Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Anda mencocokkan sebuah model Autoregressive \(AR\left( 1 \right)\) terhadap data berikut ini
\({y_1} = 2,0\)
\({y_2} = – 1,8\)
\({y_3} = 1,4\)
\({y_4} = – 2,0\)
\({y_5} = 1,2\)
Anda memilih \({\varepsilon _1} = 0\), \(\mu = 0\), \({\rho _1} = 0,5\) sebagai nilai awal.
Hitunglah nilai dari jumlah kuadrat dari fungsi \(S = \sum {{{\left[ {\left. {{\varepsilon _t}} \right|{\varepsilon _1} = 0,\mu = 0,{\rho _1} = 0,5} \right]}^2}} \), yaitu nilai dari \(\sum\nolimits_{i = 1}^5 {\varepsilon _i^2} \) (dibulatkan 2 desimal)
- 25,26
- 18,87
- 20,42
- 28,21
- 26,63
| Diketahui | \({y_1} = 2,0;{y_2} = – 1,8;{y_3} = 1,4;{y_4} = – 2,0;{y_5} = 1,2\) \({\varepsilon _1} = 0\); \(\mu = 0\); \({\rho _1} = 0,5\) | ||||||||||||||||||||||||||||||||||
| Rumus yang digunakan | Untuk \(AR\left( p \right)\) \({y_t} = {\phi _1}{y_{t – 1}} + {\phi _2}{y_{t – 2}} + \cdots + {\phi _p}{y_{t – p}} + \delta + {\varepsilon _t}\) \(\mu = \frac{\delta }{{1 – {\phi _1} – {\phi _2} – \cdots – {\phi _p}}}\) \({\rho _k} = \phi _1^k\) \({\varepsilon _t} = {y_t} – {\hat y_t} = {y_t} – {\phi _1}{y_{t – 1}} – {\phi _2}{y_{t – 2}} – \cdots – {\phi _p}{y_{t – p}} – \delta \) | ||||||||||||||||||||||||||||||||||
| Proses pengerjaan | \({\rho _1} = {\phi _1}\) \({\phi _1} = 0,5\) | ||||||||||||||||||||||||||||||||||
| \(\mu = \frac{\delta }{{1 – {\phi _1}}}\) \(0 = \frac{\delta }{{1 – 0,5}}\) \(\delta = 0\) | |||||||||||||||||||||||||||||||||||
Diperoleh model \(AR\left( 1 \right)\)
\({y_t} = 0,5{y_{t – 1}} + {\varepsilon _t}\)
Jadi diperoleh \(\sum\nolimits_{i = 1}^5 {\varepsilon _i^2} = 25,26\) | |||||||||||||||||||||||||||||||||||
| Jawaban | a. 25,26 | ||||||||||||||||||||||||||||||||||