Pembahasan Ujian PAI: A50 – No. 11 – Juni 2016

By Fery Widhiatmoko On Mar 28, 2019

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Metoda Statistika
Periode Ujian	:	Juni 2016
Nomor Soal	:	11

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

Kunci Jawaban & Pembahasan

Fery Widhiatmoko 564 posts 11 comments

Merupakan lulusan Magister Matematika minat Aktuaria dari Universitas Gadjah Mada.

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