Pembahasan Ujian PAI: A60 – No. 9 – November 2016

1. \({T_x}\) dan \({T_y}\) adalah “independent”
2. Fungsi “survival” untuk \(\left( x \right)\) mengikuti \({l_x} = 100\left( {95 – x} \right)\), \(0 \le x \le 95\)
3. Fungsi “survival” untuk \(\left( y \right)\) berdasarkan konstan “force of mortality”, \({\mu _{y + t}} = \mu \), \(t \ge 0\)
4. \(n < 95 – x\)

• \({}_n{q_{xy}} = \int\limits_0^n {{}_t{p_{xy}} \cdot {\mu _{x + t}}dt} \)
• \({}_t{p_{xy}} = {}_t{p_x} \cdot {}_t{p_y}\)
• \({}_t{p_x} = \frac{{{l_{x + t}}}}{{{l_x}}}\)
• \({}_t{p_x} = \exp \left( { – \int_0^t {\mu \left( {x + s} \right)ds} } \right)\)
• \({\mu _x} = – \frac{{d\left( {{l_x}} \right)}}{{dt}} \cdot \frac{1}{{{l_x}}}\)

Peluang \(\left( x \right)\) meninggal dalam waktu n tahun dan meninggal sebelum \(\left( y \right)\) \({}_n{q_{xy}} = \int\limits_0^n {\left( {\frac{{95 – x – t}}{{95 – x}}} \right) \cdot \exp \left( { – \mu t} \right) \cdot \left( {\frac{1}{{95 – x – t}}} \right)dt} \) \({}_n{q_{xy}} = \int\limits_0^n {\left( {\frac{{\exp \left( { – \mu t} \right)}}{{95 – x}}} \right)dt} \) \({}_n{q_{xy}} = \frac{1}{{95 – x}} \cdot \frac{1}{\mu }\left[ {1 – \exp \left( { – \mu n} \right)} \right] = \frac{{1 – \exp \left( { – \mu n} \right)}}{{\mu \left( {95 – x} \right)}}\)