Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | Juni 2015 |
| Nomor Soal | : | 28 |
SOAL
Misalkan Anda melakukan smoothing deret waktu \({y_t}\) menggunakan metode exponential smoothing 2-parameter dari Holt:
| \(t\) | \({y_t}\) | \({\tilde y_t}\) | \({r_t}\) |
| 1995 | 120,50 | 117,50 | 12,00 |
| 1996 | 135,00 | 130,88 | 12,96 |
| 1997 | 147,70 | 144,80 | 13,64 |
| 1998 | 146,60 | \({\tilde y_{1998}}\) | \({r_{1998}}\) |
Hitunglah forecast 2-periode \({\hat y_{2000}}\) dengan terlebih dahulu melengkapi table di atas dengan deret exponential 2-parameter dari Holt
- Lebih kecil dari 166
- Paling sedikit 166, tetapi lebih kecil dari 172
- Paling sedikit 172, tetapi lebih kecil dari 176
- Paling sedikit 176, tetapi lebih kecil dari 180
- Paling sedikit 180
| Diketahui | smoothing deret waktu \({y_t}\) menggunakan metode exponential smoothing 2-parameter dari Holt:| \(t\) | \({y_t}\) | \({\tilde y_t}\) | \({r_t}\) | | 1995 | 120,50 | 117,50 | 12,00 | | 1996 | 135,00 | 130,88 | 12,96 | | 1997 | 147,70 | 144,80 | 13,64 | | 1998 | 146,60 | \({\tilde y_{1998}}\) | \({r_{1998}}\) |
|
| Rumus yang digunakan | \({{\tilde y}_t} = \alpha {y_t} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{t – 1}} + {r_{t – 1}}} \right)\)
\({r_t} = \gamma \left( {{{\tilde y}_t} – {{\tilde y}_{t – 1}}} \right) + \left( {1 – \gamma } \right){r_{t – 1}}\)
\({{\hat y}_{T + l}} = {{\tilde y}_T} + l{r_T}\) |
| Proses pengerjaan | Mencari \(\alpha \) dan \(\gamma \)
\({{\tilde y}_{1996}} = \alpha {y_{1996}} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{1995}} + {r_{1995}}} \right)\)
\(130.88 = 135\alpha + \left( {1 – \alpha } \right)\left( {117.5 + 12} \right)\)
\(130.88 = 135\alpha + 129.5 – 129.5\alpha \)
\(5.5\alpha = 1.38\)
\(\alpha = 0.2509091 \approx 0.25\)
\({r_{1996}} = \gamma \left( {{{\tilde y}_{1996}} – {{\tilde y}_{1995}}} \right) + \left( {1 – \gamma } \right){r_{1995}}\)
\(12.96 = \gamma \left( {130.88 – 117.5} \right) + \left( {1 – \gamma } \right)12\)
\(12.96 = 13.38\gamma + 12 – 12\gamma \)
\(1.38\gamma = 0.96\)
\(\gamma = 0.6956522 \approx 0.7\) |
| Melengkapi tabel \({\tilde y_{1998}}\) dan \({r_{1998}}\)
\({{\tilde y}_{1998}} = \alpha {y_{1998}} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{1997}} + {r_{1997}}} \right)\)
\({{\tilde y}_{1998}} = 0.25\left( {146.6} \right) + 0.75\left( {144.8 + 13.64} \right)\)
\({{\tilde y}_{1998}} = 155.48\)
\({r_{1998}} = \gamma \left( {{{\tilde y}_{1998}} – {{\tilde y}_{1997}}} \right) + \left( {1 – \gamma } \right){r_{1997}}\)
\({r_{1998}} = 0.7\left( {155.48 – 144.8} \right) + 0.3\left( {13.64} \right)\)
\({r_{1998}} = 11.568\) |
| \({{\hat y}_{1998 + 2}} = {{\tilde y}_{1998}} + 2{r_{1998}}\)
\({{\hat y}_{1998 + 2}} = 155.48 + 2 \cdot 11.568\)
\({{\hat y}_{1998 + 2}} = 178.616\) |
| Jawaban | d. Paling sedikit 176, tetapi lebih kecil dari 180 |
