Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan dan ialah peubah acak saling bebas dengan
\({\mu _X} = 1,{\mu _Y} = – 1,\sigma _X^2 = \frac{1}{2},\sigma _Y^2 = 2\) Hitung \(E[{(X + 1)^2}{(Y – 1)^2}].\)- 1
- 9/2
- 16
- 17
- 27
PEMBAHASAN
| Kalkulasi | \(E[{(X + 1)^2}{(Y – 1)^2}] = E[{(X + 1)^2}]E[{(Y – 1)^2}]\) \(E[{(X + 1)^2}] = E[{X^2} + 2X + 1]\) \(E[{(X + 1)^2}] = E[{X^2}] + 2E[X] + 1\) \(E[{(X + 1)^2}] = (\sigma _X^2 + \mu _X^2) + 2{\mu _X} + 1\) \(E[{(X + 1)^2}] = \left( {\frac{1}{2} + 1} \right) + 2(1) + 1\) \(E[{(X + 1)^2}] = 4,5\) \(E[{(Y – 1)^2}] = E[{Y^2} – 2Y + 1]\) \(E[{(Y – 1)^2}] = E[{Y^2}] – 2E[Y] + 1\) \(E[{(Y – 1)^2}] = (\sigma _Y^2 + \mu _Y^2) – 2{\mu _Y} + 1\) \(E[{(Y – 1)^2}] = (2 + {( – 1)^2}) – 2( – 1) + 1\) \(E[{(Y – 1)^2}] = 6\) \(E[{(X + 1)^2}{(Y – 1)^2}] = \left( {4,5} \right)\left( 6 \right)\) \(E[{(X + 1)^2}{(Y – 1)^2}] = 27\) |
| Jawaban | e. 27 |