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 |
| \(t\) |
\({y_t}\) |
\({y_t} – \bar y\) |
| 1 |
980 |
-15 |
| 2 |
1.020 |
25 |
| 3 |
960 |
-40 |
| 4 |
1.030 |
35 |
| 5 |
985 |
-12 |
|
| 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 |
