Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan peubah acak kontinyu mempunyai fungsi kepadatan peluang :
\(f(x) = \frac{{\Gamma (a + b)}}{{\Gamma (a)\Gamma (b)}}{x^{(a – 1)}}{(1 – x)^{(b – 1)}}\) ,0 < x < 1, dan a >0 & b >0Jika b = 6 dan a = 5. Tentukan ekspektasi dari \({(1 – X)^{ – 4}}\) !
- 42
- 63
- 210
- 252
- 315
PEMBAHASAN
| Step 1 | a = 5 dan b = 6 \(f(x) = \frac{{\Gamma (5 + 6)}}{{\Gamma (5)\Gamma (6)}}{x^{(5 – 1)}}{(1 – x)^{(6 – 1)}}\) \(f(x) = \frac{{\Gamma (11)}}{{\Gamma (5)\Gamma (6)}}{x^4}{(1 – x)^5}\) \(f(x) = \frac{{10!}}{{4!5!}}{x^4}{(1 – x)^5}\) \(f(x) = 1260{x^4}{(1 – x)^5}\) |
| Step 2 | \(E[{(1 – X)^{ – 4}}] = \int\limits_0^1 {{{(1 – x)}^{ – 4}}} f(x)dx\) \(E[{(1 – X)^{ – 4}}] = \int\limits_0^1 {{{(1 – x)}^{ – 4}}} 1260{x^4}{(1 – x)^5}dx\) \(E[{(1 – X)^{ – 4}}] = 1260\int\limits_0^1 {(1 – x)} {x^4}dx\) \(E[{(1 – X)^{ – 4}}] = 1260\int\limits_0^1 {({x^4} – {x^5})} dx\) \(E[{(1 – X)^{ – 4}}] = 1260\left( {\frac{{{x^5}}}{5} – \frac{{{x^6}}}{6}} \right)\left| \begin{array}{l}1\\0\end{array} \right.\) \(E[{(1 – X)^{ – 4}}] = 1260\left( {\frac{1}{5} – \frac{1}{6}} \right)\) \(E[{(1 – X)^{ – 4}}] = 1260\frac{1}{{30}}\) \(E[{(1 – X)^{ – 4}}] = 42\) |
| Jawaban | a. 42 |