Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | A60 – Matematika Aktuaria |
| Periode Ujian | : | November 2017 |
| Nomor Soal | : | 28 |
SOAL
Untuk suatu “special 3-year temporary life annuity-due” pada (x) , diberikan sebagai berikut:
(i)
| t | Annuity Payment | \({P_{x + t}}\) |
| 0 | 15 | 0,95 |
| 1 | 20 | 0,90 |
| 2 | 25 | 0,85 |
(ii)
Hitunglah variansi dari “present value random variable” untuk anuitas ini
- 91
- 102
- 114
- 127
- 139
PEMBAHASAN
| Diketahui | \({(I\ddot Y)_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \left\{ \begin{array}{l}15,K(x) = 0\\ 15 + 20v,K(x) = 1\\15 + 20v + 25{v^2},K(x) = 2,3,4,…\end{array} \right.\) \({(I\ddot Y)_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \left\{ \begin{array}{l}15,K(x) = 0\\{\rm{33,86792}},K(x) = 1\\{\rm{56,11784}}\,,K(x) = 2,3,4,…\end{array} \right.\) |
| Rumus | \(Var\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] -E{\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right]^2}\) |
| Step 1 | \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 15P(K(x) = 0) + \left( {33,86792} \right)P(K(x) = 1) + \left( {56,11784} \right)P(K(x) = 2)\) \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 15\,{q_x} + \left( {33,86792} \right){p_x}\,{q_{x + 1}}\, + \left( {56,11784} \right){}_2{p_x}\) \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 15\,(1 – 0,95) + \left( {33,86792} \right)(0,95)(1 – 0,9) + \left( {56,11784} \right){p_{x\,}}\,{p_{x + 1}}\) \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 0,75 + 3,21745 + \left( {56,11784} \right)(0,95)(0,90)\,\) \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 3,96745 + 47,98075\,\) \(E\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 51,9482\) |
| Step 2 | \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = {15^2}P(K(x) = 0) + {\left( {33,86792} \right)^2}P(K(x) = 1) + {\left( {56,11784} \right)^2}P(K(x) = 2)\) \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = {15^2}\,{q_x} + {\left( {33,86792} \right)^2}{p_x}\,{q_{x + 1}}\, + {\left( {56,11784} \right)^2}{}_2{p_x}\) \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = {15^2}\,(1 – 0,95) + {\left( {33,86792} \right)^2}(0,95)(1 – 0,9) + {\left( {56,11784} \right)^2}{p_{x\,}}\,{p_{x + 1}}\) \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = 11,25 + 108,96842 + {\left( {56,11784} \right)^2}(0,95)(0,90)\) \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = 120,21842 + 2.692,57623\) \(E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] = 2.812,79465\) |
| Step 3 | \(Var\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = E\left[ {{{\left( {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)}^2}} \right] – E{\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right]^2}\) \(Var\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 2.812,79465 – {\left( {51,9482} \right)^2}\) \(Var\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] = 114,17916676\) \(Var\left[ {{{(I\ddot Y)}_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right] \cong 114\) |
| Jawaban | c. 114 |