Pembahasan Ujian PAI: A60 - No. 23 - November 2016

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 \(y = 60\)

\(\frac{{47}}{{160}}\)
\(\frac{3}{8}\)
\(\frac{{13}}{{40}}\)
\(\frac{{31}}{{80}}\)
\(\frac{2}{5}\)

[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]

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. \(\frac{{47}}{{160}}\)