Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
November 2015 |
| Nomor Soal |
: |
26 |
SOAL
Pada distribusi gamma dengan dua parameter didefinisikan oleh Probability Density function (PDF) sebagai berikut:
\(f\left( t \right) = \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{t^{\alpha – 1}}{e^{\frac{{ – t}}{\beta }}},t > 0,\alpha > 0,\beta > 0\)
Dengan mean dan variance-nya adalah \(\mu = \beta \alpha \) dan \({\sigma ^2} = {\beta ^2}\alpha \).
Jika diketahui sampel 10 data kematian tikus laboratorium (dalam hari) adalah 2; 3; 5; 6; 7; 8; 10; 11; 11; 12; hitunglah estimasi dari \(\alpha \) dan \(\beta \) dengan metode moments (dibulatkan 3 desimal).
- \(\alpha = 1,050\) dan \(\beta = 7,500\)
- \(\alpha = 1,512\) dan \(\beta = 6,463\)
- \(\alpha = 1,473\) dan \(\beta = 5,090\)
- \(\alpha = 1,322\) dan \(\beta = 8,254\)
- \(\alpha = 1,032\) dan \(\beta = 7,367\)
| Diketahui |
PDF distribusi gamma dengan mean \(\mu = \beta \alpha \) dan variance-nya \({\sigma ^2} = {\beta ^2}\alpha \) adalah \(f\left( t \right) = \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{t^{\alpha – 1}}{e^{\frac{{ – t}}{\beta }}},t > 0,\alpha > 0,\beta > 0\) |
| Rumus yang digunakan |
\({\mu _{_k}}^\prime = E\left( {{T^k}} \right)\) dan \({m_{_k}}^\prime = \frac{1}{n}\sum\limits_{i = 1}^n {T_i^k} \) |
| Proses pengerjaan |
\({\mu _{_1}}^\prime = E\left( T \right) = \mu = \alpha \beta = {m_{_1}}^\prime = \frac{1}{n}\sum\limits_{i = 1}^n {{T_i}} \)
\({\mu _{_2}}^\prime = E\left( {{T^2}} \right) = {\sigma ^2} + {\mu ^2} = \alpha {\beta ^2} + {\alpha ^2}{\beta ^2} = {m_{_2}}^\prime = \frac{1}{n}\sum\limits_{i = 1}^n {T_i^2} \)
Sehingga, \(\alpha \beta = \bar T\) dan \(\alpha {\beta ^2} + {\alpha ^2}{\beta ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {T_i^2} \) |
|
Dari hasil di atas di peroleh estimasi \(\hat \beta = \frac{{\bar T}}{{\hat \alpha }}\) disubstitusi ke persamaan 2
\(\hat \alpha {\left( {\frac{{\bar T}}{{\hat \alpha }}} \right)^2} + {{\hat \alpha }^2}{\left( {\frac{{\bar T}}{{\hat \alpha }}} \right)^2} = \frac{1}{n}\sum\limits_{i = 1}^n {T_i^2} \)
\(\hat \alpha = \frac{{{{\bar T}^2}}}{{\frac{1}{n}\sum\limits_{i = 1}^n {T_i^2} – {{\bar T}^2}}}\)
\(= \frac{{{{\bar T}^2}}}{{\sum\limits_{i = 1}^n {T_i^2} – n{{\bar T}^2}}}\) |
|
\(\bar T = \frac{{2 + 3 + 5 + 6 + 7 + 8 + 10 + 11 + 11 + 12}}{{10}} = 7,5\)
\(\sum\limits_{i = 1}^n {T_i^2} = {2^2} + {3^2} + {5^2} + {6^2} + {7^2} + {8^2} + {10^2} + {11^2} + {11^2} + {12^2} = 673\)
\(\hat \alpha = \frac{{n{{\bar T}^2}}}{{\sum\limits_{i = 1}^n {T_i^2} – n{{\bar T}^2}}}\)
\(= \frac{{10 \cdot {{\left( {7,5} \right)}^2}}}{{673 – 10 \cdot {{\left( {7,5} \right)}^2}}}\)
\(= \frac{{1125}}{{221}} = 5,090497738\)
\(\hat \beta = \frac{{\bar T}}{{\hat \alpha }}\)
\(= \frac{{7,25}}{{\frac{{1125}}{{221}}}}\)
\(= \frac{{221}}{{150}} = 1,4733333\) |
| Jawaban |
c. \(\alpha = 1,473\) dan \(\beta = 5,090\) |
