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 \(S = \sum {{{\left[ {{\varepsilon _t}\left| {{\varepsilon _1} = 0;\mu = 0;{\rho _1} = 0,5} \right.} \right]}^2}} \)
- 2
- 12
- 15
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- 27
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 |