Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | November 2015 |
| Nomor Soal | : | 25 |
SOAL
Untuk sebuah deret waktu \({y_t}\), diketahui:
| \(t\) | \({y_t}\) | \({y_t} – \bar y\) |
| 1 | 980 | -15 |
| 2 | 1.020 | 25 |
| 3 | 960 | -40 |
| 4 | 1.030 | 35 |
| 5 | 985 | -12 |
Hitunglah estimasi fungsi autokorelasi parsial (partial autocorrelation) pada time displacement \(k = 2\) (dibulatkan 2 desimal)
- 0,41
- -0,63
- -0,46
- -0,84
- 0,35
| Diketahui |
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| Rumus yang digunakan | Autocorrelation: \({r_k} = \frac{{\sum\nolimits_{i = 1}^{n – k} {\left( {{y_t} – \bar y} \right)\left( {{y_{i + k}} – \bar y} \right)} }}{{\sum\nolimits_{i = 1}^n {{{\left( {{y_t} – \bar y} \right)}^2}} }}\) Partial Autocorrelation: \({{\hat \varphi }_{11}} = {r_1}\) \({{\hat \varphi }_{22}} = \frac{{{r_2} – r_1^2}}{{1 – r_1^2}}\) \({{\hat \varphi }_{kj}} = {{\hat \varphi }_{k – 1,j}} – {\varphi _{kk}}{\varphi _{k – 1,k – j}}\) \(k = 2, \ldots ;j = 1,2, \ldots ,k – 1\) \({{\hat \varphi }_{kk}} = \frac{{{r_k} – \sum\nolimits_{j = 1}^{k – 1} {{\varphi _{k – 1,j}}{r_{k – j}}} }}{{1 – \sum\nolimits_{j = 1}^{k – 1} {{\varphi _{k – 1,j}}{r_j}} }}\) \(k = 3, \ldots \) | ||||||||||||||||||
| Proses Pengerjaan | \({r_1} = \frac{{\sum\nolimits_{i = 1}^4 {\left( {{y_t} – \bar y} \right)\left( {{y_{i + 1}} – \bar y} \right)} }}{{\sum\nolimits_{i = 1}^5 {{{\left( {{y_t} – \bar y} \right)}^2}} }}\) \(= \frac{{\left( { – 15} \right)\left( {25} \right) + \left( {25} \right)\left( { – 40} \right) + \left( { – 40} \right)\left( {35} \right) + \left( {35} \right)\left( { – 12} \right)}}{{{{\left( { – 15} \right)}^2} + {{\left( {25} \right)}^2} + {{\left( { – 40} \right)}^2} + {{\left( {35} \right)}^2} + {{\left( { – 12} \right)}^2}}}\) \(= – \frac{{1065}}{{1273}} = – 0,836606\) | ||||||||||||||||||
| \({r_2} = \frac{{\sum\nolimits_{i = 1}^3 {\left( {{y_t} – \bar y} \right)\left( {{y_{i + 2}} – \bar y} \right)} }}{{\sum\nolimits_{i = 1}^5 {{{\left( {{y_t} – \bar y} \right)}^2}} }}\) \(= \frac{{\left( { – 15} \right)\left( { – 40} \right) + \left( {25} \right)\left( {35} \right) + \left( { – 40} \right)\left( { – 12} \right)}}{{{{\left( { – 15} \right)}^2} + {{\left( {25} \right)}^2} + {{\left( { – 40} \right)}^2} + {{\left( {35} \right)}^2} + {{\left( { – 12} \right)}^2}}}\) \(= \frac{{1955}}{{3819}} = 0,511914\) | |||||||||||||||||||
| \({{\hat \varphi }_{22}} = \frac{{{r_2} – r_1^2}}{{1 – r_1^2}}\) \(= \frac{{\frac{{1955}}{{3819}} – {{\left( { – \frac{{1065}}{{1273}}} \right)}^2}}}{{1 – {{\left( { – \frac{{1065}}{{1273}}} \right)}^2}}}\) \(= – 0,626467\) | |||||||||||||||||||
| Jawaban | b. -0,63 |