Pokazać,że współczynnik korelacji nie zmienia się przy przekształceniach liniowych tzn.:
\(\displaystyle{ \rho (X,Y)=\rho (aX+b,cY+d)}\)
dzieki
wspolczynnik korelacji
wspolczynnik korelacji
\(\displaystyle{ \rho_{X,Y}= \frac{cov(X,Y)}{s_{x}s_{y}} = \frac{E(X-EX)(Y-EY)}{ \sqrt{D^{2}XD^{2}Y} }
\\ \rho_{aX+b,cY+d}= \frac{E(aX+b-E(aX+b))(cY+d-E(cY+d))}{ \sqrt{D^{2}(aX+b)D^{2}(cY+d)} }=*
\\ D^{2}(aX+b)=E(aX+b)^{2} - (E(aX+b))^{2}=E(a^{2}X^{2}+2abX+b^{2})-(aEX+b)^{2}=a^{2}EX^{2}+2abEX+b^{2}-a^{2}(EX)^{2}-2abEX-b^{2}=a^{2}EX^{2}-a^{2}(EX)^{2}=a^{2}(EX^{2}-(EX)^{2})=a^{2}D^{2}X
\\ *=\frac{E(aX+b-E(aX+b))(cY+d-E(cY+d))}{ac \sqrt{D^{2}XD^{2}Y} }=\frac{E(aX-aEX)(cY-cEY)}{ac \sqrt{D^{2}XD^{2}Y}}=\frac{acE(X-EX)(Y-EY)}{ac\sqrt{D^{2}XD^{2}Y}}=\rho_{X,Y}}\)
\\ \rho_{aX+b,cY+d}= \frac{E(aX+b-E(aX+b))(cY+d-E(cY+d))}{ \sqrt{D^{2}(aX+b)D^{2}(cY+d)} }=*
\\ D^{2}(aX+b)=E(aX+b)^{2} - (E(aX+b))^{2}=E(a^{2}X^{2}+2abX+b^{2})-(aEX+b)^{2}=a^{2}EX^{2}+2abEX+b^{2}-a^{2}(EX)^{2}-2abEX-b^{2}=a^{2}EX^{2}-a^{2}(EX)^{2}=a^{2}(EX^{2}-(EX)^{2})=a^{2}D^{2}X
\\ *=\frac{E(aX+b-E(aX+b))(cY+d-E(cY+d))}{ac \sqrt{D^{2}XD^{2}Y} }=\frac{E(aX-aEX)(cY-cEY)}{ac \sqrt{D^{2}XD^{2}Y}}=\frac{acE(X-EX)(Y-EY)}{ac\sqrt{D^{2}XD^{2}Y}}=\rho_{X,Y}}\)