Pembahasan-Soal-Ujian-Profesi-Aktuaris-1024x481 (3) pai

Pembahasan Ujian PAI: A60 – No. 28 – Mei 2018

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 $E\left[ Z \right]$

  1. \({}_{\left. {10} \right|}{\bar A_x} + {}_{\left. {20} \right|}{\bar A_x} – {}_{\left. {30} \right|}{\bar A_x}\)
  2. \({\bar A_x} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
  3. \({}_{10}{E_x} \cdot {\bar A_x} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
  4. \({}_{10}{E_x} \cdot {\bar A_{x + 10}} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
  5. \({}_{10}{E_x}\left[ {{{\bar A}_{x + 10}} + {}_{10}{E_{x + 10}} \cdot {{\bar A}_{x + 20}} – {}_{10}{E_{x + 20}} \cdot {{\bar A}_{x + 30}}} \right]\)
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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. \({}_{10}{E_x} \cdot {\bar A_{x + 10}} + {}_{20}{E_x} \cdot {\bar A_{x + 20}} – 2{}_{30}{E_x} \cdot {\bar A_{x + 30}}\)
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