Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui tiga hasil pengamatan sebagai berikut:
0,70 0,82 0,92
Anda mencocokkan sebuah distribusi dengan fungsi kepadatan (density function) berikut ini terhadap data:
\(f\left( x \right) = \left( {p + 1} \right){x^p},\) \(0 < x < 1,p > – 1\)
Hitunglah estimasi maximum likelihood atas \(p\) (dibulatkan 2 desimal).
- 2,12
- 2,67
- 3,70
- 4,32
- 6,81
| Diketahui | tiga hasil pengamatan sebagai berikut: 0,70; 0,82; 0,92 \(f\left( x \right) = \left( {p + 1} \right){x^p},\) \(0 < x < 1,p > – 1\) |
| Rumus yang digunakan | \(L\left( p \right) = \prod\limits_{i = 1}^n {f\left( {{x_i}} \right)} \) |
| Proses pengerjaan | \(L\left( p \right) = \prod\limits_{i = 1}^n {f\left( {{x_i}} \right)} \) \(= \prod\limits_{i = 1}^n {\left( {p + 1} \right){x_i}^p} \) \(= {\left( {p + 1} \right)^n} \cdot {\left( {\prod\limits_{i = 1}^n {{x_i}} } \right)^p}\) \(\ln \left[ {L\left( p \right)} \right] = n \cdot \ln \left( {p + 1} \right) + p \cdot \ln \left( {\prod\limits_{i = 1}^n {{x_i}} } \right)\) \(= n \cdot \ln \left( {p + 1} \right) + p \cdot \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} \) \(\frac{{d\ln \left[ {L\left( p \right)} \right]}}{{dp}} = \frac{n}{{\hat p + 1}} + \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} = 0\) \(– n = \hat p \cdot \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} + \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} \) \(\hat p = \frac{{ – n – \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} }}{{\sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} }}\) |
| Sehingga, \(\hat p = \frac{{ – n – \sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} }}{{\sum\limits_{i = 1}^n {\ln \left( {{x_i}} \right)} }}\) \(= \frac{{ – 3 – \left( {\ln \left( {0,7} \right) + \ln \left( {0,82} \right) + \ln \left( {0,92} \right)} \right)}}{{\ln \left( {0,7} \right) + \ln \left( {0,82} \right) + \ln \left( {0,92} \right)}}\) \(= 3,698457\) | |
| Jawaban | c. 3,70 |