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 |
