Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Fungsi pembangkit momen dari distribusi gabungan suatu peubah acak dan adalah \({M_{X,Y}}(t{}_1,{t_2}) = \frac{1}{{3(1 – {t_2})}} + \frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}}\) untuk 2 < 1. Hitung \(Var[x]\)
- 1/18
- 1/9
- 1/6
- 2/9
- 1/3
PEMBAHASAN
\(E[X] = {M_x}'(t){|_{t = 0}}\) \(E[X] = \)\(\frac{{\partial (\frac{1}{{3(1 – {t_2})}} + \frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}})}}{{\partial {t_1}}}\) | t1 = 0, t2 = 0\(E[X] = \frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}}\) | t1 = 0, t2 = 0
\(E[X] = \frac{2}{3}\) \(E[{X^2}] = \frac{{{\partial ^2}(\frac{1}{{3(1 – {t_2})}} + \frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}})}}{{\partial {t_1}^2}}\) | t1 = 0, t2 = 0
\(E[{X^2}] = \frac{{\partial (\frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}})}}{{\partial {t_1}}}\) | t1 = 0, t2 = 0
\(E[{X^2}] = \frac{2}{3}{e^{{t_1}}}\frac{2}{{(2 – {t_2})}}\) | t1 = 0, t2 = 0
\(E[{X^2}] = \frac{2}{3}\) \(Var[X] = \frac{2}{3} – {\left( {\frac{2}{3}} \right)^2}\) \(Var[X] = \frac{2}{9}\)
Jawaban pada pilihan : D. 2/9