Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Pemodelan dan Teori Risiko |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 19 |
SOAL
Diketahui informasi sebagai berikut untuk suatu kelompok pemegang polis:
- Besaran klaim berdistribusi seragam tetapi tidak melebihi suatu limit \(\theta \)
- Distribusi prior dari \(\theta \) ialah
\(\begin{array}{*{20}{c}} {\pi \left( \theta \right) = \frac{{500}}{{{\theta ^2}}},}&{\theta > 500} \end{array}\) - Dua klaim saling bebas sebesar 400 dan 600 telah diamati
Hitung peluang bahwa klaim selanjutnya akan lebih besar dari 550. (Cari jawaban paling mendekati)
- 0,19
- 0,22
- 0,15
- 0,24
- 0,31
| Diketahui | - Besaran klaim berdistribusi seragam tetapi tidak melebihi suatu limit \(\theta \)
- Distribusi prior dari \(\theta \) ialah
\(\begin{array}{*{20}{c}} {\pi \left( \theta \right) = \frac{{500}}{{{\theta ^2}}},}&{\theta > 500} \end{array}\) - Dua klaim saling bebas sebesar 400 dan 600 telah diamati
|
| Rumus yang digunakan | Uniform: \({f\left( x \right) = \frac{1}{{b – a}},}\) \({S\left( x \right) = \Pr \left( {X > x} \right) = \frac{{b – x}}{{b – a}}}\)
\(\pi \left( {\left. \theta \right|x} \right) = \frac{{f\left( {x;\theta } \right)}}{{f\left( x \right)}} = \frac{{f\left( {\left. x \right|\theta } \right)\pi \left( \theta \right)}}{{\int_0^\infty {f\left( {x;\theta } \right)d\theta } }}\)
\(\Pr \left( {\left. Y \right|X} \right) = \int\limits_0^\infty {\Pr \left( {\left. y \right|\theta } \right)\pi \left( {\left. \theta \right|x} \right)d\theta } \) |
| Proses pengerjaan | Model distribusi adalah \(\begin{array}{*{20}{c}} {f\left( {\left. x \right|\theta } \right) = \frac{1}{\theta },}&{0 < x < \theta } \end{array}\) Model posteriornya adalah
\(\begin{array}{*{20}{c}} {\pi \left( {\left. \theta \right|x = 400,600} \right) = \frac{{f\left( {\left. {x = 400,600} \right|\theta } \right)\pi \left( \theta \right)}}{{\int_0^\infty {f\left( {x = 400,600;\theta } \right)d\theta } }} \propto \frac{1}{\theta } \cdot \frac{1}{\theta } \cdot \frac{{500}}{{{\theta ^2}}} \propto \frac{1}{{{\theta ^4}}},}&{\theta > 600} \end{array}\) |
| Karena salah satu observasi memiliki nilai 600 maka nilai parameter harus lebih besar dari 600 Sehingga nilai konstan diperoleh dari \(\int_{600}^\infty {{\theta ^{ – 4}}d\theta } = \frac{1}{{3{{\left( {600} \right)}^3}}}\) sehingga model posterior lengkapnya adalah \(\begin{array}{*{20}{c}} {\pi \left( {\left. \theta \right|x = 400,600} \right) = 3{{\left( {600} \right)}^3}{\theta ^{ – 4}},}&{\theta > 600} \end{array}\) |
| \(\Pr \left( {\left. {{X_3} > 550} \right|400,600} \right) = \int_{600}^\infty {\Pr \left( {\left. {{X_3} > 550} \right|\theta } \right)\pi \left( {\left. \theta \right|x = 400,600} \right)d\theta } \)
\(\Pr \left( {\left. {{X_3} > 550} \right|400,600} \right) = \int_{600}^\infty {\frac{{\theta – 550}}{\theta }\left[ {3{{\left( {600} \right)}^3}{\theta ^{ – 4}}} \right]d\theta } \)
\(\Pr \left( {\left. {{X_3} > 550} \right|400,600} \right) = 3{\left( {600} \right)^3}\int_{600}^\infty {\frac{{\theta – 550}}{{{\theta ^5}}}d\theta } \)
\(\Pr \left( {\left. {{X_3} > 550} \right|400,600} \right) = 3{\left( {600} \right)^3}\left[ {\int_{600}^\infty {\frac{1}{{{\theta ^4}}}d\theta } – \int_{600}^\infty {\frac{{550}}{{{\theta ^5}}}d\theta } } \right]\)
\(\Pr \left( {\left. {{X_3} > 550} \right|400,600} \right) = 0.3125\) |
| Jawaban | e. 0,31 |
