Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Dari sebuah fungsi kepadatan gabungan (joint density function) dari \({T_x}\) dan \({T_y}\) berikut:
\({f_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right) = \frac{4}{{{{\left( {1 + {t_x} + 2{t_y}} \right)}^3}}}\), untuk \({t_x} > 0\) dan \({t_y} > 0\)
Tentukan \({}_n{q_{xy}}\)
- \(\frac{1}{{1 + 3n}}\)
- \(\frac{1}{{1 + n}}\)
- \(\frac{n}{{1 + n}}\)
- \(\frac{{3n}}{{1 + 3n}}\)
- \(\frac{{5n}}{{1 + 5n}}\)
| Diketahui | Dari sebuah fungsi kepadatan gabungan (joint density function) dari \({T_x}\) dan \({T_y}\) berikut: \({f_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right) = \frac{4}{{{{\left( {1 + {t_x} + 2{t_y}} \right)}^3}}}\), untuk \({t_x} > 0\) dan \({t_y} > 0\) |
| Rumus yang digunakan | \({}_t{p_{xy}} = {S_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right) = \Pr \left( {{T_x} \ge n{\rm{ dan }}{T_y} \ge n} \right) = \int\limits_n^\infty {\int\limits_n^\infty {{f_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right)d{t_x}d{t_y}} } \) |
| Proses pengerjaan | \({}_t{p_{xy}} = {S_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right) = \Pr \) (\({{T_x} \ge n{\rm{ }}}\) dan \({{\rm{ }}{T_y} \ge n}\)) \({}_t{p_{xy}} = \int\limits_n^\infty {\int\limits_n^\infty {{f_{{T_x},{T_y}}}\left( {{t_x},{t_y}} \right)d{t_x}} d{t_y}} \) \({}_t{p_{xy}} = \int\limits_n^\infty {\int\limits_n^\infty {\frac{4}{{{{\left( {1 + {t_x} + 2{t_y}} \right)}^3}}}d{t_x}} d{t_y}} \) \({{}_t{p_{xy}} = \int\limits_n^\infty {4\int\limits_{1 + n + 2{t_y}}^{\infty + 2{t_y}} {\frac{1}{{{u^3}}}du} d{t_y}} }\) misal \({u = 1 + {t_x} + 2{t_y} \Rightarrow du = d{t_x}}\) \({}_t{p_{xy}} = \int\limits_n^\infty {4\left[ { – \frac{1}{{2{u^2}}}} \right]_{1 + n + 2{t_y}}^{\infty + 2{t_y}}d{t_y}} \) \({}_t{p_{xy}} = \int\limits_n^\infty {\frac{2}{{{{\left( {1 + n + 2{t_y}} \right)}^2}}}d{t_y}} \) \({{}_t{p_{xy}} = \int\limits_{1 + 3n}^{\infty + n} {\frac{1}{{{v^2}}}dv} }\) misal \({v = 1 + n + 2{t_y} \Rightarrow dv = 2d{t_y}}\) \({}_t{p_{xy}} = \left[ { – \frac{1}{x}} \right]_{1 + 3n}^{\infty + n}\) \({}_t{p_{xy}} = \frac{1}{{1 + 3n}}\) \({}_t{q_{xy}} = 1 – {}_t{p_{xy}} = 1 – \frac{1}{{1 + 3n}} = \frac{{3n}}{{1 + 3n}}\) |
| Jawaban | d. \(\frac{{3n}}{{1 + 3n}}\) |