Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | Agustus 2023 |
| Nomor Soal | : | 4 |
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
ii. \(a,b,c,d,e,f\) adalah konstanta
iii. \(Y = a + b{Z_1} + c{Z_2}\) dan \(X = d + e{Z_1} + f{Z_2}\) |
| 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 | \(E\left( {Y|X} \right) = E\left[ {a + b{Z_1} + c{Z_2}} \right] + \frac{{E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right] – \;E\left[ {a + b{Z_1} + c{Z_2}} \right]E\left[ {d + e{Z_1} + f{Z_2}} \right]}}{{Var\left[ {d + e{Z_1} + f{Z_2}} \right]}}\left( {X – E\left( {d + e{Z_1} + f{Z_2}} \right)} \right)\;\) - \(E\left[ {a + b{Z_1} + c{Z_2}} \right] = a + bE\left[ {{Z_1}} \right] + c\left[ {{Z_2}} \right] \to \;{Z_1}\) dan \({Z_2}\) berdistribusi normal (0,1)
\(= a + b\left( 0 \right) + c\left( 0 \right)\)
\(= a\) - \(E\left[ {d + e{Z_1} + f{Z_2}} \right] = d + eE\left[ {{Z_1}} \right] + f\left[ {{Z_2}} \right] \to \;{Z_1}\) dan \({Z_2}\) berdistribusi normal (0,1)
\(= d + e\left( 0 \right) + f\left( 0 \right)\)
\(= d\) - \(E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right] = ad + aeE\left[ {{Z_1}} \right] + afE\left[ {{Z_2}} \right] + bdE\left[ {{Z_1}} \right] + beE{\left[ {{Z_1}} \right]^2} + bf\;E\left[ {{Z_1}{Z_2}} \right] + cd + cdE\left[ {{Z_1}{Z_2}} \right] + cfE{\left[ {{Z_2}} \right]^2}\)
\(\to \;{Z_1}\) dan \({Z_2}\) berdistribusi normal (0,1)\(= ad + ae\left( 0 \right) + af\left( 0 \right) + bd\left( 0 \right) + be\left( {1 + 0} \right) + bf\left( 0 \right) + cd + cd\left( 0 \right) + cf\left( {1 + 0} \right)\)
\(= ad + be + cf\) \(E\left( {Y|X} \right) = a + \frac{{ad + be + cf – \;ad}}{{{e^2}Var\left[ {{Z_1}] + {f^2}Var[{Z_2}} \right]}}\left( {X – d} \right)\)
\(E\left( {Y|X} \right) = a + \frac{{be + cf}}{{{e^2} + {f^2}}}\left( {X – d} \right)\) |
| Jawaban | e. \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} + {f^2}} \right)} \right]\left( {X – d} \right)\) |
