Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2017 |
| Nomor Soal | : | 25 |
SOAL
Diberikan sebagai berikut:
- \({P_x} = 0,090\)
- “Net Premium Reserve” pada akhir tahun ke n untuk suatu asuransi “fully discrete whole life” dengan benefit 1 pada (x) adalah 0,563
- \({P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }} = 0,00864\)
Hitunglah \({P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}\)
- 0,008
- 0,024
- 0,040
- 0,065
- 0,085
| Diketahui | \({}_n{V_x} = 0,563\) \({P_x} = 0,090\) \({P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }} = 0,00864\) |
| Rumus | Retrospektif, \({}_n{V_x} = {P_x}\,{\ddot S_{x:\left. {\overline {\, n \,}}\! \right| }} – {}_n{K_x}\) \({\ddot S_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{}_n{E_x}}}\,\) \({}_n{K_x} = \frac{{{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{}_n{E_x}}}\) \({P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} = \frac{{{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) |
| Step 1 | \({}_n{V_x} = {P_x}\,{\ddot S_{x:\left. {\overline {\, n \,}}\! \right| }} – {}_n{K_x}\) \({}_n{V_x} = {P_x}\,\left( {\frac{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{}_n{E_x}}}} \right) – \left( {\frac{{{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{}_n{E_x}}}} \right)\) \({}_n{V_x} = {P_x}\,\left( {\frac{1}{{\frac{{{}_n{E_x}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}}}} \right) – \left( {\frac{{{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{}_n{E_x}}}\left( {\frac{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}} \right)} \right)\) \({}_n{V_x} = {P_x}\,\left( {\frac{1}{{{P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}}}} \right) – \left( {\frac{{{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\left( {\frac{1}{{\frac{{{}_n{E_x}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}}}} \right)} \right)\) \({}_n{V_x} = {P_x}\,\left( {\frac{1}{{{P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}}}} \right) – \left( {{P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}\left( {\frac{1}{{{P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}}}} \right)} \right)\) \({}_n{V_x} = \frac{{{P_x} – {P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{{P_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}}}\,\) \(0,563 = \frac{{0,090 – {P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}}}{{0,00864}}\,\) \({P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} = 0,090 – 0,563\left( {0,00864} \right)\,\) \({P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} = {\rm{0}}{\rm{,08513568}}\) \({P_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} \cong {\rm{0}}{\rm{,085}}\) |
| Jawaban | e. 0,085 |