Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | A60 – Matematika Aktuaria |
| Periode Ujian | : | Mei 2017 |
| Nomor Soal | : | 2 |
SOAL
Hitunglah nilai dari \({}_{n – 1}{V_{x:\left. {\overline {\,n \,}}\! \right| }}\), jika diberikan \({A_{x:\left. {\overline {\,n \,}}\! \right| }} = 0,20\,\,dan\,\,d = 0,08\)
- 0,85
- 0,90
- 0,95
- 1,00
- 1,05
PEMBAHASAN
| Rumus | \({}_t{V_{x:\left. {\overline {\, n \,}}\! \right| }} = {A_{x + t:\left. {\overline {\, {n – t} \,}}\! \right| }} – {P_{x:\left. {\overline {\, n \,}}\! \right| }}{\ddot a_{x + t:\left. {\overline {\, {n – t} \,}}\! \right| }}\) \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = {A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}\) \({A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} = \sum\limits_{k = 0}^{n – 1} {{v^{k + 1}}\,{}_k{p_x}\,{q_{x + k}}} \) \({A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }} = \sum\limits_{t = 1}^n {{v^t}{}_tp{}_x} \) \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = 1 – d{\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }}\) \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }} – d}}{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) \(d = 1 – v\) |
| Step 1 | \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = {A_{\mathop x\limits^| + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} + {A_{x + n – 1:\mathop {\left. {\overline {\, 1 \,}}\! \right| }\limits^| }}\) |
| \({A_{\mathop x\limits^| + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = \sum\limits_{k = 0}^0 {{v^{k + 1}}\,{}_k{p_{x + n – 1}}\,{q_{x + n – 1 + k}}} \) \({A_{\mathop x\limits^| + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = v\,{q_{x + n – 1}}\) | |
| \({A_{x + n – 1:\mathop {\left. {\overline {\, 1 \,}}\! \right| }\limits^| }} = \sum\limits_{t = 1}^1 {{v^t}{}_tp{}_{x + n – 1}} \) \({A_{x + n – 1:\mathop {\left. {\overline {\, 1 \,}}\! \right| }\limits^| }} = v\,{p_{x + n – 1}}\) | |
| \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = v\,{q_{x + n – 1}} + v\,{p_{x + n – 1}}\) \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = v\,\left( {1 – {p_{x + n – 1}} + {p_{x + n – 1}}} \right)\) \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = v\,\) | |
| Step 2 | \({A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = 1 – (0,08){\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\) \(v = 1 – (0,08){\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\) \({\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = \frac{{1 – v}}{d}\) \({\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = \frac{d}{d}\) \({\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} = 1\) |
| Step 3 | \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = {A_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }} – {P_{x:\left. {\overline {\, n \,}}\! \right| }}{\ddot a_{x + n – 1:\left. {\overline {\, 1 \,}}\! \right| }}\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = v – {P_{x:\left. {\overline {\, n \,}}\! \right| }}\,\,(1)\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – d) – – \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – d) – – \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{\left( {\frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}} \right)}}\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – d) – – \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}\,d}}{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – 0,08) – – \frac{{0,2(0,08)}}{{1 – 0,2}}\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = (1 – 0,08) – 0,02\) \({}_{n – 1}{V_{x:\left. {\overline {\, n \,}}\! \right| }} = 0,90\) |
| Jawaban | b. 0,90 |
