Pembahasan-Soal-Ujian-Profesi-Aktuaris

Pembahasan Ujian PAI: A60 – No. 10 – Mei 2018

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A60 - Matematika AktuariaAktuariaEdukasiPembahasan Ujian Profesi Aktuaris (Persatuan Aktuaris Indonesia / PAI)

Pembahasan Soal Ujian Profesi Aktuaris

Institusi : Persatuan Aktuaris Indonesia (PAI)
Mata Ujian : Matematika Aktuaria
Periode Ujian : Mei 2018
Nomor Soal : 10

SOAL

Untuk sebuah asuransi dwiguna (endowment insurance) dengan 15 kali pembayaran yang berkelanjutan secara penuh (fully continuous) selama 25 tahun senilai 1000 untuk (35), diketahui:

  1. \({\mu _{35 + t}} = 0,03\) untuk \(t \ge 0\)
  2. \(\delta = 0,05\)
  3. \(1000{\bar A_{\mathop {35}\limits^1 :\left. {\overline {\,  {25} \,}}\! \right| }} = 324,25\)
  4. \({\bar a_{35:\left. {\overline {\,  {25} \,}}\! \right| }} = 8,7351\)

Hitunglah \(_5V\) net premium reserve pada tahun ke-5!

  1. 139,03
  2. 149,65
  3. 152,17
  4. 154,23
  5. 163,31
[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]
Diketahui
  • \({\mu _{35 + t}} = 0,03\)
  • \(\delta = 0,05\)
  • \(1000{\bar A_{\mathop {35}\limits^1 :\left. {\overline {\,  {25} \,}}\! \right| }} = 324,25\)
  • \({\bar a_{35:\left. {\overline {\,  {25} \,}}\! \right| }} = 8,7351\)
Step 1 \(\,_5^{15}{V_{35:\left. {\overline {\,  {25} \,}}\! \right| }} = P \cdot \frac{{{{\bar a}_{35:\left. {\overline {\,  5 \,}}\! \right| }}}}{{_5{E_x}}} – \frac{{{{\bar A}_{\mathop {35}\limits^1 :\left. {\overline {\,  5 \,}}\! \right| }}}}{{_5{E_x}}}\) ………(*)
Step 2
  • \({\bar a_{35:\left. {\overline {\, 5 \,}}\! \right| }} = \int_0^5 {{e^{ – \delta t}} \cdot {e^{ – \mu t}}dt} \) \(= \int_0^5 {{e^{ – \left( {0,05t} \right)}} \cdot {e^{ – \left( {0,03t} \right)}}dt} = \int_0^5 {{e^{ – 0,08t}}dt} \) \(= \frac{1}{{0,08}}\left[ {1 – {e^{ – 0,08t}}\,\,\left| {_{_0}^{^5}} \right.} \right]\) \(= \frac{1}{{0,08}}\left[ {\left( {1 – {e^{ – 0,08\left( 5 \right)}}} \right) – \left( {1 – {e^0}} \right)} \right]\) \(= 4,120999425\)
  • \({\bar A_{35:\left. {\overline {\, 5 \,}}\! \right| }} = 1 – \delta \cdot {\bar a_{35:\left. {\overline {\, 5 \,}}\! \right| }}\) \(= 1 – 0,05\left( {4,120999425} \right)\) \(= 0,793950028\)
  • \(_5{E_x} = {e^{ – \delta t}} \cdot {e^{ – \mu t}}\) \(= {e^{ – 0,05\left( 5 \right)}} \cdot {e^{ – 0,03\left( 5 \right)}} = 0,6703\)
  • \({\bar A_{\mathop {35}\limits^1 :\left. {\overline {\,  5 \,}}\! \right| }} = {\bar A_{35:\left. {\overline {\,  5 \,}}\! \right| }}{ – _5}{E_x}\) \(= 0,793950028 – 0,6703 = 0,12365\)
  • \({\bar A_{35:\left. {\overline {\,  {25} \,}}\! \right| }} = {e^{ – \delta t}} \cdot {e^{ – \mu t}} + {\bar A_{\mathop {35}\limits^1 :\left. {\overline {\,  {25} \,}}\! \right| }}\) \(= {e^{ – 0,08\left( {25} \right)}} + 0,32425\) \(= 0,459585283\)

Maka \(P = \frac{{{{\bar A}_{35:\left. {\overline {\,  {25} \,}}\! \right| }}}}{{{{\bar a}_{35:\left. {\overline {\, {25} \,}}\! \right| }}}} = \frac{{0,459585283}}{{8,7351}} = 0,0526136\)

Step 3 (*)

\(1000\,_5^{15}{V_{35:\left. {\overline {\, {25} \,}}\! \right| }} = 1000\left[ {P \cdot \frac{{{{\bar a}_{35:\left. {\overline {\, 5 \,}}\! \right| }}}}{{_5{E_x}}} – \frac{{{{\bar A}_{\mathop {35}\limits^1 :\left. {\overline {\, 5 \,}}\! \right| }}}}{{_5{E_x}}}} \right]\)

\(= 1000\left[ {0,0526136 \cdot \frac{{4,120999425}}{{0,6703}} – \frac{{0,12365}}{{0,6703}}} \right]\) \(= 138,998\)

Jawaban yang paling mendekati adalah 139,03
Jawaban  a. 139,03
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