Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | Juni 2015 |
| Nomor Soal | : | 2 |
SOAL
Diketahui survival function dari \(X\) adalah \(\begin{array}{*{20}{c}} {S\left( x \right) = {e^{ – x}}\left( {x + 1} \right),}&{x \ge 0} \end{array}\)
Tentukan \(E\left[ X \right]\)
- 0,25
- 1
- 0,5
- 2
- 0
| Diketahui | \(\begin{array}{*{20}{c}} {S\left( x \right) = {e^{ – x}}\left( {x + 1} \right),}&{x \ge 0} \end{array}\) |
| Rumus yang digunakan | \(E\left[ X \right] = \int\limits_0^\infty {S\left( x \right)dx} \) |
| Proses pengerjaan | \({E\left[ X \right] = \int\limits_0^\infty {{e^{ – x}}\left( {x + 1} \right)dx} }\) misal \({u = – x \Leftrightarrow du = – dx}\)
\({E\left[ X \right] = \int\limits_0^\infty {{e^u}\left( {u – 1} \right)du} }\) misal \({v = {e^u} \Leftrightarrow dv}\) \(= {e^u}du\)
\(E\left[ X \right] = \int\limits_0^\infty {\left[ {\ln \left( v \right) – 1} \right]dv} \)
\(E\left[ X \right] = \left. {\left[ {v\ln \left( v \right) – v – v} \right]} \right|_0^\infty = \left. {\left[ {v\ln \left( v \right) – 2v} \right]} \right|_0^\infty \)
\(E\left[ X \right] = \left. {\left[ {{e^{ – x}}\ln \left( {{e^{ – x}}} \right) – 2{e^{ – x}}} \right]} \right|_0^\infty \)
\(E\left[ X \right] = 2\) |
| Jawaban | d. 2 |
