Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Seorang aktuaris mengamati lima buah besaran klaim: 11,0; 15,2; 18,0; 21,0; dan 25,8. Tentukan parameter \(\mu \) dari fungsi kepadatan di bawah ini:
\(\begin{array}{*{20}{c}} {f\left( x \right) = \frac{1}{{\sqrt {2\pi x} }}\exp \left[ { – \frac{1}{{2x}}{{\left( {x – \mu } \right)}^2}} \right],}&{x > 0,}&{\mu > 0} \end{array}\)
- 12,64
- 17,64
- 14,54
- 12,85
- 16,74
| Diketahui |
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| Rumus yang digunakan | \(\begin{array}{*{20}{c}} {L\left( \mu \right) = \prod\limits_{i = 1}^n {f\left( {{x_i};\mu } \right)} ,}&{\frac{{d\ln \left[ {L\left( \theta \right)} \right]}}{{d\theta }} = 0} \end{array}\) |
| Proses pengerjaan | \(\frac{1}{{\sqrt {2\pi x} }}\) merupakan faktor pengali dan bernilai konstan sehingga bisa dihiraukan Dengan \({x_i}\) sebanyak 5 observasi maka \(L\left( \mu \right) = \prod\limits_{i = 1}^n {f\left( {{x_i};\mu } \right)} = \prod\limits_{i = 1}^n {\exp \left[ { – \frac{1}{{2{x_i}}}{{\left( {{x_i} – \mu } \right)}^2}} \right]} \) \(L\left( \mu \right) = \exp \left[ { – \sum\limits_{i = 1}^n {\frac{1}{{2{x_i}}}{{\left( {{x_i} – \mu } \right)}^2}} } \right]\) \(\ln \left[ {L\left( \mu \right)} \right] = – \sum\limits_{i = 1}^n {\frac{1}{{2{x_i}}}{{\left( {{x_i} – \mu } \right)}^2}} \) \(\frac{{d\ln \left[ {L\left( \mu \right)} \right]}}{{d\mu }} = \sum\limits_{i = 1}^n {\frac{{{x_i} – \mu }}{{{x_i}}}} = \sum\limits_{i = 1}^n {\frac{{{x_i}}}{{{x_i}}}} – \mu \sum\limits_{i = 1}^n {\frac{1}{{{x_i}}}} = 0\) \(0 = n – \mu \sum\limits_{i = 1}^n {\frac{1}{{{x_i}}}} \) \(\mu = \frac{n}{{\sum\limits_{i = 1}^n {\frac{1}{{{x_i}}}} }} = \frac{5}{{\frac{1}{{11.0}} + \frac{1}{{15.2}} + \frac{1}{{18.0}} + \frac{1}{{21.0}} + \frac{1}{{25.8}}}} = 16.743\) |
| Jawaban | e. 16,74 |