Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | November 2015 |
| Nomor Soal | : | 30 |
SOAL
Untuk model double decrement di bawah ini
- \({}_tp_{30}^{‘\left( d \right)} = 1 – \frac{t}{{55}},0 \le t \le 55\)
- \({}_tp_{30}^{‘\left( w \right)} = 1 – \frac{t}{{30}},0 \le t \le 30\)
Hitunglah \(\mu _{30 + 15}^{\left( \tau \right)}\) (dibulatkan 5 desimal)
- 0,01678
- 0,02000
- 0,07500
- 0,09167
- 0,20556
| Diketahui | \({}_tp_{30}^{‘\left( d \right)} = 1 – \frac{t}{{55}},0 \le t \le 55\) dan \({}_tp_{30}^{‘\left( w \right)} = 1 – \frac{t}{{30}},0 \le t \le 30\) |
| Rumus yang digunakan | \({}_tp_x^{\left( \tau \right)} = \prod\limits_{i = j}^m {{}_tp_x^{‘\left( j \right)}} \)
\(\mu _{x + t}^{\left( \tau \right)} = – \frac{1}{{{}_tp_x^{\left( \tau \right)}}} \cdot \frac{d}{{dt}}{}_tp_x^{\left( \tau \right)}\)
\(= – \frac{d}{{dt}}\ln {}_tp_x^{\left( \tau \right)}\) |
| Proses pengerjaan | \({}_tp_{30}^{\left( \tau \right)} = {}_tp_{30}^{‘\left( d \right)} \cdot {}_tp_{30}^{‘\left( w \right)}\)
\(= \left( {1 – \frac{t}{{55}}} \right) \cdot \left( {1 – \frac{t}{{30}}} \right)\)
\(= 1 – \frac{{17}}{{330}}t + \frac{1}{{1650}}{t^2}\)
\(= {t^2} – 85t + 1650\) |
| \(\mu _{40 + t}^{\left( \tau \right)} = – \frac{1}{{{}_tp_x^{\left( \tau \right)}}} \cdot \frac{d}{{dt}}{}_tp_x^{\left( \tau \right)}\)
\(= \frac{1}{{{t^2} – 85t + 1650}} \cdot \frac{d}{{dt}}\left( { – {t^2} + 85t – 1650} \right)\)
\(= \frac{{85 – 2t}}{{{t^2} – 85t + 1650}}\)
\(\mu _{30 + 15}^{\left( \tau \right)} = \frac{{85 – 2\left( {15} \right)}}{{{{\left( {15} \right)}^2} – 85\left( {15} \right) + 1650}}\)
\(= 0,0916666667\) |
| Jawaban | d. 0,09167 |
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