Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2017 |
| Nomor Soal | : | 26 |
SOAL
Untuk suatu asuransi “fully continuous whole life” dengan benefit 1:
(i) \({\mu _x} = 0,04\,,\,x > 0\) (ii) \(\delta = 0,08\) (iii) L adalah variabel acak “loss-at-issue” pada “net premium”
Hitunglah Var(L)
- 1/10
- 1/5
- 1/4
- 1/3
- 1/2
| Rumus | \(Var[L] = \frac{{Var({{\bar Z}_x})}}{{{{(\delta {{\bar a}_x})}^2}}}\) \(Var({\bar Z_x}) = {}^2{\bar A_x} – {\bar A_x}^2\) \({\bar A_x} = 1 – \delta {\bar a_x}\) |
| Step 1 | \(Var({\bar Z_x}) = {}^2{\bar A_x} – {\bar A_x}^2\)
\({\bar A_x} = \int\limits_0^\infty {{v^t}\,{f_{Tx}}(t)\,\,dt} \)
\({\bar A_x} = \int\limits_0^\infty {{e^{ – \delta t}}\,{}_t{p_x}\,{\mu _{x + t}}\,\,dt} \)
Dengan asumsi Constant Force of Mortality \({\bar A_x} = \int\limits_0^\infty {{e^{ – \delta t}}\,{\mu _x}{e^{ – {\mu _x}t}}\,\,dt} \) \({\bar A_x} = {\mu _x}\int\limits_0^\infty {{e^{ – (\delta + {\mu _x})t}}\,\,dt} \) \({\bar A_x} = \frac{{{\mu _x}}}{{ – (\delta + {\mu _x})}}\left( {{e^{ – (\delta + {\mu _x})(\infty )}} – {e^{ – (\delta + {\mu _x})(0)}}} \right)\) \({\bar A_x} = \frac{{{\mu _x}}}{{ – (\delta + {\mu _x})}}\left( {0 – 1} \right)\) \({\bar A_x} = \frac{{{\mu _x}}}{{\delta + {\mu _x}}}\) \({\bar A_x} = \frac{{0,04}}{{0,08 + 0,04}}\) \({\bar A_x} = \frac{1}{3}\) |
| \({}^2{\bar A_x} = \int\limits_0^\infty {{v^{2t}}\,{f_{Tx}}(t)\,\,dt} \)
\({}^2{\bar A_x} = \int\limits_0^\infty {{e^{ – 2\delta t}}\,{}_t{p_x}\,{\mu _{x + t}}\,\,dt} \)
Dengan asumsi Constant Force of Mortality \({}^2{\bar A_x} = \int\limits_0^\infty {{e^{ – 2\delta t}}\,{\mu _x}{e^{ – {\mu _x}t}}\,\,dt} \) \({}^2{\bar A_x} = {\mu _x}\int\limits_0^\infty {{e^{ – (2\delta + {\mu _x})t}}\,\,dt} \) \({}^2{\bar A_x} = \frac{{{\mu _x}}}{{ – (2\delta + {\mu _x})}}\left( {{e^{ – (2\delta + {\mu _x})(\infty )}} – {e^{ – (2\delta + {\mu _x})(0)}}} \right)\) \({}^2{\bar A_x} = \frac{{{\mu _x}}}{{ – (2\delta + {\mu _x})}}\left( {0 – 1} \right)\) \({}^2{\bar A_x} = \frac{{{\mu _x}}}{{2\delta + {\mu _x}}}\) \({}^2{\bar A_x} = \frac{{0,04}}{{(2)0,08 + 0,04}}\) \({}^2{\bar A_x} = \frac{1}{5}\) | |
| \(Var({\bar Z_x}) = \frac{1}{5} – {\left( {\frac{1}{3}} \right)^2}\) \(Var({\bar Z_x}) = \frac{4}{{45}}\) | |
| Step 2 | \({\bar A_x} = 1 – \delta {\bar a_x}\) \(\frac{1}{3} = 1 – \delta {\bar a_x}\) \(\delta {\bar a_x} = 1 – \frac{1}{3}\) \(\delta {\bar a_x} = \frac{2}{3}\) |
| Step 3 | \(Var[L] = \frac{{Var({{\bar Z}_x})}}{{{{(\delta {{\bar a}_x})}^2}}}\) \(Var[L] = \frac{{\left( {\frac{4}{{45}}} \right)}}{{{{\left( {\frac{2}{3}} \right)}^2}}}\) \(Var[L] = \frac{1}{5}\) |
| Jawaban | b. 1/5 |