Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | A60 – Matematika Aktuaria |
| Periode Ujian | : | Mei 2017 |
| Nomor Soal | : | 5 |
SOAL
Suatu unit “continuously-operation air conditioning” mempunyai waktu hidup berdistribusi “exponential” dengan “mean” 4 tahun. Ketika unit “fail” harus diganti dengan biaya 1000, yang dianggap sebagai “unit of money”. Anggap \(\overline Z \) menyatakan “present value” variable acak untuk setiap pembayaran unit pada saat terjadi “fail”. Dengan menggunakan “effective annual interest rate 5%” hitunglah
\(Var[\overline Z ]\)
- 0,00918
- 0,01918
- 0,02918
- 0,03918
- 0,04918
PEMBAHASAN
| Diketahui | \({T_x}\, \sim eksponensial\,\left( {\lambda = \frac{1}{4}} \right)\)
\(\overline Z = {v^{{T_x}}}\,,\,{T_x} > 0\) |
| Step 1 | \(E[\overline Z ] = \int\limits_0^\infty {{v^t}\,{f_{{T_x}}}(t)\,dt} \)
\(E[\overline Z ] = \int\limits_0^\infty {{e^{ – \delta t}}\,\lambda {e^{ – \lambda t}}\,dt} \)
\(E[\overline Z ] = \lambda \int\limits_0^\infty {{e^{ – (\delta + \lambda )t}}\,dt} \)
\(E[\overline Z ] = \frac{\lambda }{{ – (\delta + \lambda )}}{e^{ – (\delta + \lambda )t}}\left| {_0^\infty } \right.\)
\(E[\overline Z ] = \frac{\lambda }{{ – (\delta + \lambda )}}(0 – 1)\)
\(E[\overline Z ] = \frac{\lambda }{{(\delta + \lambda )}}\) |
| \(\lambda = \frac{1}{4}\delta = \ln (1 + 0,05)\)
\(E[\overline Z ] = \frac{{\frac{1}{4}}}{{\left( {ln(1,05) + \frac{1}{4}} \right)}}\)
\(E[\overline Z ] = {\rm{0,83671}}\) |
| Step 2 | \(E[{\overline Z ^2}] = \int\limits_0^\infty {{v^2}^t\,{f_{{T_x}}}(t)\,dt} \)
\(E[{\overline Z ^2}] = \int\limits_0^\infty {{e^{ – 2\delta t}}\,\lambda {e^{ – \lambda t}}\,dt} \)
\(E[{\overline Z ^2}] = \lambda \int\limits_0^\infty {{e^{ – (2\delta + \lambda )t}}\,dt} \)
\(E[{\overline Z ^2}] = \frac{\lambda }{{ – (2\delta + \lambda )}}{e^{ – (2\delta + \lambda )t}}\left| {_0^\infty } \right.\)
\(E[{\overline Z ^2}] = \frac{\lambda }{{ – (2\delta + \lambda )}}(0 – 1)\)
\(E[{\overline Z ^2}] = \frac{\lambda }{{(2\delta + \lambda )}}\) |
| \(\lambda = \frac{1}{4}\delta = \ln (1 + 0,05)\)
\(E[{\overline Z ^2}] = \frac{{\frac{1}{4}}}{{\left( {2ln(1,05) + \frac{1}{4}} \right)}}\)
\(E[{\overline Z ^2}] = {\rm{0,71926}}\) |
| Step 3 | \(Var[\overline Z ] = E[{\overline Z ^2}] – {\left( {E[\overline Z ]} \right)^2}\)
\(Var[\overline Z ] = 0,71926 – {\left( {0,83671} \right)^2}\)
\(Var[\overline Z ] = 0,0191763759\)
\(Var[\overline Z ] \cong 0,01918\) |
| Jawaban | b. 0,01918 |
