Pembahasan Ujian PAI: A70 - No. 6 - November 2016

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Permodelan dan Teori Risiko
Periode Ujian	:	November 2016
Nomor Soal	:	6

SOAL

Total kerugian pada tahun \(k\) , \(k = 1,2,3, \ldots ,n\) memiliki fungsi distribusi dengan fungsi kepadatan (density function) sebagai berikut:

\(f\left( {x;k} \right) = \frac{{{k^k}{x^{k – 1}}\exp \left[ { – \frac{{kx}}{\theta }} \right]}}{{\Gamma \left( k \right){\theta ^k}}}\) ; \(x > 0\)

Misalkan \({X_k}\) adalah total kerugian yang diamati pada tahun \(k\). Ada mengamati \({X_1} = 7.000\), \({X_2} = 7.500\), \({X_3} = 8.000\), \({X_4} = 8.500\), \({X_5} = 9.000\) Tentukan estimasi maximum likelihood untuk \(\theta \)

7.667
7.833
8.000
8.167
8.333

[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]

Diketahui	Total kerugian pada tahun \(k\) , \(k = 1,2,3, \ldots ,n\) memiliki fungsi distribusi dengan fungsi kepadatan (density function) sebagai berikut: \(f\left( {x;k} \right) = \frac{{{k^k}{x^{k – 1}}\exp \left[ { – \frac{{kx}}{\theta }} \right]}}{{\Gamma \left( k \right){\theta ^k}}}\) ; \(x > 0\) Misalkan \({X_k}\) adalah total kerugian yang diamati pada tahun \(k\). Ada mengamati \({X_1} = 7.000\), \({X_2} = 7.500\), \({X_3} = 8.000\), \({X_4} = 8.500\), \({X_5} = 9.000\)
Rumus yang digunakan	\(\Gamma \left( n \right) = \left( {n – 1} \right)!\) \(L\left( \theta \right) = \prod\limits_{i = 1}^n {f\left( {{x_i};\theta } \right)} \Rightarrow \frac{{d\ln \left[ {L\left( \theta \right)} \right]}}{{d\theta }} = 0\)
Proses pengerjaan	\(f\left( {{x_1} = 7000;k = 1} \right) = \frac{{\exp \left[ { – \frac{{7000}}{\theta }} \right]}}{\theta }\) \(f\left( {{x_2} = 7500;k = 2} \right) = \frac{{\left( 4 \right)\left( {7500} \right)\exp \left[ { – \frac{{\left( 2 \right)\left( {7500} \right)}}{\theta }} \right]}}{{\Gamma \left( 2 \right){\theta ^2}}} = \frac{{30,000\exp \left[ { – \frac{{15,000}}{\theta }} \right]}}{{{\theta ^2}}}\) \(f\left( {{x_3} = 8000;k = 3} \right) = \frac{{\left( {27} \right)\left( {{{8000}^2}} \right)\exp \left[ { – \frac{{\left( 3 \right)\left( {8000} \right)}}{\theta }} \right]}}{{\Gamma \left( 3 \right){\theta ^3}}} = \frac{{864 \times {{10}^6}\exp \left[ { – \frac{{24,000}}{\theta }} \right]}}{{{\theta ^3}}}\) \(f\left( {{x_3} = 8500;k = 4} \right) = \frac{{\left( {256} \right)\left( {{{8500}^3}} \right)\exp \left[ { – \frac{{\left( 4 \right)\left( {8500} \right)}}{\theta }} \right]}}{{\Gamma \left( 4 \right){\theta ^4}}}\frac{{2.620267 \times {{10}^{13}}\exp \left[ { – \frac{{34,000}}{\theta }} \right]}}{{{\theta ^4}}}\) \(f\left( {{x_3} = 9000;k = 5} \right) = \frac{{\left( {3125} \right)\left( {{{9000}^4}} \right)\exp \left[ { – \frac{{\left( 5 \right)\left( {9000} \right)}}{\theta }} \right]}}{{\Gamma \left( 5 \right){\theta ^5}}} = \frac{{8.542969 \times {{10}^{17}}\exp \left[ { – \frac{{45,000}}{\theta }} \right]}}{{{\theta ^5}}}\)
	\(\ln \left[ {L\left( \theta \right)} \right] = – \frac{{7000}}{\theta } – \ln \left( \theta \right) – \frac{{15,000}}{\theta } – \ln \left( {{\theta ^2}} \right)\) \(– \frac{{24,000}}{\theta } – \ln \left( {{\theta ^3}} \right) – \frac{{34,000}}{\theta } – \ln \left( {{\theta ^4}} \right) – \frac{{45,000}}{\theta } – \ln \left( {{\theta ^5}} \right)\) \(\frac{{d\ln \left[ {L\left( \theta \right)} \right]}}{{d\theta }} = \frac{{7000}}{{{\theta ^2}}} – \frac{1}{\theta } + \frac{{15,000}}{{{\theta ^2}}} – \frac{{2\theta }}{{{\theta ^2}}} + \frac{{24,000}}{{{\theta ^2}}} – \frac{{3\theta }}{{{\theta ^3}}} + \frac{{34,000}}{{{\theta ^2}}} – \frac{{4\theta }}{{{\theta ^4}}} + \frac{{45,000}}{{{\theta ^2}}} – \frac{{5\theta }}{{{\theta ^5}}} = 0\) \(\frac{{125,000}}{{{\theta ^2}}} – \frac{{15}}{\theta } = 0\) \(\hat \theta = \frac{{125,000}}{{15}} = 8,333.33\)
Jawaban	E. 8.333