Pembahasan Soal Ujian Profesi Aktuaris
Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian | : | A20 – Probabilitas dan Statistika |
Periode Ujian | : | Juni 2016 |
Nomor Soal | : | 18 |
SOAL
Misalkan\(f({x_1},{x_2}) = 21x_1^2x_2^3\,,\,0 < {x_1} < {x_2} < 1\) , dan 0 untuk lainnya, adalah fungsi kepadatan peluang gabungan antara \({X_1}\) dan \({X_2}\) . Berapa rataan dari \({X_1}\) kondisional diberikan \({X_2} = {x_2}\,,\,0 < {x_2} < 1\,?\)
- 21/32
- 11/32
- 3/8
- 23/35
- 1/7
Rumus | \(Rataan\,bersyarat\,{X_2} = {x_2}\,\)
\(E[E[{X_1}|{X_2} = {x_2}]] = E[{X_1}]\) |
Step 1 | \(f[{x_1}] = \int\limits_{{x_1}}^1 {21x_1^2x_2^3\,d} {x_2}\)
\(f[{x_1}] = 21x_1^2\left( {\frac{{1 – x_1^4}}{4}} \right)\)
\(f[{x_1}] = \frac{{21}}{4}\left( {x_1^2 – x_1^6} \right)\) |
Step 2 | \(E[{X_1}] = \int\limits_0^1 {{x_1}\left( {\frac{{21}}{4}\left( {x_1^2 – x_1^6} \right)} \right)} \,d{x_1}\)
\(E[{X_1}] = \frac{{21}}{4}\int\limits_0^1 {\left( {x_1^3 – x_1^7} \right)} d{x_1}\)
\(E[{X_1}] = \frac{{21}}{4}\left( {\frac{1}{4} – \frac{1}{8}} \right)\)
\(E[{X_1}] = \frac{{21}}{4}\left( {\frac{1}{8}} \right)\)
\(E[{X_1}] = \frac{{21}}{{32}}\) |
Jawaban | a. 21/32 |