Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diberikan beberapa informasi berikut:
i. \({Z_1}\) dan \({Z_2}\) merupakan peubah acak berdistribusi normal (0,1) yang saling bebas
ii. \(a,b,c,d,e,f\) adalah konstanta
iii. \(Y = a + b{Z_1} + c{Z_2}\) dan \(X = d + e{Z_1} + f{Z_2}\)
Tentukan \(E\left( {Y|X} \right)\)
a. \(a\)
b. \(a + \left( {b + c} \right)\left( {X – d} \right)\)
c. \(a + \left( {be + cf} \right)\left( {X – d} \right)\)
d. \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} – {f^2}} \right)} \right]\left( {X – d} \right)\)
e. \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} + {f^2}} \right)} \right]\left( {X – d} \right)\)
| Diketahui | Diberikan beberapa informasi berikut: i. \({Z_1}\) dan \({Z_2}\) merupakan peubah acak berdistribusi normal (0,1) yang saling bebas |
| Rumus yang digunakan | \(E\left[ {{X^2}} \right] = Var\left[ X \right] + {\left( {E\left[ X \right]} \right)^2} = \;{\sigma ^2} + {\mu ^2}\) \(Var\left[ {aX + bY + c} \right] = \;{a^2}Var\left( X \right) + {b^2}Var\left( Y \right)\) \(E\left[ {aX + bY + c} \right] = aE\left[ X \right] + bE\left[ Y \right] + c\) \(Cov\left( {X,Y} \right) = E\left[ {XY} \right] – E\left[ X \right]E\left[ Y \right]\) \(E\left[ {Y{\rm{|}}X} \right] = E\left[ Y \right] + \;\frac{{Cov\left( {X,Y} \right)}}{{Var\left[ X \right]}}\left( {X – E\left( X \right)} \right)\) |
| Proses pengerjaan | \(E\left( {Y|X} \right) = E\left[ {a + b{Z_1} + c{Z_2}} \right] + \frac{{E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right] – \;E\left[ {a + b{Z_1} + c{Z_2}} \right]E\left[ {d + e{Z_1} + f{Z_2}} \right]}}{{Var\left[ {d + e{Z_1} + f{Z_2}} \right]}}\left( {X – E\left( {d + e{Z_1} + f{Z_2}} \right)} \right)\;\)
\(E\left( {Y|X} \right) = a + \frac{{ad + be + cf – \;ad}}{{{e^2}Var\left[ {{Z_1}] + {f^2}Var[{Z_2}} \right]}}\left( {X – d} \right)\) |
| Jawaban | e. \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} + {f^2}} \right)} \right]\left( {X – d} \right)\) |