Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2016 |
| Nomor Soal | : | 15 |
SOAL
Untuk “two lives” (50) dan (60) dengan “independent future lifetime”:
- \({\mu _{50 + t}} = 0,002t\), \(t > 0\)
- \({\mu _{60 + t}} = 0,003t\), \(t > 0\)
Hitunglah \({}_{20}q_{50:60}^1 – {}_{20}{q_{50:\mathop {60}\limits^2 }}\)
- 0,17
- 0,18
- 0,30
- 0,31
- 0,37
| Diketahui | Untuk “two lives” (50) dan (60) dengan “independent future lifetime”:
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| Rumus yang digunakan |
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| Proses pengerjaan | \({f_{{T_{50}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {{\mu _{50 + s}}ds} } \right] \cdot {\mu _{50 + t}}\) \({f_{{T_{50}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {0.002sds} } \right] \cdot 0.002t = 0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]\) \({f_{{T_{60}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {{\mu _{60 + s}}ds} } \right] \cdot {\mu _{60 + t}}\) \({f_{{T_{50}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {0.003sds} } \right] \cdot 0.003t = 0.003t \cdot \exp \left[ { – 0.0015{t^2}} \right]\) \({}_{20}q_{50:60}^1 = P\left( {{T_{50}} \le 20{\rm{ dan }}{T_{50}} < {T_{60}}} \right)\) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\int\limits_t^\infty {{f_{{T_{60}},{T_{50}}}}\left( {s,t} \right)ds} dt} \) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\int\limits_t^\infty {{f_{{T_{60}}}}\left( s \right) \cdot {f_{{T_{50}}}}\left( t \right)ds} dt} \) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\int\limits_t^\infty {\left( {0.003s \cdot \exp \left[ { – 0.0015{s^2}} \right]} \right) \cdot \left( {0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]} \right)ds} dt} \) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\left( {0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]} \right)\int\limits_t^\infty {\left( {0.003s \cdot \exp \left[ { – 0.0015{s^2}} \right]} \right) \cdot ds} dt} \) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\left( {0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]} \right) \cdot \exp \left[ { – 0.0015{t^2}} \right]dt} \) \({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\left( {0.002t \cdot \exp \left[ { – 0.0025{t^2}} \right]} \right)dt} \) \({}_{20}q_{50:60}^1 = 0.252848\) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = P\left( {{T_{60}} \le 20{\rm{ dan }}{T_{50}} < {T_{60}}} \right)\) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\int\limits_0^t {{f_{{T_{50}},{T_{60}}}}\left( {s,t} \right)ds} dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\int\limits_0^t {{f_{{T_{50}}}}\left( s \right) \cdot {f_{{T_{60}}}}\left( t \right)ds} dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\int\limits_0^t {\left( {0.002s \cdot \exp \left[ { – 0.001{s^2}} \right]} \right) \cdot \left( {0.003t \cdot \exp \left[ { – 0.0015{t^2}} \right]} \right)ds} dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\left( {0.003t \cdot \exp \left[ { – 0.0015{t^2}} \right]} \right)\int\limits_0^t {\left( {0.002s \cdot \exp \left[ { – 0.001{s^2}} \right]} \right) \cdot ds} dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\left( {0.003t \cdot \exp \left[ { – 0.0015{t^2}} \right]} \right) \cdot \left( {1 – \exp \left[ { – 0.001{t^2}} \right]} \right)dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\left( {0.003t \cdot \exp \left[ { – 0.0015{t^2}} \right]} \right)dt} – \int\limits_0^{20} {\left( {0.003t \cdot \exp \left[ { – 0.0025{t^2}} \right]} \right)dt} \) \({}_{20}{q_{50:\mathop {60}\limits^2 }} = 0.451188 – 0.379272 = 0.071917\) \({}_{20}q_{50:60}^1 – {}_{20}{q_{50:\mathop {60}\limits^2 }} = 0.252848 – 0.071917 = 0.180931\) |
| Jawaban | B. 0,18 |
