Answer
Jan 17'24

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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