Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diberikan forecast error 3 langkah ke depan berdasarkan ARIMA model
\({e_T}(3) = 0,34{\varepsilon _{T + 3}} + 0,26{\varepsilon _{T + 2}} – 0,55{\varepsilon _{T + 1}}\)Diketahui pula, variance dari forecast error adalah 0,89.
Hitunglah variance dari error, \({\sigma _\varepsilon }^2\)
- 0,89
- 1,10
- 1,83
- 2,15
- 2,50
| Diketahui | \({e_T}(3) = {\psi _0}{\varepsilon _{T + 3}} + {\psi _1}{\varepsilon _{T + 2}} + {\psi _2}{\varepsilon _{T + 1}}\) \(= 0,34{\varepsilon _{T + 3}} + 0,26{\varepsilon _{T + 2}} – 0,55{\varepsilon _{T + 1}}\) variance dari forecast error = 0,89 |
| Rumus yang digunakan | \(Var\left[ {{e_T}\left( l \right)} \right] = \left( {{\psi _0}^2 + {\psi _1}^2 + \ldots + {\psi _{l – 1}}^2} \right)\sigma _\varepsilon ^2\) |
| Proses Pengerjaan | \(Var\left[ {{e_T}\left( 3 \right)} \right] = \left( {{\psi _0}^2 + {\psi _1}^2 + {\psi _2}^2} \right)\sigma _\varepsilon ^2\) \(\sigma _\varepsilon ^2 = \frac{{Var\left[ {{e_T}\left( 3 \right)} \right]}}{{{\psi _0}^2 + {\psi _1}^2 + {\psi _2}^2}}\) \(= \frac{{0,89}}{{{{0,34}^2} + {{0,26}^2} + {{\left( { – 0,55} \right)}^2}}}\) \(= 1,8324\) |
| Jawaban | c. 1,83 |