Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Untuk sebuah model double decrement:
- \({}_tp_{40}^{‘\left( 1 \right)} = 1 – \frac{t}{{65}},0 \le t \le 65\)
- \({}_tp_{40}^{‘\left( 2 \right)} = 1 – \frac{t}{{30}},0 \le t \le 30\)
Hitunglah \(\mu _{40 + 15}^{\left( \tau \right)}\) (dibulatkan 3 desimal)
- 0,058
- 0,067
- 0,075
- 0,080
- 0,087
| Diketahui | \({}_tp_{40}^{‘\left( 1 \right)} = 1 – \frac{t}{{65}},0 \le t \le 65\) dan \({}_tp_{40}^{‘\left( 2 \right)} = 1 – \frac{t}{{30}},0 \le t \le 30\) |
| Rumus yang digunakan | \({}_tp_x^{\left( \tau \right)} = \prod\limits_{i = 1}^m {{}_tp_x^{‘\left( i \right)}} \) \(\mu _{x + t}^{\left( \tau \right)} = – \frac{1}{{{}_tp_x^{\left( \tau \right)}}} \cdot \frac{d}{{dt}}{}_tp_x^{\left( \tau \right)}\) \(= – \frac{d}{{dt}}\ln {}_tp_x^{\left( \tau \right)}\) |
| Proses pengerjaan | \({}_tp_{40}^{\left( \tau \right)} = {}_tp_{40}^{‘\left( 1 \right)} \cdot {}_tp_{40}^{‘\left( 2 \right)}\) \(= \left( {1 – \frac{t}{{65}}} \right) \cdot \left( {1 – \frac{t}{{30}}} \right)\) \(= 1 – \frac{{19}}{{390}}t + \frac{1}{{1950}}{t^2}\) \(= {t^2} – 95t + 1950\) |
| \(\mu _{40 + t}^{\left( \tau \right)} = – \frac{1}{{{}_tp_x^{\left( \tau \right)}}} \cdot \frac{d}{{dt}}{}_tp_x^{\left( \tau \right)}\) \(= \frac{1}{{{t^2} – 95t + 1950}} \cdot \frac{d}{{dt}}\left( { – {t^2} + 95t – 1950} \right)\) \(= \frac{{95 – 2t}}{{{t^2} – 95t + 1950}}\) \(\mu _{40 + 15}^{\left( \tau \right)} = \frac{{95 – 2\left( {15} \right)}}{{{{\left( {15} \right)}^2} – 95\left( {15} \right) + 1950}}\) \(= 0,08666667\) | |
| Jawaban | b. 0,087 |