Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui model AR(1) dengan data sebagai berikut:
\({y_1} = 2,0 ; {y_2} = – 1,7 ; {y_3} = 1,5 ; {y_4} = – 2,0 ; {y_5} = 1,5\)Diberikan nilai awal \({\varepsilon _1} = 0 ; \mu = 0 ; {\rho _1} = 0,5\)
Tentukan nilai dari fungsi Sum of Square \(S = \sum {{{\left[ {{\varepsilon _t}\left| {{\varepsilon _1} = 0;\mu = 0;{\rho _1} = 0,5} \right.} \right]}^2}} \)
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| Diketahui | \({y_1} = 2,0 ; {y_2} = – 1,7 ; {y_3} = 1,5 ; {y_4} = – 2,0 ; {y_5} = 1,5\) \({\varepsilon _1} = 0 ; \mu = 0 ; {\rho _1} = 0,5\) | ||||||||||||||||||||||||||||||||||
| 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 \(\sum\nolimits_{i = 1}^5 {\varepsilon _i^2} = 26,63\) | |||||||||||||||||||||||||||||||||||
| Jawaban | e. 27 | ||||||||||||||||||||||||||||||||||