Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui suatu proses autoregressive-moving average ARMA (1,1)
\({y_t} = 0.9{y_{t – 1}} + 2 + {\varepsilon _t} – 0.2{\varepsilon _{t – 1}}\)
Hitunglah nilai \({\rho _2}\)
- 0,72
- 0,74
- 0,76
- 0,78
- 0,80
| Diketahui | ARMA (1,1) \({y_t} = 0.9{y_{t – 1}} + 2 + {\varepsilon _t} – 0.2{\varepsilon _{t – 1}}{\rm{ (*)}}\) dari (*) diperoleh \(\mu = 2,{\rm{ }}\phi = 0.9,{\rm{ }}\theta = – 0.2\) |
| Rumus yang digunakan | \({\rho _x}(h) = \phi {\rho _x}(h – 1)\) \({\rho _x}(1) = \frac{{(\theta + \phi )(1 + \theta \phi )}}{{1 + 2\theta \phi + {\theta ^2}}}\) \({\rho _x}(0) = 1\) |
| Proses pengerjaan | \({\rho _x}(h) = \phi {\rho _x}(h – 1)\) \({\rho _x}(2) = \phi {\rho _x}(2 – 1) = \phi {\rho _x}(1)\) Sehingga, \({\rho _x}(2) = \phi (\frac{{(\theta + \phi )(1 + \theta \phi )}}{{1 + 2\theta \phi + {\theta ^2}}})\) \(= (0.9)(\frac{{( – 0.2 + 0.9) + (1 + ( – 0.2)(0.9)}}{{1 + 2( – 0.2)(0.9) + {{( – 0.2)}^2}}})\) \(= 0.759\) \(= 0.76\) |
| Jawaban | c. 0,76 |