Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
Periode Ujian |
: |
Juni 2016 |
Nomor Soal |
: |
5 |
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 >0
Jika 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 |