Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
April 2019 |
| Nomor Soal |
: |
21 |
SOAL
Sebuah asuransi berjangka diskrit 3 tahun (3-years fully discrete term insurance) dengan manfaat sebesar 1.000 milik seorang berusia 40, dengan double decremenet model sebagai berikut:
-
| \(x\) |
\(l_x^{\left( \tau \right)}\) |
\(d_x^{\left( 1 \right)}\) |
\(d_x^{\left( 2 \right)}\) |
| 40 |
2.000 |
20 |
60 |
| 41 |
– |
30 |
50 |
| 42 |
– |
40 |
– |
- Decrement 1 adalah kematian. Decrement 2 adalah withdrawal
- Tidak ada manfaat yang dibayarkan untuk withdrawal
- \(i = 0,05\)
Tentukanlah premi netto tahunan (level annual net premium) untuk asuransi ini (gunakan pembulatan terdekat)
- 14,3
- 14,7
- 15,1
- 15,5
- 15,7
| Diketahui |
Sebuah asuransi berjangka diskrit 3 tahun (3-years fully discrete term insurance) dengan manfaat sebesar 1.000 milik seorang berusia 40, dengan double decremenet model sebagai berikut:
-
| \(x\) |
\(l_x^{\left( \tau \right)}\) |
\(d_x^{\left( 1 \right)}\) |
\(d_x^{\left( 2 \right)}\) |
| 40 |
2.000 |
20 |
60 |
| 41 |
– |
30 |
50 |
| 42 |
– |
40 |
– |
- Decrement 1 adalah kematian. Decrement 2 adalah withdrawal
- Tidak ada manfaat yang dibayarkan untuk withdrawal
- \(i = 0,05\)
|
| Rumus yang digunakan |
\({l_{x + 1}^{\left( \tau \right)} = l_x^{\left( \tau \right)} – \sum\nolimits_{j = 1}^n {d_x^{\left( j \right)}} ,}\) \({q_x^{\left( j \right)} = \frac{{d_x^{\left( j \right)}}}{{l_x^{\left( \tau \right)}}},}\) \({p_x^{\left( \tau \right)} = \frac{{l_{x + 1}^{\left( \tau \right)}}}{{l_x^{\left( \tau \right)}}},}\) \({{}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} }\)
\(P_{x:\overline {\left. n \right|} }^1 = \frac{{A_{x:\overline {\left. n \right|} }^1}}{{{{\ddot a}_{x:\overline {\left. n \right|} }}}}\)
\(A_{x:\overline {\left. n \right|} }^1 = b\sum\limits_{k = 0}^{n – 1} {{v^{k + 1}}{}_k{p_x}{q_{x + k}}} \)
\({{\ddot a}_{x:\overline {\left. n \right|} }} = \sum\limits_{k = 0}^{n – 1} {{v^k}{}_k{p_x}} \) |
| Proses pengerjaan |
\(l_{41}^{\left( \tau \right)} = l_{40}^{\left( \tau \right)} – d_{40}^{\left( 1 \right)} – d_{40}^{\left( 2 \right)} = 2000 – 20 – 60 = 1920\)
\(l_{42}^{\left( \tau \right)} = l_{41}^{\left( \tau \right)} – d_{41}^{\left( 1 \right)} – d_{41}^{\left( 2 \right)} = 1920 – 30 – 50 = 1840\) |
| \(A_{40:\overline {\left. 3 \right|} }^1 = b\sum\limits_{k = 0}^2 {{v^{k + 1}}{}_k{p_{40}}{q_{40 + k}}} = 1000\left( {vq_{40}^{\left( 1 \right)} + {v^2}p_{40}^{\left( \tau \right)}q_{41}^{\left( 1 \right)} + {v^2}{}_2p_{40}^{\left( \tau \right)}q_{42}^{\left( 1 \right)}} \right)\)
\(A_{40:\overline {\left. 3 \right|} }^1 = 1000\left[ {\frac{{d_{40}^{\left( 1 \right)}}}{{l_{40}^{\left( \tau \right)}}}v + \left( {\frac{{l_{41}^{\left( \tau \right)}}}{{l_{40}^{\left( \tau \right)}}}} \right)\left( {\frac{{d_{41}^{\left( 1 \right)}}}{{l_{41}^{\left( \tau \right)}}}} \right){v^2} + \left( {\frac{{l_{41}^{\left( \tau \right)}}}{{l_{40}^{\left( \tau \right)}}}} \right)\left( {\frac{{l_{42}^{\left( \tau \right)}}}{{l_{41}^{\left( \tau \right)}}}} \right)\left( {\frac{{d_{42}^{\left( 1 \right)}}}{{l_{42}^{\left( \tau \right)}}}} \right){v^2}} \right]\)
\(A_{40:\overline {\left. 3 \right|} }^1 = 1000\left[ {\frac{{d_{40}^{\left( 1 \right)}}}{{l_{40}^{\left( \tau \right)}}}v + \frac{{d_{41}^{\left( 1 \right)}}}{{l_{40}^{\left( \tau \right)}}}{v^2} + \frac{{d_{42}^{\left( 1 \right)}}}{{l_{40}^{\left( \tau \right)}}}{v^3}} \right]\)
\(A_{40:\overline {\left. 3 \right|} }^1 = \frac{{1000}}{{2000}}\left[ {\frac{{20}}{{1.05}} + \frac{{30}}{{{{1.05}^2}}} + \frac{{40}}{{{{1.05}^3}}}} \right] = 40.4060\) |
| \({{\ddot a}_{40:\overline {\left. 3 \right|} }} = \sum\limits_{k = 0}^2 {{v^k}{}_k{p_{40}}} = 1 + vp_{40}^{\left( \tau \right)} + {v^2}{}_2p_{40}^{\left( \tau \right)}\)
\({{\ddot a}_{40:\overline {\left. 3 \right|} }} = 1 + \frac{{l_{41}^{\left( \tau \right)}}}{{l_{40}^{\left( \tau \right)}}}v + \left( {\frac{{l_{41}^{\left( \tau \right)}}}{{l_{40}^{\left( \tau \right)}}}} \right)\left( {\frac{{l_{42}^{\left( \tau \right)}}}{{l_{41}^{\left( \tau \right)}}}} \right){v^2}\)
\({{\ddot a}_{40:\overline {\left. 3 \right|} }} = 1 + \frac{{1920}}{{2000\left( {1.05} \right)}} + \frac{{1840}}{{2000\left( {{{1.05}^2}} \right)}} = 2.74875\) |
| \(P_{40:\overline {\left. 3 \right|} }^1 = \frac{{A_{40:\overline {\left. 3 \right|} }^1}}{{{{\ddot a}_{40:\overline {\left. 3 \right|} }}}} = \frac{{40.4060}}{{2.74875}} = 14.700\) |
| Jawaban |
b. 14,7 |
