Pembahasan Soal Ujian Profesi Aktuaris
Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian | : | Metoda Statistika |
Periode Ujian | : | April 2019 |
Nomor Soal | : | 12 |
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 \(AR\left( 1 \right)\)
\({y_t} = 0,5{y_{t – 1}} + {\varepsilon _t}\)
\(t\) | \({y_t}\) | \({\hat y_t}\) | \({\varepsilon _t}\) | \(\varepsilon _t^2\) | 1 | 2.00 | | 0.00 | 0.00 | 2 | -1.70 | 1.00 | -2.70 | 7.29 | 3 | 1.50 | -0.85 | 2.35 | 5.52 | 4 | 2.00 | 0.75 | -2.75 | 7.56 | 5 | 1.50 | -1.00 | 2.50 | 6.25 | Total | 26.63 | Jadi, diperoleh \(\sum\nolimits_{i = 1}^5 {\varepsilon _i^2} = 26,63\) |
Jawaban | e. 27 |