Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Permodelan dan Teori Risiko |
| Periode Ujian |
: |
Agustus 2019 |
| Nomor Soal |
: |
27 |
SOAL
Besaran klaim berdistribusi eksponensial dengan rata-rata \(\lambda \). \(\lambda \) bervariasi untuk setiap pemegang polis dan mengikuti distribusi Pareto dengan parameter \(\alpha = 5\) dan \(\theta \). Variansi dari besar klaim sebesar 9,75. Hitung besar \(\theta \)
- 4
- 5
- 6
- 7
- 8
| Diketahui |
- Besaran klaim berdistribusi eksponensial dengan rata-rata \(\lambda \).
- \(\lambda \) bervariasi untuk setiap pemegang polis dan mengikuti distribusi Pareto dengan parameter \(\alpha = 5\) dan \(\theta \).
- Variansi dari besar klaim sebesar 9,75.
|
| Rumus yang digunakan |
\(Var\left[ X \right] = E\left[ {{X^2}} \right] – {\left( {E\left[ {{X^2}} \right]} \right)^2} = E\left[ {Var\left( {\left. X \right|Y} \right)} \right] + Var\left( {E\left[ {\left. X \right|Y} \right]} \right)\)
\(= E\left[ {{Y^2}} \right] + Var\left[ Y \right]\)
Eksponensial: \(E\left[ X \right] = \lambda \) dan \(Var\left( X \right) = {\lambda ^2}\)
Pareto: \(E\left[ X \right] = \frac{\theta }{{\alpha – 1}}\) dan \(Var\left( X \right) = \frac{{2{\theta ^2}}}{{\left( {\alpha – 1} \right)\left( {\alpha – 2} \right)}} – {\left( {\frac{\theta }{{\alpha – 1}}} \right)^2}\) |
| Proses pengerjaan |
Sesuai distribusi Eksponensial diperoleh
\(Var\left( {\left. X \right|\lambda } \right) = {\lambda ^2}\) dan \(E\left[ {\left. X \right|\lambda } \right] = \lambda \) |
|
Berdasarkan variansi dari besar klaim
\(Var\left[ X \right] = E\left[ {Var\left( {\left. X \right|\lambda } \right)} \right] + Var\left( {E\left[ {\left. X \right|\lambda } \right]} \right)\)
\(9.75 = E\left[ {{\lambda ^2}} \right] + Var\left( \lambda \right)\)
\(9.75 = \frac{{2{\theta ^2}}}{{\left( {\alpha – 1} \right)\left( {\alpha – 2} \right)}} + \left[ {\frac{{2{\theta ^2}}}{{\left( {\alpha – 1} \right)\left( {\alpha – 2} \right)}} – {{\left( {\frac{\theta }{{\alpha – 1}}} \right)}^2}} \right]\)
\(9.75 = \frac{{2{\theta ^2}}}{{\left( {5 – 1} \right)\left( {5 – 2} \right)}} + \left[ {\frac{{2{\theta ^2}}}{{\left( {5 – 1} \right)\left( {5 – 2} \right)}} – {{\left( {\frac{\theta }{{5 – 1}}} \right)}^2}} \right]\)
\(9.75 = \frac{{2{\theta ^2}}}{{12}} + \left( {\frac{{2{\theta ^2}}}{{12}} – \frac{{{\theta ^2}}}{{16}}} \right) = \frac{{13}}{{48}}{\theta ^2}\)
\(\theta = \sqrt {\frac{{9.75\left( {48} \right)}}{{13}}} = 6\) |
| Jawaban |
c. 6 |
