Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui fungsi force of mortality \(\mu \left( x \right) = \frac{1}{{x + 1}},x \ge 0\).
Tentukan PDF (probability density function) dari X.
- \(f\left( x \right) = \frac{1}{{x + 1}},x \ge 0\)
- \(f\left( x \right) = \frac{x}{{x + 1}},x \ge 0\)
- \(f\left( x \right) = {e^{x + 1}},x \ge 0\)
- \(f\left( x \right) = \frac{{x + 1}}{x},x \ge 0\)
- \(f\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}},x \ge 0\)
| Diketahui | \(\mu \left( x \right) = \frac{1}{{x + 1}},x \ge 0\) |
| Rumus yang digunakan | \(f\left( x \right) = \mu \left( x \right) \cdot \exp \left[ { – \int_0^x {\mu \left( t \right)dt} } \right]\) |
| Proses pengerjaan | \(f\left( x \right) = \mu \left( x \right) \cdot \exp \left[ { – \int_0^x {\mu \left( t \right)dt} } \right]\) \(= \frac{1}{{x + 1}} \cdot \exp \left[ { – \int_0^x {\frac{1}{{t + 1}}dt} } \right]\) \(= \frac{1}{{x + 1}} \cdot \exp \left[ { – \ln \left( {x + 1} \right)} \right]\) \(= \frac{1}{{{{\left( {x + 1} \right)}^2}}}\) |
| Jawaban | e. \(f\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}},x \ge 0\) |