Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
Mei 2017 |
| Nomor Soal |
: |
21 |
SOAL
Untuk “two lives” (50) dan (60) dengan “independent future lifetime”:
- \(\begin{array}{*{20}{c}}{{\mu _{50 + t}} = 0,002t}&{t > 0}\end{array}\)
- \(\begin{array}{*{20}{c}}{{\mu _{60 + t}} = 0,00046t}&{t > 0}\end{array}\)
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”:
- \(\begin{array}{*{20}{c}}{{\mu _{50 + t}} = 0,002t}&{t > 0}\end{array}\)
- \(\begin{array}{*{20}{c}}{{\mu _{60 + t}} = 0,00046t}&{t > 0}\end{array}\)
|
| Rumus yang digunakan |
\({f_{{T_x}}}\left( t \right) = {}_t{p_x} \cdot {\mu _{x + t}} = \exp \left[ { – \int\limits_0^t {{\mu _{x + s}}ds} } \right]{\mu _{x + t}}\) |
| Proses pengerjaan |
- Pdf dari \({T_{50}}\)
\({f_{{T_{50}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {{\mu _{50 + s}}ds} } \right]{\mu _{50 + t}} = \exp \left[ { – \int\limits_0^t {0.002sds} } \right]0.002t\)
\({f_{{T_{50}}}}\left( t \right) = 0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]\)
- Pdf dari \({T_{60}}\)
\({f_{{T_{60}}}}\left( t \right) = \exp \left[ { – \int\limits_0^t {{\mu _{60 + s}}ds} } \right]{\mu _{60 + t}} = \exp \left[ { – \int\limits_0^t {0.00046sds} } \right]0.00046t\)
\({f_{{T_{60}}}}\left( t \right) = 0.00046t \cdot \exp \left[ { – 0.00023{t^2}} \right]\)
\({}_{20}q_{50:60}^1 = \Pr \left( {{T_{50}} \le 20{\rm{ dan }}{T_{50}} < {T_{60}}} \right) = \int\limits_0^{20} {\int\limits_t^\infty {{f_{{T_{60}}}}\left( s \right){f_{{T_{50}}}}\left( t \right)ds} dt} \)
\({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\int\limits_t^\infty {\left( {0.00046s \cdot \exp \left[ { – 0.00023{s^2}} \right]} \right)\left( {0.002t \cdot \exp \left[ { – 0.001{t^2}} \right]} \right)ds} dt} \)
\({}_{20}q_{50:60}^1 = \int\limits_0^{20} {\left( {0.002t \cdot \exp \left[ { – 0.00123{t^2}} \right]} \right)dt} \)
\({}_{20}q_{50:60}^1 = \frac{{200}}{{246}}\left( {1 – \exp \left[ { – 0.00123\left( {{{20}^2}} \right)} \right]} \right) = 0.315933\)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = \Pr \left( {{T_{60}} \le 20{\rm{ dan }}{T_{50}} < {T_{60}}} \right) = \int\limits_0^{20} {\int\limits_0^t {{f_{{T_{50}}}}\left( s \right){f_{{T_{60}}}}\left( t \right)ds} dt} \)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\int\limits_0^t {\left( {0.002s \cdot \exp \left[ { – 0.001{s^2}} \right]} \right)\left( {0.00046t \cdot \exp \left[ { – 0.00023{t^2}} \right]} \right)ds} dt} \)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\left( {0.00046t \cdot \exp \left[ { – 0.00023{t^2}} \right]} \right)\left( {1 – \exp \left[ { – 0.001{t^2}} \right]} \right)dt} \)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\left( {0.00046t \cdot \exp \left[ { – 0.00023{t^2}} \right]} \right)dt} – \int\limits_0^{20} {\left( {0.00046t \cdot \exp \left[ { – 0.00123{t^2}} \right]} \right)dt} \)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = \left( {1 – \exp \left[ { – 0.00023{t^2}} \right]} \right) + \frac{{46}}{{246}}\left( {\exp \left[ { – 0.00123\left( {{{20}^2}} \right)} \right] – 1} \right)\)
\({}_{20}{q_{50:\mathop {60}\limits^2 }} = 0.015230\)
\({}_{20}q_{50:60}^1 – {}_{20}{q_{50:\mathop {60}\limits^2 }} = 0.315933 – 0.015230 = 0.300703\) |
| Jawaban |
C. 0,30 |
