excans:Bd3576e403

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.