Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
Mei 2018 |
| Nomor Soal |
: |
28 |
SOAL
Peubah acak nilai tunai untuk (x) dapat dinyatakan sebagai:
\(Z = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 0\\ {{v^{{T_x}}}}\\ {2{v^{{T_x}}}}\\ 0 \end{array}}&{\begin{array}{*{20}{c}} {{T_x} \le 10}\\ {10 < {T_x} \le 20}\\ {20 < {T_x} \le 30}\\ {lainnya} \end{array}} \end{array}} \right.\)
Dari pilihan-pilihan berikut, manakah ekspresi yang tepat untuk menggambarkan $E\left[ Z \right]$
- \({}_{\left. {10} \right|}{\bar A_x} + {}_{\left. {20} \right|}{\bar A_x} – {}_{\left. {30} \right|}{\bar A_x}\)
- \({\bar A_x} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
- \({}_{10}{E_x} \cdot {\bar A_x} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
- \({}_{10}{E_x} \cdot {\bar A_{x + 10}} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
- \({}_{10}{E_x}\left[ {{{\bar A}_{x + 10}} + {}_{10}{E_{x + 10}} \cdot {{\bar A}_{x + 20}} – {}_{10}{E_{x + 20}} \cdot {{\bar A}_{x + 30}}} \right]\)
| Diketahiui |
Peubah acak nilai tunai untuk (x) dapat dinyatakan sebagai:
\(Z = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 0\\ {{v^{{T_x}}}}\\ {2{v^{{T_x}}}}\\ 0 \end{array}}&{\begin{array}{*{20}{c}} {{T_x} \le 10}\\ {10 < {T_x} \le 20}\\ {20 < {T_x} \le 30}\\ {lainnya} \end{array}} \end{array}} \right.\) |
| Rumus yang digunakan |
\({}_{\left. n \right|}{\bar A_x} = {}_n{E_x} \cdot {\bar A_{x + n}}\) |
| Proses pengerjaan |
\(E\left[ Z \right] = {}_{\left. {10} \right|}{\bar A_x} + {}_{\left. {20} \right|}{\bar A_x} – 2{}_{\left. {30} \right|}{\bar A_x}\)
Dengan menggunakan \({}_{\left. n \right|}{\bar A_x} = {}_n{E_x} \cdot {\bar A_{x + n}}\) maka diperoleh
\(E\left[ Z \right] = {}_{\left. {10} \right|}{{\bar A}_x} + {}_{\left. {20} \right|}{{\bar A}_x} – 2{}_{\left. {30} \right|}{{\bar A}_x}\)
\(E\left[ Z \right] = {}_{10}{E_x} \cdot {{\bar A}_{x + 10}} + {}_{20}{E_x} \cdot {{\bar A}_{x + 20}} – 2\left( {{}_{30}{E_x} \cdot {{\bar A}_{x + 30}}} \right)\) |
| Jawaban |
d. \({}_{10}{E_x} \cdot {\bar A_{x + 10}} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\) |
