Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2016 |
| Nomor Soal |
: |
23 |
SOAL
Untuk “two lives” \(\left( x \right)\) dan \(\left( y \right)\) dengan “independent future lifetimes”
- \({\mu _x} = \frac{2}{{100 – x}}\), \(x < 100\)
- \({\mu _y} = \frac{3}{{100 – y}}\), \(y < 100\)
Hitunglah \({}_{20}{q_{x\mathop y\limits^2 }}\) untuk \(x = 60\) dan \(y = 60\)
- \(\frac{{47}}{{160}}\)
- \(\frac{3}{8}\)
- \(\frac{{13}}{{40}}\)
- \(\frac{{31}}{{80}}\)
- \(\frac{2}{5}\)
| Diketahui |
Untuk “two lives” \(\left( x \right)\) dan \(\left( y \right)\) dengan “independent future lifetimes”
- \({\mu _x} = \frac{2}{{100 – x}}\), \(x < 100\)
- \({\mu _y} = \frac{3}{{100 – y}}\), \(y < 100\)
|
| Rumus yang digunakan |
- \({}_t{q_{x\mathop y\limits^2 }} = {F_T}\left( t \right) = \int\limits_0^t {{}_s{p_y} \cdot {}_s{q_x} \cdot {\mu _{y + s}}ds} \)
- \({}_t{p_{xy}} = {}_t{p_x} \cdot {}_t{p_y}\)
- Hukum de Moivre
\({\mu _x} = \frac{\alpha }{{\omega – x}}\)
\({}_t{p_x} = {\left( {\frac{{\omega – x – t}}{{\omega – x}}} \right)^\alpha }\)
|
| Proses pengerjaan |
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \int\limits_0^{20} {{{\left( {\frac{{40 – t}}{{40}}} \right)}^3} \cdot \left( {1 – {{\left( {\frac{{40 – t}}{{40}}} \right)}^2}} \right) \cdot \frac{3}{{40 – t}}dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \int\limits_0^{20} {\frac{{3{{\left( {40 – t} \right)}^2}}}{{{{40}^3}}} \cdot \left( {1 – {{\left( {\frac{{40 – t}}{{40}}} \right)}^2}} \right)dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{3}{{{{40}^5}}}\int\limits_0^{20} {{{\left( {40 – t} \right)}^2} \cdot \left( {{{40}^2} – {{\left( {40 – t} \right)}^2}} \right)dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{3}{{{{40}^5}}}\int\limits_0^{20} {{{\left( {40 – t} \right)}^2} \cdot \left( {80t – {t^2}} \right)dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{3}{{{{40}^5}}}\int\limits_0^{20} {\left( {1600 – 80t + {t^2}} \right) \cdot \left( {80t – {t^2}} \right)dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{3}{{{{40}^5}}}\int\limits_0^{20} {\left( {128,000t – 8,000{t^2} + 160{t^3} – {t^4}} \right)dt} \)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{3}{{{{40}^5}}}\left[ {64,000{{\left( {20} \right)}^2} – \frac{{8,000}}{3}{{\left( {20} \right)}^3} + 40{{\left( {20} \right)}^4} – \frac{1}{5}{{\left( {20} \right)}^5}} \right]\)
\({}_{20}{q_{60:\mathop {60}\limits^2 }} = \frac{{47}}{{160}}\) |
| Jawaban |
A. \(\frac{{47}}{{160}}\) |
