Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan X1 dan X2 adalah dua peubah acak saling bebas berdistribusi normal dengan rata-rata dan variansi 1. Jika [c|X1− X2|] = 1, maka nilai c adalah
- \(\sqrt \pi \)
- \(\frac{1}{{\sqrt \pi }}\)
- \(\frac{{\sqrt {2\pi } }}{4}\)
- \(\frac{2}{{\sqrt \pi }}\)
- \(\frac{{\sqrt \pi }}{2}\)
PEMBAHASAN
| Diketahui | X1 dan X2 ~ Normal (1,1) |
| Misalkan | Y = X1− X2 Y ~ Normal (0,2)\({f_Y}(y) = \frac{1}{{\sqrt 2 \sqrt {2\pi } }}{e^{ – (\frac{{{y^2}}}{{2(2)}})}}\) \({f_Y}(y) = \frac{1}{{2\sqrt \pi }}{e^{ – \frac{{{y^2}}}{4}}}, – \infty < y < \infty \) |
| Step 1 | \(E[c|X1 – X2|] = 1\) \(cE[|Y|] = 1\) \(c = \frac{1}{{E[|Y|]}}\) |
| Step 2 | \(E[|Y|] = 2\int\limits_0^\infty y (\frac{1}{{2\sqrt \pi }}{e^{ – \frac{{{y^2}}}{4}}})dy\) \(E[|Y|] = \frac{2}{{2\sqrt \pi }}\int\limits_0^\infty {y({e^{ – \frac{{{y^2}}}{4}}}} )dy\)Misalkan \(t = \frac{{{y^2}}}{4},y = 2\sqrt t ,\frac{{dy}}{{dt}} = {t^{ – (\frac{1}{2})}}\) \(E[|Y|] = \frac{1}{{\sqrt \pi }}\int\limits_0^\infty {2\sqrt t ({e^{ – t}}} )\frac{{dt}}{{\sqrt t }}\) \(E[|Y|] = \frac{2}{{\sqrt \pi }}(( – 1)(0 – 1))\) \(E[|Y|] = \frac{2}{{\sqrt \pi }}\) |
| Step 3 | \(c = \frac{1}{{E[|Y|]}}\) \(c = \frac{1}{{\frac{2}{{\sqrt \pi }}}}\) \(c = \frac{{\sqrt \pi }}{2}\) |
| Jawaban | e. \(\frac{{\sqrt \pi }}{2}\) |