Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Permodelan dan Teori Risiko |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 8 |
SOAL
Untuk setiap pemegang polis asuransi, besar kerugian \({X_1}, \ldots ,{X_n}\), bersyarat \(\Theta \), saling bebas dan identik dengan rata-rata
\(\begin{array}{*{20}{c}} {\mu \left( \theta \right) = E\left[ {\left. {{X_j}} \right|\Theta = \theta } \right],}&{j = 1,2, \ldots ,n} \end{array}\)
dan variansi
\(\begin{array}{*{20}{c}} {v\left( \theta \right) = Var\left[ {\left. {{X_j}} \right|\Theta = \theta } \right],}&{j = 1,2, \ldots ,n} \end{array}\)
Diketahui
- Buhlman credibility mengestiamsi \({X_5}\) berdasarkan \({X_1}, \ldots ,{X_4}\) adalah \(Z = 0,4\)
- Nilai ekspektasi dari proses variansi ialah 8
Hitung \(Cov\left( {{X_i},{X_j}} \right),i \ne j\)
- \(– \frac{1}{3}\)
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
- \(\frac{4}{3}\)
- 2
| Diketahui | Untuk setiap pemegang polis asuransi, besar kerugian \({X_1}, \ldots ,{X_n}\), bersyarat \(\Theta \), saling bebas dan identik dengan rata-rata \(\begin{array}{*{20}{c}} {\mu \left( \theta \right) = E\left[ {\left. {{X_j}} \right|\Theta = \theta } \right],}&{j = 1,2, \ldots ,n} \end{array}\)
dan variansi \(\begin{array}{*{20}{c}} {v\left( \theta \right) = Var\left[ {\left. {{X_j}} \right|\Theta = \theta } \right],}&{j = 1,2, \ldots ,n} \end{array}\)
Diketahui - Buhlman credibility mengestiamsi \({X_5}\) berdasarkan \({X_1}, \ldots ,{X_4}\) adalah \(Z = 0,4\)
- Nilai ekspektasi dari proses variansi ialah 8
|
| Rumus yang digunakan | \(\begin{array}{*{20}{c}} {k = \frac{v}{a},}&{Z = \frac{n}{{n + k}}} \end{array}\)
\(Cov\left( {{X_i},{X_j}} \right) = E\left[ {{X_i}{X_j}} \right] – E\left[ {{X_i}} \right]E\left[ {{X_j}} \right]\)
\(Cov\left( {{X_i},{X_j}} \right) = E\left[ {E\left( {\left. {{X_i}{X_j}} \right|\theta } \right)} \right] – E\left[ {E\left( {\left. {{X_i}} \right|\theta } \right)} \right]E\left[ {E\left( {\left. {{X_j}} \right|\theta } \right)} \right]\)
\(Cov\left( {{X_i},{X_j}} \right) = E\left[ {\mu {{\left( \theta \right)}^2}} \right] – E{\left[ {\mu \left( \theta \right)} \right]^2}\)
\(Cov\left( {{X_i},{X_j}} \right) = Var\left[ {\mu \left( \theta \right)} \right] = a\) |
| Proses pengerjaan | Diberikan \(n = 4\), \(v = 8\), dan \(Z = 0.4\). Maka diperoleh
\(Z = \frac{n}{{n + k}} = \frac{n}{{n + \frac{v}{a}}}\)
\(0.4 = \frac{4}{{4 + \frac{8}{a}}}\)
\(a = \frac{4}{3}\) |
| \(Cov\left( {{X_i},{X_j}} \right) = Var\left[ {\mu \left( \theta \right)} \right] = a = \frac{4}{3}\) |
| Jawaban | d. \(\frac{4}{3}\) |
