Pembahasan Ujian PAI: A50 – No. 15 – November 2018

\(Y = Min\left( {{X_1},{X_2}} \right)\), \(Z = Max\left( {{X_1},{X_2}} \right)\)

\({F_{X,Y}}\left( {x,y} \right) = {F_X}\left( x \right) \cdot {F_Y}\left( y \right){\rm{ }}\) dan \({{\rm{S}}_{X,Y}}\left( {x,y} \right) = {S_X}\left( x \right) \cdot {S_Y}\left( y \right)\)
\(F\left( x \right) = P\left( {X \le x} \right)\) dan \(S\left( x \right) = 1 – F\left( x \right) = P\left( {X > x} \right)\)
\(Y = Min\left( {{X_1},{X_2}} \right) \to {F_Y}\left( y \right) = P\left( {Y \le y} \right) = 1 – P\left( {Y > y} \right)\)
\(Z = Max\left( {{X_1},{X_2}} \right) \to {F_Z}\left( y \right) = P\left( {Z \le z} \right)\)

\({F_Y}\left( y \right) = P\left( {Y \le y} \right) = 1 – P\left( {Y > y} \right)\)

\(= 1 – P\left( {{X_1} > x,{X_2} > x} \right)\)
\(= 1 – \left[ {P\left( {{X_1} > x} \right)P\left( {{X_2} > x} \right)} \right]\)

\(1 – {F_Y}\left( y \right) = P\left( {{X_1} > x} \right)P\left( {{X_2} > x} \right)\)
\({S_Y}\left( y \right) = {S_{{X_1}}}\left( x \right){S_{{X_2}}}\left( x \right)\)

\({F_Z}\left( z \right) = P\left( {Z \le z} \right)\)

\(= P\left( {{X_1} \le x,{X_2} \le x} \right)\)
\(= P\left( {{X_1} \le x} \right)P\left( {{X_2} \le x} \right)\)

\({F_Z}\left( z \right) = {F_{{X_1}}}\left( x \right){F_{X2}}\left( x \right)\)