Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | April 2019 |
| Nomor Soal | : | 30 |
SOAL
Diberikan beberapa informasi berikut:
- \({Z_1}\) dan \({Z_2}\) merupakan peubah acak berdistribusi Normal (0,1) yang saling bebas
- \(a,b,c,d,e,f\) adalah konstanta
- \(Y = a + b{Z_1} + c{Z_2}\) dan \(X = d + e{Z_1} + f{Z_2}\)
Tentukan \(E\left( {Y|X} \right)\)
- \(a\)
- \(a + \left( {b + c} \right)\left( {X – d} \right)\)
- \(a + \left( {be + cf} \right)\left( {X – d} \right)\)
- \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} – {f^2}} \right)} \right]\left( {X – d} \right)\)
- \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} + {f^2}} \right)} \right]\left( {X – d} \right)\)
| Diketahui | - \({Z_1}\) dan \({Z_2}\) merupakan peubah acak berdistribusi Normal (0,1) yang saling bebas
(NB: \({\mu _{{Z_i}}} = E\left[ {{Z_i}} \right] = 0\) dan \(\sigma _{{Z_i}}^2 = Var\left[ {{Z_i}} \right] = 1\)) - \(a,b,c,d,e,f\) adalah konstanta
- \(Y = a + b{Z_1} + c{Z_2}\) dan \(X = d + e{Z_1} + f{Z_2}\)
|
| Rumus yang digunakan | \(E\left[ {Y|X} \right] = E\left[ Y \right] + \frac{{Cov\left( {X,Y} \right)}}{{Var\left[ X \right]}}\left( {X – E\left[ X \right]} \right)\)
\({\mathop{\rm cov}} \left( {X,Y} \right) = E\left[ {XY} \right] – E\left[ X \right]E\left[ Y \right]\)
\(E\left[ {aX + bY} \right] = aE\left[ X \right] + bE\left[ Y \right]\)
\(Var\left[ {aX + bY + c} \right] = {a^2}Var\left[ X \right] + {b^2}Var\left[ Y \right]\)
\(E\left[ {{X^2}} \right] = Var\left[ X \right] + {\left( {E\left[ X \right]} \right)^2} = {\sigma ^2} + {\mu ^2}\) |
| Proses pengerjaan | \(E\left[ {Y|X} \right] = E\left[ Y \right] + \frac{{E\left[ {XY} \right] – E\left[ X \right]E\left[ Y \right]}}{{Var\left( X \right)}}\left( {X – E\left[ X \right]} \right)\)
\(= E\left[ {a + b{Z_1} + c{Z_2}} \right] + \frac{{E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right]}}{{Var\left[ {d + e{Z_1} + f{Z_2}} \right]}}\left( {X – E\left[ {d + e{Z_1} + f{Z_2}} \right]} \right)\)
Kita selesaikan secara terpisah |
| \(E\left[ {a + b{Z_1} + c{Z_2}} \right] = a + bE\left[ {{Z_1}} \right] + cE\left[ {{Z_2}} \right] = a + b\left( 0 \right) + c\left( 0 \right) = a\) |
| \(E\left[ {d + e{Z_1} + f{Z_2}} \right] = d + eE\left[ {{Z_1}} \right] + fE\left[ {{Z_2}} \right] = d + e\left( 0 \right) + f\left( 0 \right) = d\) |
| \(\left( {E\left[ {a + b{Z_1} + c{Z_2}} \right]} \right)\left( {E\left[ {d + e{Z_1} + f{Z_2}} \right]} \right) = ad\) |
| \(Var\left[ {d + e{Z_1} + f{Z_2}} \right] = {e^2}Var\left[ {{Z_1}} \right] + {f^2}Var\left[ {{Z_2}} \right] = {e^2}\left( 1 \right) + {f^2}\left( 1 \right) = {e^2} + {f^2}\) |
| \(E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right]\)
\(= E\left[ {ad + ae{Z_1} + af{Z_2} + bd{Z_1} + beZ_1^2 + bf{Z_1}{Z_2} + cd{Z_2} + ce{Z_1}{Z_2} + cfZ_2^2} \right]\)
\(= ad + aeE\left[ {{Z_1}} \right] + afE\left[ {{Z_2}} \right] + bdE\left[ {{Z_1}} \right] + beE\left[ {Z_1^2} \right] + bfE\left[ {{Z_1}} \right]E\left[ {{Z_2}} \right] + cdE\left[ {{Z_2}} \right] + ceE\left[ {{Z_1}} \right]E\left[ {{Z_2}} \right] + cfE\left[ {Z_2^2} \right]\)
\(= ad + ae\left( 0 \right) + af\left( 0 \right) + bd\left( 0 \right) + be\left( 1 \right) + bf\left( 0 \right)\left( 0 \right) + cd\left( 0 \right) + ce\left( 0 \right)\left( 0 \right) + cf\left( 1 \right)\)
\(= ad + be + cf\) |
| \(E\left[ {Y|X} \right] = E\left[ {a + b{Z_1} + c{Z_2}} \right] + \)
\(\frac{{E\left[ {\left( {a + b{Z_1} + c{Z_2}} \right)\left( {d + e{Z_1} + f{Z_2}} \right)} \right] – \left( {E\left[ {a + b{Z_1} + c{Z_2}} \right]} \right)\left( {E\left[ {d + e{Z_1} + f{Z_2}} \right]} \right)}}{{Var\left[ {d + e{Z_1} + f{Z_2}} \right]}}\left( {X – E\left[ {d + elatex {Z_1} + f{Z_2}} \right]} \right)\)
\(= a + \frac{{ad + be + cf – ad}}{{{e^2} + {f^2}}}\left( {X – d} \right)\)
\(= a + \frac{{be + cf}}{{{e^2} + {f^2}}}\left( {X – d} \right)\) |
| Jawaban | e. \(a + \left[ {\left( {be + cf} \right)/\left( {{e^2} + {f^2}} \right)} \right]\left( {X – d} \right)\) |
