Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
Metoda Statistika |
Periode Ujian |
: |
November 2014 |
Nomor Soal |
: |
28 |
SOAL
Dari sebuah time series, Anda diberikan data sebagai berikut:
\(t\) |
\({y_t}\) |
\({y_t} – \bar y\) |
1 |
984 |
-16 |
2 |
1023 |
23 |
3 |
965 |
-35 |
4 |
1040 |
40 |
5 |
988 |
-12 |
Perkirakan fungsi autokorelasi parsial pada saat perpindahan \(k = 2\)
- – 0,46
- – 0,16
- 0,51
- 0,84
- Tidak ada jawaban yang benar
Diketahui |
Dari sebuah time series, Anda diberikan data sebagai berikut:
\(t\) |
\({y_t}\) |
\({y_t} – \bar y\) |
1 |
984 |
-16 |
2 |
1023 |
23 |
3 |
965 |
-35 |
4 |
1040 |
40 |
5 |
988 |
-12 |
|
Rumus yang digunakan |
Autocorrelation
\({r_k} = \frac{{\sum\nolimits_{t = 1}^{T – k} {\left( {{y_t} – \bar y} \right)\left( {{y_{t + k}} – \bar y} \right)} }}{{\sum\nolimits_{t = 1}^T {{{\left( {{y_t} – \bar y} \right)}^2}} }}\) dengan \(\bar y = \frac{1}{T}\sum\limits_{t = 1}^T {{y_t}} \)
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 |
- \({r_1} = \frac{{\sum\limits_{t = 1}^4 {\left( {{y_t} – \bar y} \right)\left( {{y_{t + 1}} – \bar y} \right)} }}{{\sum\limits_{t = 1}^5 {{{\left( {{y_t} – \bar y} \right)}^2}} }}\)
\({r_1} = \frac{{\left( { – 16} \right)\left( {23} \right) + \left( {23} \right)\left( { – 35} \right) + \left( { – 35} \right)\left( {40} \right) + \left( {40} \right)\left( { – 12} \right)}}{{{{\left( { – 16} \right)}^2} + {{\left( {23} \right)}^2} + {{\left( { – 35} \right)}^2} + {{\left( {40} \right)}^2} + {{\left( { – 12} \right)}^2}}}\)
\({r_1} = – \frac{{3053}}{{3754}}\)
- \({r_2} = \frac{{\sum\limits_{t = 1}^3 {\left( {{y_t} – \bar y} \right)\left( {{y_{t + 2}} – \bar y} \right)} }}{{\sum\limits_{t = 1}^5 {{{\left( {{y_t} – \bar y} \right)}^2}} }}\)
\({r_2} = \frac{{\left( { – 16} \right)\left( { – 35} \right) + \left( {23} \right)\left( {40} \right) + \left( { – 35} \right)\left( { – 12} \right)}}{{{{\left( { – 16} \right)}^2} + {{\left( {23} \right)}^2} + {{\left( { – 35} \right)}^2} + {{\left( {40} \right)}^2} + {{\left( { – 12} \right)}^2}}}\)
\({r_2} = \frac{{950}}{{1877}}\)
- \({{\hat \varphi }_{22}} = \frac{{{r_2} – r_1^2}}{{1 – r_1^2}}\)
\({{\hat \varphi }_{22}} = \frac{{\frac{{950}}{{1877}} – {{\left( { – \frac{{3053}}{{3754}}} \right)}^2}}}{{1 – {{\left( { – \frac{{3053}}{{3754}}} \right)}^2}}}\)
\({{\hat \varphi }_{22}} = – 0.458579\)
|
Jawaban |
A. – 0,46 |