Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan \({Y_1} < {Y_2}\) ialah statistik terurut dari suatu sampel acak berukuran 2 dari distribusi \(N(0,{\sigma ^2})\) . Hitung \(E({Y_1})\)
- \(\frac{{2\sigma }}{{\sqrt \pi }}\)
- \(\frac{\sigma }{{\sqrt \pi }}\)
- -\(\frac{{3\sigma }}{{\sqrt \pi }}\)
- -\(\frac{{2\sigma }}{{\sqrt \pi }}\)
- -\(\frac{\sigma }{{\sqrt \pi }}\)
[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]
PEMBAHASAN
| Rumus | \({f_Y}(y) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ – \frac{{{y^2}}}{{{\sigma ^2}}}\left( {\frac{1}{2}} \right)}}\) |
| Step 1 | \({f_{{Y_1},{Y_2}}}({y_1},{y_2}) = 2!{f_{{Y_1}}}({y_1}){f_{{Y_2}}}({y_2})\) \({f_{{Y_1},{Y_2}}}({y_1},{y_2}) = 2!\left( {\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ – \frac{{{y_1}^2}}{{{\sigma ^2}}}\left( {\frac{1}{2}} \right)}}} \right)\left( {\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ – \frac{{{y_2}^2}}{{{\sigma ^2}}}\left( {\frac{1}{2}} \right)}}} \right)\) \({f_{{Y_1},{Y_2}}}({y_1},{y_2}) = \frac{2}{{{\sigma ^2}2\pi }}\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2 + {y_2}^2} \right)}}} \right)\) \({f_{{Y_1},{Y_2}}}({y_1},{y_2}) = \frac{1}{{{\sigma ^2}\pi }}\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2 + {y_2}^2} \right)}}} \right),\, – \infty < {y_1} < \infty , – \infty < {y_2} < \infty \) |
| Step 2 | \(E[{Y_1}] = \int\limits_{ – \infty }^\infty {\int\limits_{ – \infty }^{{y_2}} {{y_1}} } {f_{{Y_1},{Y_2}}}({y_1},{y_2})\,d{y_1}\,d{y_2}\) \(E[{Y_1}] = \int\limits_{ – \infty }^\infty {\int\limits_{ – \infty }^{{y_2}} {{y_1}} } \frac{1}{{{\sigma ^2}\pi }}\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2 + {y_2}^2} \right)}}} \right)\,d{y_1}\,d{y_2}\) \(E[{Y_1}] = \frac{1}{{{\sigma ^2}\pi }}\int\limits_{ – \infty }^\infty {\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_2}^2} \right)}}} \right)\left[ {\int\limits_{ – \infty }^{{y_2}} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2} \right)}}} \right)\,d{y_1}\,} \right]} d{y_2}\) |
| Subsitusi: \(\,u = y_1^2\) \(du = 2{y_1}d{y_1}\) \(\int\limits_{ – \infty }^{{y_2}} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2} \right)}}} \right)\,d{y_1} = \int\limits_\infty ^{{y_2}^2} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( u \right)}}} \right)\,\frac{{du}}{{2{y_1}}}\) \(\int\limits_{ – \infty }^{{y_2}} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2} \right)}}} \right)\,d{y_1} = \int\limits_\infty ^{{y_2}^2} {\frac{1}{2}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( u \right)}}} \right)\,du\) \(\int\limits_{ – \infty }^{{y_2}} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2} \right)}}} \right)\,d{y_1} = \frac{1}{2}\left( { – 2{\sigma ^2}} \right){e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_2}^2 – (\infty )} \right)}}\) \(\int\limits_{ – \infty }^{{y_2}} {{y_1}} \left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_1}^2} \right)}}} \right)\,d{y_1} = \left( { – {\sigma ^2}} \right)\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_2}^2} \right)}}} \right)\) | |
| \(E[{Y_1}] = \frac{1}{{{\sigma ^2}\pi }}\int\limits_{ – \infty }^\infty {\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_2}^2} \right)}}} \right)\left[ {\left( { – {\sigma ^2}} \right)\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {{y_2}^2} \right)}}} \right)\,} \right]} d{y_2}\) \(E[{Y_1}] = \frac{{ – {\sigma ^2}}}{{{\sigma ^2}\pi }}\int\limits_{ – \infty }^\infty {\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {2{y_2}^2} \right)}}} \right)} d{y_2}\) \(E[{Y_1}] = \frac{{ – 1}}{\pi }\int\limits_{ – \infty }^\infty {\left( {{e^{ – \left( {\frac{1}{{2{\sigma ^2}}}} \right)\left( {2{y_2}^2} \right)}}} \right)} d{y_2}\) –> Diarahkan ke fungsi densitas distribusi normal \((0,\frac{{{\sigma ^2}}}{2})\) | |
| Step 3 | \(N(0,\frac{{{\sigma ^2}}}{2})\,menjadi\,N(0,{\sigma ^2})\,\)
\(\frac{1}{2}\left( {\frac{1}{{\frac{{{\sigma ^2}}}{2}}}} \right) = \frac{1}{{{\sigma ^2}}}\)
\(E[{Y_1}] = \left( { – \frac{1}{\pi }} \right)\left( {\frac{\sigma }{{\sqrt 2 }}} \right)\left( {\sqrt {2\pi } } \right)\left( 1 \right)\) \(\to \) (1) ialah hasil integral fungsi densitas distribusi normal \(E[{Y_1}] = \left( { – \frac{\sigma }{{\sqrt \pi }}} \right)\) |
| Jawaban | e. \(\frac{\sigma }{{\sqrt \pi }}\) |
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