Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2017 |
| Nomor Soal | : | 24 |
SOAL
Diberikan sebagai berikut:
- \({\mu _{x + t}} = 0,01\,\) \(0 \le t < 5\)
- \({\mu _{x + t}} = 0,02\,\) \(5 \le t\)
- \(\delta = 0,06\)
Hitunglah nilai dari \({\bar a_x}\)
- 12,5
- 13,0
- 13,4
- 13,9
- 14,3
| Rumus | \({\bar a_x} = \int\limits_0^\infty {{v^t}\,\,{}_t{p_x}\,\,dt} \) \({}_t{p_x} = \exp [ – \int\limits_0^t {{\mu _{x + s}}\,ds]} \) |
| Step 1 | Untuk \(0 \le t < 5\) \({}_t{p_x} = \exp [ – \int\limits_0^t {0,01\,\,ds]} \) \({}_t{p_x} = \exp [ – 0,01(t – 0)]\) \({}_t{p_x} = \exp [ – 0,01t]\) |
| \({\bar a_x} = \int\limits_0^5 {{v^t}\,\,{}_t{p_x}\,\,dt} \) \({\bar a_x} = \int\limits_0^5 {{e^{ – \delta t}}\,\,{}_t{p_x}\,\,dt} \) \({\bar a_x} = \int\limits_0^5 {{e^{ – 0,06t}}\,\,{e^{ – 0,01t}}\,\,dt} \) \({\bar a_x} = \int\limits_0^5 {{e^{ – (0,06 + 0,01)t}}\,\,dt} \) \({\bar a_x} = \frac{1}{{ – 0,07}}\left( {{e^{ – 0,07(5)}} – {e^{ – 0,07(0)}}} \right)\) \({\bar a_x} = \frac{1}{{ – 0,07}}\left( {{e^{ – 0,35}} – 1} \right)\) \({\bar a_x} = \frac{1}{{0,07}}\left( {1 – {e^{ – 0,35}}} \right)\) \({\bar a_x} \cong 4,21874\) | |
| Untuk \(5 \le t\) | |
| Step 2 | \({}_t{p_x} = \exp \left[ { – \left( {\int\limits_0^5 {0,01\,\,ds + } \int\limits_5^t {0,02\,\,ds} } \right)} \right]\) \({}_t{p_x} = \exp \left[ { – \left( {0,01(5 – 0) + 0,02(t – 5)} \right)} \right]\) \({}_t{p_x} = \exp \left[ { – \left( {0,05 + 0,02t – 0,1} \right)} \right]\) \({}_t{p_x} = \exp \left[ {0,05 – 0,02t} \right]\) |
| \({\bar a_x} = \int\limits_5^\infty {{v^t}\,\,{}_t{p_x}\,\,dt} \) \({\bar a_x} = \int\limits_0^\infty {{e^{ – \delta t}}\,\,{}_t{p_x}\,\,dt} \) \({\bar a_x} = \int\limits_5^\infty {{e^{ – 0,06t}}\,\,{e^{0,05 – 0,02t}}\,\,dt} \) \({\bar a_x} = \int\limits_5^\infty {{e^{ – 0,06t}}\,\,{e^{0,05}}\,{e^{ – 0,02t}}\,\,dt} \) \({\bar a_x} = {e^{0,05}}\int\limits_5^\infty {{e^{ – (0,06 + 0,02)t}}\,\,dt} \) \({\bar a_x} = \frac{{{e^{0,05}}}}{{ – (0,08)}}\left( {{e^{ – (0,08)(\infty )}} – {e^{ – (0,08)(5)}}} \right)\) \({\bar a_x} = \frac{{{e^{0,05}}}}{{ – (0,08)}}\left( {0 – {e^{ – (0,4)}}} \right)\) \({\bar a_x} = \frac{{{e^{0,05}}}}{{(0,08)}}\left( {{e^{ – (0,4)}}} \right)\) \({\bar a_x} \cong 8,80860\) | |
| Step 3 | \({\bar a_x} = \int\limits_0^5 {{v^t}\,\,{}_t{p_x}\,\,dt} + \int\limits_5^\infty {{v^t}\,\,{}_t{p_x}\,\,dt} \) \({\bar a_x} = 4,21874 + 8,80860\) \({\bar a_x} = 13,02734\) \({\bar a_x} \cong 13,0\) |
| Jawaban | b. 13,0 |
