Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Permodelan dan Teori Risiko |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 5 |
SOAL
Diketahui:
- Frekuensi klaim tahunan suatu asuransi kebakaran mengikuti distribusi Poisson dengan rataan \(\lambda \)
- Distribusi prior dari \(\lambda \) memiliki fungsi kepadatan peluang sebagai berikut:
\(\pi \left( \lambda \right) = \left( {0,4} \right)\frac{1}{6}{e^{ – \frac{\lambda }{6}}} + \left( {0,6} \right)\frac{1}{{12}}{e^{ – \frac{\lambda }{{12}}}},\lambda > 0\)
Terhadap suatu pemegang polis, 10 klaim berhasil diobservasi di tahun pertama. Hitung ekspektasi Bayesian banyaknya klaim di tahun 2
- 9,6
- 9,7
- 9,8
- 9,9
- 10,0
| Diketahui |
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| Rumus yang digunakan | \(\begin{array}{*{20}{c}} {p\left( {x,\lambda } \right) = p\left( {\left. x \right|\lambda } \right)\pi \left( \lambda \right),}&{p\left( x \right) = \int\limits_0^\infty {p\left( {x,\lambda } \right)d\lambda } } \end{array}\)
\(E\left[ {\left. Y \right|x} \right] = \int\limits_0^\infty {E\left[ {\left. Y \right|\lambda } \right] \cdot \pi \left( {\left. \lambda \right|x} \right)d\lambda } = \frac{{\int\limits_0^\infty {E\left[ {\left. Y \right|\lambda } \right] \cdot p\left( {\left. x \right|\lambda } \right)d\lambda } }}{{p\left( x \right)}}\)
Distribusi gamma \(\begin{array}{*{20}{c}} {f\left( x \right) = \frac{{{x^{\alpha – 1}}{e^{ – \frac{x}{\theta }}}}}{{\Gamma \left( \alpha \right){\theta ^\alpha }}},}&{\Gamma \left( n \right) = \left( {n – 1} \right)!} \end{array}\) |
| Proses pengerjaan | Untuk 10 klaim mengikuti distribusi Poisson \(p\left( {\left. {x = 10} \right|\lambda } \right) = \frac{{{\lambda ^{10}}}}{{10!}}{e^{ – \lambda }}\) |
| Joint Distribution \(p\left( {x = 10,\lambda } \right) = p\left( {\left. {x = 10} \right|\lambda } \right)\pi \left( \lambda \right) = \left( {\frac{{{\lambda ^{10}}}}{{10!}}{e^{ – \lambda }}} \right)\left( {\left( {0,4} \right)\frac{1}{6}{e^{ – \frac{\lambda }{6}}} + \left( {0,6} \right)\frac{1}{{12}}{e^{ – \frac{\lambda }{{12}}}}} \right)\) \(p\left( {x = 10,\lambda } \right) = 8{\lambda ^{10}}{e^{ – \frac{{7\lambda }}{6}}} + 6{\lambda ^{10}}{e^{ – \frac{{13\lambda }}{{12}}}}\) | |
| Ekspektasi Bayesian
\(E\left[ {\left. Y \right|x = 10} \right] = \int\limits_0^\infty {E\left[ {\left. Y \right|\lambda } \right] \cdot \pi \left( {\left. \lambda \right|x = 10} \right)d\lambda } = \frac{{\int\limits_0^\infty {E\left[ {\left. Y \right|\lambda } \right] \cdot p\left( {\left. {x = 10} \right|\lambda } \right)d\lambda } }}{{p\left( {x = 10} \right)}}\) \(E\left[ {\left. Y \right|x = 10} \right] = \frac{{\int\limits_0^\infty {\lambda \cdot \left( {8{\lambda ^{10}}{e^{ – \frac{{7\lambda }}{6}}} + 6{\lambda ^{10}}{e^{ – \frac{{13\lambda }}{{12}}}}} \right)d\lambda } }}{{\int\limits_0^\infty {\left( {8{\lambda ^{10}}{e^{ – \frac{{7\lambda }}{6}}} + 6{\lambda ^{10}}{e^{ – \frac{{13\lambda }}{{12}}}}} \right)d\lambda } }} = \frac{{\int\limits_0^\infty {\left( {8{\lambda ^{11}}{e^{ – \frac{{7\lambda }}{6}}} + 6{\lambda ^{11}}{e^{ – \frac{{13\lambda }}{{12}}}}} \right)d\lambda } }}{{\int\limits_0^\infty {\left( {8{\lambda ^{10}}{e^{ – \frac{{7\lambda }}{6}}} + 6{\lambda ^{10}}{e^{ – \frac{{13\lambda }}{{12}}}}} \right)d\lambda } }}\) Kita bisa ubah fungsi di atas ke model disrtibusi Gamma. Kita tentukan nilai konstan \(\frac{1}{{\Gamma \left( \alpha \right){\theta ^\alpha }}}\) sehingga diperoleh integral-nya menjadi \(\Gamma \left( \alpha \right){\theta ^\alpha }\) dengan \(\alpha = \) pangkat dari variabel random ditambah 1 dan \(\theta = \) kebalikan dari nilai pangkat eksponen
Sehingga hasil \(E\left[ {\left. Y \right|x = 10} \right]\) adalah \(\frac{{8\Gamma \left( {12} \right){{\left( {\frac{6}{7}} \right)}^{12}} + 6\Gamma \left( {12} \right){{\left( {\frac{{12}}{{13}}} \right)}^{12}}}}{{8\Gamma \left( {11} \right){{\left( {\frac{6}{7}} \right)}^{11}} + 6\Gamma \left( {11} \right){{\left( {\frac{{12}}{{13}}} \right)}^{11}}}} = \frac{{\Gamma \left( {12} \right)\left[ {8{{\left( {\frac{6}{7}} \right)}^{12}} + 6{{\left( {\frac{{12}}{{13}}} \right)}^{12}}} \right]}}{{\Gamma \left( {11} \right)\left[ {8{{\left( {\frac{6}{7}} \right)}^{11}} + 6{{\left( {\frac{{12}}{{13}}} \right)}^{11}}} \right]}} = 11\frac{{3.5543}}{{3.9554}} = 9.8847\) |
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| Jawaban | d. 9,9 |
