Pembahasan Ujian PAI: A20 - No. 17 - Juni 2016

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	A20 – Probabilitas dan Statistika
Periode Ujian	:	Juni 2016
Nomor Soal	:	17

SOAL

Misalkan \({X_1},{X_2}\) mempunyai fungsi kepadatan peluang gabungan \(h({x_1},{x_2}) = 8{x_1}{x_2}\,,\,0 < {x_1} < {x_2} < 1\,\,,\,\) dan \(h({x_1},{x_2}) = 0\,\) untuk \({x_1},{x_2}\) lainnya. Cari peluang gabungan antara \({Y_1},{Y_2}\,\) dimana \({Y_1} = \frac{{{X_1}}}{{{X_2}}}\,\) dan \(\,{Y_2} = {X_2}\). Petunjuk : Gunakan pertidaksamaan \(0 < {y_1}{y_2} < {y_2} < 1\) dalam memetakan S (bidang dimana x terdefinisi) ke (bidang dimana y terdefinisi) dengan Jacobian Matrix !

\(8{y_1}y_2^3\)
\(8y_1^3y_2^3\)
\(8y_1^2y_2^3\)
\(8{y_1}y_2^2\)
\(8{y_1}y_2^4\)

Kunci Jawaban & Pembahasan

Diketahui	\(h({x_1},{x_2}) = 8{x_1}{x_2}\,,\,0 < {x_1} < {x_2} < 1\,\) \({Y_1} = \frac{{{X_1}}}{{{X_2}}}\,dan\,\,{Y_2} = {X_2}\)
Step 1	\(Transformasi:\,{X_1} = {Y_1}{X_2}\,\,dan\,\,{X_2} = {Y_2}\) \(maka\) \({X_1} = {Y_1}{Y_2}\,\,dan\,\,{X_2} = {Y_2}\)
Step 2	\(Jacobian\,Matrix = \left\| \begin{array}{l} \frac{{\partial {X_1}}}{{\partial {Y_1}}}\,\,\,\,\,\frac{{\partial {X_1}}}{{\partial {Y_2}}}\\ \frac{{\partial {X_2}}}{{\partial {Y_1}}}\,\,\,\,\,\frac{{\partial {X_2}}}{{\partial {Y_1}}} \end{array} \right\|\) \(Jacobian\,Matrix = \left\| \begin{array}{l} {y_2}\,\,\,\,\,\,\,\,{y_1}\\ 0\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right\|\) \(Jacobian\,Matrix = {y_2}(1) – {y_1}(0)\) \(Jacobian\,Matrix = {y_2}\)
Step 3	\(h({y_1},{y_2}) = h({x_1},{x_2})(Jacobian\,Matrix)\) \(h({y_1},{y_2}) = 8{x_1}{x_2}({y_2})\) \(h({y_1},{y_2}) = 8({y_1}{y_2})({y_2})({y_2})\) \(h({y_1},{y_2}) = 8{y_1}y_2^3\)
Jawaban	a. \(8{y_1}y_2^3\)

Pembahasan Ujian PAI: A20 – No. 17 – Juni 2016