Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diberikan
\({\lambda _X}\left( x \right) = k,\) \(x > 0\)Tentukanlah \({m_a}\), tingkat kematian central dalam interval \(\left( {a,a + 1} \right)\)
- \(k\)
- \(\frac{k}{2}\)
- \(\frac{k}{{2a}}\)
- \(\frac{{2k}}{{2a + 1}}\)
- 0
| Diketahui | \({\lambda _X}\left( x \right) = k,\) \(x > 0\) \({m_a}\) dengan interval \(\left( {a,a + 1} \right)\) |
| Rumus yang digunakan | \(S\left( x \right) = \exp \left( { – \int\limits_0^x {\lambda \left( y \right)dy} } \right)\) \({m_a} = \frac{{\int\limits_a^{a + 1} {S\left( y \right)\lambda \left( y \right)} dy}}{{\int\limits_a^{a + 1} {S\left( y \right)} dy}}\) |
| Proses pengerjaan | \(S\left( x \right) = \exp \left( { – \int\limits_0^x {\lambda \left( y \right)dy} } \right)\) \(= \exp \left( { – \int\limits_0^x {kdy} } \right)\) \(= \exp \left( { – kx} \right)\) |
| \(\int\limits_a^{a + 1} {S\left( y \right)\lambda \left( y \right)} dy = \int\limits_a^{a + 1} {{e^{ – ky}} \cdot k} dy\), misal \(u = – ky\) maka \(du = – kdy\) \(= k\left[ {\int\limits_{ – ka}^{ – k\left( {a + 1} \right)} { – \frac{{{e^u}}}{k}du} } \right]\) \(= – \int\limits_{ – ka}^{ – k\left( {a + 1} \right)} {{e^u}du} \) \(= – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}}\) | |
| \(\int\limits_a^{a + 1} {S\left( y \right)} dy = \int\limits_a^{a + 1} {{e^{ – ky}}} dy\), misal \(u = – ky\) maka \(du = – kdy\) \(= \int\limits_{ – ka}^{ – k\left( {a + 1} \right)} { – \frac{{{e^u}}}{k}du} \) \(= \frac{{ – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}}}}{k}\) | |
| \({m_a} = \frac{{\int\limits_a^{a + 1} {S\left( y \right)\lambda \left( y \right)} dy}}{{\int\limits_a^{a + 1} {S\left( y \right)} dy}} = \frac{{ – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}}}}{{\frac{{ – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}}}}{k}}}\) \(= \frac{{ – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}} \cdot k}}{{ – {e^{ – k\left( {a + 1} \right)}} + {e^{ – ka}}}}\) \(= k\) | |
| Jawaban | a. \(k\) |