Revision as of 17:04, 15 January 2024 by Admin (Created page with "You are given that mortality follows Gompertz Law with <math>B=0.00027</math> and <math>c=1.1</math>. Calculate <math>f_{50}(10)</math>. <ul class="mw-excansopts"><li> <math>\quad 0.048</math></li><li> <math>\quad 0.050</math></li><li> <math>\quad 0.052</math></li><li> <math>\quad 0.054</math></li><li> <math>\quad 0.056</math></li></ul>")
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ABy Admin
Jan 15'24

Exercise

You are given that mortality follows Gompertz Law with [math]B=0.00027[/math] and [math]c=1.1[/math]. Calculate [math]f_{50}(10)[/math].

  • [math]\quad 0.048[/math]
  • [math]\quad 0.050[/math]
  • [math]\quad 0.052[/math]
  • [math]\quad 0.054[/math]
  • [math]\quad 0.056[/math]
ABy Admin
Jan 15'24

Answer: A

[math]f_{x}(t)=-\frac{d}{d t} S_{x}(t)=-\frac{d}{d t}\left(e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)}\right)[/math]

[math]=-e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot\left(-\frac{B}{\ln c} \cdot c^{x}\right) \cdot c^{t} \cdot \ln c[/math]

[math]=e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot B c^{x+t}[/math]

[math]=0.00027 \times 1.1^{x+t} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{x}\right)\left(1.1^{t}-1\right)}[/math]

[math]f_{50}(10)=0.00027 \times 1.1^{50+10} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}=0.04839[/math]

Alternative Solution:

[math]f_{x}(t)={ }_{t} p_{x} \cdot \mu_{x+t}[/math]

Then we can use the formulas given for Makeham with [math]A=0, B=0.00027[/math] and [math]c=1.1[/math]

[math]f_{x}(t)=\left(e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}\right)\left(0.00027 \times 1.1^{50+10}\right)=0.04839[/math]

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

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

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

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

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