Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Pemodelan dan Teori Risiko |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
15 |
SOAL
Diketahui:
- Frekuensi banyaknya klaim mengikuti distribusi Poisson dengan rata-rata \(\lambda \)
- Severity klaim mengikuti distribusi lognormal dengan parameter \(\mu \) dan \(\sigma \)
- Frekuensi banyaknya klaim dan nilai severity klaim saling bebas
- Distribusi prior memiliki fungsi peluang kepadatan gabungan
\(\begin{array}{*{20}{c}} {f\left( {\lambda ,\mu ,\sigma } \right) = 2\sigma ,}&{0 < \lambda < 1,}&{0 < \mu < 1,}&{0 < \sigma < 1} \end{array}\)
Hitung standard kredibilitas Buhlman \(k\) untuk aggregate loss
- 1,69
- 2,69
- 4,69
- 6,69
- 8,69
| Diketahui |
- Frekuensi banyaknya klaim mengikuti distribusi Poisson dengan rata-rata \(\lambda \)
- Severity klaim mengikuti distribusi lognormal dengan parameter \(\mu \) dan \(\sigma \)
- Frekuensi banyaknya klaim dan nilai severity klaim saling bebas
- Distribusi prior memiliki fungsi peluang kepadatan gabungan
\(\begin{array}{*{20}{c}} {f\left( {\lambda ,\mu ,\sigma } \right) = 2\sigma ,}&{0 < \lambda < 1,}&{0 < \mu < 1,}&{0 < \sigma < 1} \end{array}\)
|
| Rumus yang digunakan |
Poison : \(E\left[ N \right] = \lambda \) dan Lognormal: \(E\left[ {{X^k}} \right] = \exp \left( {k\mu + \frac{{{k^2}{\sigma ^2}}}{2}} \right)\)
\(k = \frac{v}{a}\)
\(v = E\left[ {v\left( \Theta \right)} \right] = E\left[ {Var\left( {\left. {{X_j}} \right|\Theta } \right)} \right]\) ekspektasi dari process varians
\(a = Var\left[ {\mu \left( \Theta \right)} \right] = Var\left[ {E\left( {\left. {{X_j}} \right|\Theta } \right)} \right]\) varians dari hypothetical mean |
| Proses pengerjaan |
Diperoleh Hypothetical Mean \(\lambda {e^{\mu + \frac{{{\sigma ^2}}}{2}}}\) dan Process Varians \(\lambda {e^{2\mu + 2{\sigma ^2}}}\)
Karena \(\lambda \), \(\mu \), dan \(\sigma \) saling bebas ma ka kita bisa menghitungnya secara terpisah |
| \(E\left[ {HM} \right] = E\left[ {\lambda {e^{\mu + \frac{{{\sigma ^2}}}{2}}}} \right] = E\left[ \lambda \right]E\left[ {{e^\mu }} \right]E\left[ {{e^{\frac{{{\sigma ^2}}}{2}}}} \right]\)
\(E\left[ {HM} \right] = \int\limits_0^1 {\lambda d\lambda } \int\limits_0^1 {{e^\mu }d\mu } \int\limits_0^1 {2\sigma {e^{\frac{{{\sigma ^2}}}{2}}}d\sigma } \)
\(E\left[ {HM} \right] = \left( {\frac{1}{2}} \right)\left( {e – 1} \right)\left( 2 \right)\left( {{e^{\frac{1}{2}}} – 1} \right) = 1.114686\)
\(E\left[ {H{M^2}} \right] = E\left[ {{\lambda ^2}} \right]E\left[ {{e^{2\mu }}} \right]E\left[ {{e^{{\sigma ^2}}}} \right]\)
\(E\left[ {H{M^2}} \right] = \int\limits_0^1 {{\lambda ^2}d\lambda } \int\limits_0^1 {{e^{2\mu }}d\mu } \int\limits_0^1 {2\sigma {e^{{\sigma ^2}}}d\sigma } \)
\(E\left[ {H{M^2}} \right] = \left( {\frac{1}{3}} \right)\left( {\frac{{{e^2} – 1}}{2}} \right)\left( {e – 1} \right) = 1.829700\)
\(a = Var\left[ {HM} \right] = 1.829700 – {1.114686^2} = 0.587175\) |
| \(E\left[ {PV} \right] = E\left[ \lambda \right]E\left[ {{e^{2\mu }}} \right]E\left[ {{e^{2{\sigma ^2}}}} \right]\)
\(E\left[ {PV} \right] = \int\limits_0^1 {\lambda d\lambda } \int\limits_0^1 {{e^{2\mu }}d\mu } \int\limits_0^1 {2\sigma {e^{2{\sigma ^2}}}d\sigma } \)
\(v = E\left[ {PV} \right] = \left( {\frac{1}{2}} \right)\left( {\frac{{{e^2} – 1}}{2}} \right)\left( {\frac{{{e^2} – 1}}{2}} \right) = 5.102505\) |
| \(k = \frac{v}{a} = \frac{{5.102505}}{{0.587175}} = 8.6899\) |
| Jawaban |
e. 8,69 |
