Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
2 |
SOAL
Hitunglah nilai dari \({\ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }}\), diketahui sebagai berikut:
\({\ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }} = E\left[ {{{\ddot Y}_{x:\left. {\overline {\, 4 \,}}\! \right| }}} \right]\)
| \(k\) |
\({\ddot a_{\left. {\overline {\, k \,}}\! \right| }}\) |
\({}_{\left. {k – 1} \right|}{q_x}\) |
| 1 |
1,00 |
0,33 |
| 2 |
1,93 |
0,24 |
| 3 |
2,80 |
0,16 |
| 4 |
3,62 |
0,11 |
- 2,2186
- 2,2862
- 2,1862
- 2,1268
- 2,2681
| Diketahui |
\({\ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }} = E\left[ {{{\ddot Y}_{x:\left. {\overline {\, 4 \,}}\! \right| }}} \right]\)
| \(k\) |
\({\ddot a_{\left. {\overline {\, k \,}}\! \right| }}\) |
\({}_{\left. {k – 1} \right|}{q_x}\) |
| 1 |
1,00 |
0,33 |
| 2 |
1,93 |
0,24 |
| 3 |
2,80 |
0,16 |
| 4 |
3,62 |
0,11 |
|
| Rumus yang digunakan |
n-year endowment: \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \ddot a_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {\ddot a_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }}\)
\(\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }^1 = \sum\limits_{k = 0}^{n – 1} {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_k{p_x} \cdot {q_{x + k}}} = \sum\limits_{k = 0}^{n – 1} {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} \)
\({\ddot a_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }} = {\ddot a_{\left. {\overline {\, n \,}}\! \right| }} \cdot {}_n{p_x}\)
\({}_n{p_x} = 1 – {}_n{q_x} = 1 – \left( {\prod\limits_{k = 0}^{n – 1} {{q_{x + k}}} } \right) = 1 – \sum\limits_{k = 0}^{n – 1} {{}_{\left. k \right|}{q_x}} \) |
| Proses pengerjaan |
\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }^1 + {{\ddot a}_{x:\mathop {\left. {\overline {\, 4 \,}}\! \right| }\limits^1 }}\)
\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_4{p_x} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \sum\limits_{k = 0}^{n – 1} {{}_{\left. k \right|}{q_x}} } \right)\)
\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {{{\ddot a}_{\left. {\overline {\, 1 \,}}\! \right| }} \cdot {}_{\left. 0 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 2 \,}}\! \right| }} \cdot {}_{\left. 1 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 3 \,}}\! \right| }} \cdot {}_{\left. 2 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_{\left. 3 \right|}{q_x}} \right] + \)
\({{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \left[ {{}_{\left. 0 \right|}{q_x} + {}_{\left. 1 \right|}{q_x} + {}_{\left. 2 \right|}{q_x} + {}_{\left. 3 \right|}{q_x}} \right]} \right)\)
\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {1\left( {0.33} \right) + 1.93\left( {0.24} \right) + 2.8\left( {0.16} \right) + 3.62\left( {0.11} \right)} \right] + \)
\(3.62\left( {1 – \left[ {0.33 + 0.24 + 0.16 + 0.11} \right]} \right)\)
\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = 2.2186\) |
| Jawaban |
a. 2,2186 |
