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 !
- \(8{y_1}y_2^3\)
- \(8y_1^3y_2^3\)
- \(8y_1^2y_2^3\)
- \(8{y_1}y_2^2\)
- \(8{y_1}y_2^4\)
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. \(8{y_1}y_2^3\) |
