Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | A60 – Matematika Aktuaria |
| Periode Ujian | : | November 2017 |
| Nomor Soal | : | 2 |
SOAL
Hitunglah nilai dari \({}_{n – 1}{V_{x:\left. {\overline {\,n \,}}\! \right| }}\) , jika diberikan \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = 0,50\) dan d=0,08
- 0,80
- 0,82
- 0,84
- 0,86
- 0,90
PEMBAHASAN
| Rumus | \({}_t{V_{x:\left. {\overline {\, n \,}}\! \right| }} = {A_{x + t:\left. {\overline {\, {n – t} \,}}\! \right| }} – {P_{x:\left. {\overline {\, n \,}}\! \right| }}{\ddot a_{x + t:\left. {\overline {\, {n – t} \,}}\! \right| }}\) |
| Step 1 | \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = {A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} – {P_{x:\left. {\overline {\, n \,}}\! \right| }}{\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\) |
| Step 2 | \({A_{x:\left. {\overline {\, 1 \,}}\! \right| }} = {A_{\mathop x\limits^| :\left. {\overline {\, 1 \,}}\! \right| }} + {A_{x:\mathop {\left. {\overline {\, 1 \,}}\! \right| }\limits^| }}\) |
| \({latex A_{\mathop x\limits^| :\left. {\overline {\, 1 \,}}\! \right| }} = \sum\limits_{k = 0}^0 {{v^{k + 1}}{}_k{p_x}\,{q_{x + k}}\, = \,v\,{q_x}} \) | |
| \({A_{x:\mathop {\left. {\overline {\, 1 \,}}\! \right| }\limits^| }} = {v^{1\,}}_1{p_x}\,\, = \,\,v\,{p_x}\) | |
| \({A_{x:\left. {\overline {\, 1 \,}}\! \right| }} = v{q_x} + v{p_x}\) \({A_{x:\left. {\overline {\, 1 \,}}\! \right| }} = v(1 – {p_x}) + v{p_x}\) \({A_{x:\left. {\overline {\, 1 \,}}\! \right| }} = v – v{p_x} + v{p_x}\) \({A_{x:\left. {\overline {\, 1 \,}}\! \right| }} = v\) –> Maka \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\, = \,v\) | |
| Step 3 | \({\ddot a_{x\left. {\overline {\, {:1} \,}}\! \right| }} = \sum\limits_{t = 0}^0 {{v^t}\,{}_t{p_x}} \) \({\ddot a_{x\left. {\overline {\, {:1} \,}}\! \right| }} = 1\) –> Maka \({\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\, = 1\) |
| Step 4 | \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{\left( {\frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}} \right)}}\) \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{0,50}}{{\left( {\frac{{1 – 0,50}}{{0,08}}} \right)}}\) \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = 0,08\) |
| \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = v – (0,08)(1)\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – d) – 0,08\) \({}_{n – 1}{V_{x:\left. {\overline {\,n \,}}\! \right| }} = (1 – 0,08) – 0,08\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = 0,84\) | |
| Jawaban | c. 0,84 |
