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:
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| 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 |