Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
Juni 2015 |
| Nomor Soal |
: |
30 |
SOAL
Untuk \(S = {X_1} + {X_2} + \cdots + {X_N}\);
- \({X_1},{X_2}, \ldots \) setiap \(x\) berdistribusi eksponensial dengan rata-rata \(\theta \)
- Variabel acak \(N,{X_1},{X_2}, \ldots \) saling independent
- \(N\) berdistribusi Poisson dengan rata-rata 1; dan
- \({M_S}\left( 1 \right) = 3\)
Tentukan nilai dari \(\theta \)
- 0,50
- 0,52
- 0,54
- 0,56
- 0,58
| Diketahui |
Untuk \(S = {X_1} + {X_2} + \cdots + {X_N}\);
- \({X_1},{X_2}, \ldots \) setiap \(x\) berdistribusi eksponensial dengan rata-rata \(\theta \)
- Variabel acak \(N,{X_1},{X_2}, \ldots \) saling independent
- \(N\) berdistribusi Poisson dengan rata-rata 1; dan
- \({M_S}\left( 1 \right) = 3\)
|
| Rumus yang digunakan |
MGF Eksponensial: \({M_X}\left( t \right) = E\left( {{e^{tX}}} \right) = \frac{1}{{1 – \theta t}}\)
MGF Poisson: \({M_N}\left( t \right) = E\left( {{e^{tN}}} \right) = \exp \left[ {\lambda \cdot \left( {{e^t} – 1} \right)} \right]\)
MGF Coumpound Distribution: \({M_S}\left( t \right) = {M_N}\left[ {\ln {M_X}\left( t \right)} \right]\) |
| Proses pengerjaan |
\({M_S}\left( t \right) = {M_N}\left[ {\ln {M_X}\left( t \right)} \right]\)
\({M_S}\left( t \right) = \exp \left[ {\lambda \cdot \left( {{e^{\ln \left( {\frac{1}{{1 – \theta t}}} \right)}} – 1} \right)} \right]\)
\({M_S}\left( t \right) = \exp \left[ {\lambda \cdot \left( {\frac{1}{{1 – \theta t}} – 1} \right)} \right]\)
Karena \(\lambda = 1\) dan \({M_S}\left( 1 \right) = 3\), maka
\({M_S}\left( t \right) = \exp \left[ {\lambda \cdot \left( {\frac{1}{{1 – \theta t}} – 1} \right)} \right]\)
\({M_S}\left( 1 \right) = \exp \left[ {\frac{1}{{1 – \theta }} – 1} \right] = 3\)
\(\frac{\theta }{{1 – \theta }} = \ln \left( 3 \right)\)
\(\theta = 0.523495\) |
| Jawaban |
d. 0,52 |
