Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
April 2019 |
| Nomor Soal |
: |
9 |
SOAL
Manakah persamaan berikut ini yang benar?
- \(1 + {a_{x:\overline {\left. {n – 1} \right|} }} = \left( {1 + i} \right){a_{x:\overline {\left. n \right|} }}\)
- \({A_{x:\overline {\left. n \right|} }} = v{\ddot a_{x:\overline {\left. n \right|} }} – {a_{x:\overline {\left. {n – 1} \right|} }}\)
- \({}_{\left. n \right|}{\ddot a_x} = {a_x} – {a_{x:\overline {\left. {n – 1} \right|} }}\)
- i saja
- i dan ii
- i dan iii
- ii dan iii
- i, ii dan iii
| Diketahui |
- \(1 + {a_{x:\overline {\left. {n – 1} \right|} }} = \left( {1 + i} \right){a_{x:\overline {\left. n \right|} }}\)
- \({A_{x:\overline {\left. n \right|} }} = v{\ddot a_{x:\overline {\left. n \right|} }} – {a_{x:\overline {\left. {n – 1} \right|} }}\)
- \({}_{\left. n \right|}{\ddot a_x} = {a_x} – {a_{x:\overline {\left. {n – 1} \right|} }}\)
|
| Rumus yang digunakan |
\({{a_{x:\overline {\left. n \right|} }} = \frac{{1 – {v^n}}}{i},}\) \({{{\ddot a}_{x:\overline {\left. n \right|} }} = \frac{{1 – {v^n}}}{d},}\) \({d = \frac{i}{{1 + i}},}\)
\({d = 1 – v,}\) \({{{\ddot a}_x} = {a_x} + 1,}\) \({{{\ddot a}_{x:\overline {\left. n \right|} }} = {a_{x:\overline {\left. {n – 1} \right|} }} + 1}\) |
| Proses pengerjaan |
\(\left( {1 + i} \right){a_{x:\overline {\left. n \right|} }} = \left( {1 + i} \right) \cdot \frac{{1 – {v^n}}}{i}\)
\(\left( {1 + i} \right){a_{x:\overline {\left. n \right|} }} = \frac{{1 – {v^n}}}{d}\)
\(\left( {1 + i} \right){a_{x:\overline {\left. n \right|} }} = {\ddot a_{x:\overline {\left. n \right|} }}\) (salah) |
|
\({A_{x:\overline {\left. n \right|} }} = 1 – d \cdot {{\ddot a}_{x:\overline {\left. n \right|} }}\)
\({A_{x:\overline {\left. n \right|} }} = 1 – \left( {1 – v} \right) \cdot {{\ddot a}_{x:\overline {\left. n \right|} }}\)
\({A_{x:\overline {\left. n \right|} }} = 1 – {{\ddot a}_{x:\overline {\left. n \right|} }} + v \cdot {{\ddot a}_{x:\overline {\left. n \right|} }}\)
\({A_{x:\overline {\left. n \right|} }} = v \cdot {{\ddot a}_{x:\overline {\left. n \right|} }} – \left( {{{\ddot a}_{x:\overline {\left. n \right|} }} – 1} \right)\)
\({A_{x:\overline {\left. n \right|} }} = v \cdot {\ddot a_{x:\overline {\left. n \right|} }} – {a_{x:\overline {\left. {n – 1} \right|} }}\) (benar) |
|
\({}_{\left. n \right|}{{\ddot a}_x} = {{\ddot a}_x} – {{\ddot a}_{x:\overline {\left. n \right|} }}\)
\({}_{\left. n \right|}{{\ddot a}_x} = \left( {{a_x} + 1} \right) – \left( {{a_{x:\overline {\left. {n – 1} \right|} }} + 1} \right)\)
\({}_{\left. n \right|}{\ddot a_x} = {a_x} – {a_{x:\overline {\left. {n – 1} \right|} }}\) (benar) |
| Jawaban |
.d ii dan iii |
