excans:Bd3576e403: Difference between revisions
(Created page with "'''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)\l...") |
No edit summary |
||
| (4 intermediate revisions by the same user not shown) | |||
| Line 10: | Line 10: | ||
<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> | <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> | |||
{{soacopyright | 2024 }} | |||
{{soacopyright|2024}} | |||
{{soacopyright|2024}} | |||
{{soacopyright|2024}} | |||
Latest revision as of 02:34, 18 January 2024
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.