Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Pemodelan dan Teori Risiko |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
16 |
SOAL
Diketahui:
| Besar Klaim \(\left( X \right)\) |
Banyak Klaim |
| \(\left( {0,25} \right]\) |
25 |
| \(\left( {25,50} \right]\) |
28 |
| \(\left( {50,100} \right]\) |
15 |
| \(\left( {100,200} \right]\) |
6 |
Asumsikan setiap besar klaim di atas mengikuti distribusi uniform. Estimasi \(E\left[ {{X^2}} \right] – E\left[ {{{\left( {X \wedge 150} \right)}^2}} \right]\)
- Kurang dari 200
- Paling sedikit 200, tetapi kurang dari 300
- Paling sedikit 300, tetapi kurang dari 400
- Paling sedikit 400, tetapi kurang dari 500
- Paling sedikit 500
| Diketahui |
| Besar Klaim \(\left( X \right)\) |
Banyak Klaim |
| \(\left( {0,25} \right]\) |
25 |
| \(\left( {25,50} \right]\) |
28 |
| \(\left( {50,100} \right]\) |
15 |
| \(\left( {100,200} \right]\) |
6 |
Asumsikan setiap besar klaim di atas mengikuti distribusi uniform |
| Rumu yang digunakan |
Distribusi uniform: \(f\left( x \right) = \frac{1}{{b – a}}\)
\(E\left[ {{X^2}} \right] = \int\limits_{ – \infty }^\infty {{x^2}f\left( x \right)} \)
\(E\left[ {{{\left( {X \wedge u} \right)}^k}} \right] = \int\limits_{ – \infty }^u {{x^k}f\left( x \right)dx} + {u^k}S\left( u \right) = \int\limits_{ – \infty }^u {{x^k}f\left( x \right)dx} + {u^k}\int\limits_u^\infty {f\left( x \right)dx} \) |
| Proses pengerjaan |
\(E\left[ {{X^2}} \right] – E\left[ {{{\left( {X \wedge 150} \right)}^2}} \right] = \int_0^{200} {{x^2}f\left( x \right)dx} – \left[ {\int_0^{150} {{x^2}f\left( x \right)dx} + {{150}^2}\int_{150}^{200} {f\left( x \right)dx} } \right]\)
\(E\left[ {{X^2}} \right] – E\left[ {{{\left( {X \wedge 150} \right)}^2}} \right] = \int_{150}^{200} {\left( {{x^2} – {{150}^2}} \right)f\left( x \right)dx} \)
Karena asumsi mengikuti distribusi uniform maka untuk interval \(\left( {100,200} \right]\)\(f\left( x \right) = \frac{6}{{25 + 28 + 15 + 6}}\left( {\frac{1}{{200 – 100}}} \right) = \frac{6}{{7400}}\)
Sehingga diperoleh
\(E\left[ {{X^2}} \right] – E\left[ {{{\left( {X \wedge 150} \right)}^2}} \right] = \int_{150}^{200} {\left( {{x^2} – {{150}^2}} \right)\frac{6}{{7400}}dx} \)
\(E\left[ {{X^2}} \right] – E\left[ {{{\left( {X \wedge 150} \right)}^2}} \right] = \left. {\frac{6}{{7400}}\left( {\frac{{{x^3}}}{3} – {{150}^2} \cdot x} \right)} \right|_{150}^{200} = 337.84\) |
| Jawaban |
c. Paling sedikit 300, tetapi kurang dari 400 |
