Using the definition of Covariance, Cov(X,Y) = E[(x – My)(Y – My)], prove the followings. a. Cov(X,Y) = E(XY) - Mx Hy b. Cov(X,Y) = Cov(Y, X) C. Cov(X,X) = Var(X) d. Cov(X + Z,Y) = Cov(X,Y) + Cov(Z,Y) e. Cov(EX,Y)= XCov(X,Y) f. Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y) g. Var(EX) = Var(X) + Ej Cov(X, X;)

Answer:Step-by-step explanation:We have by definition[tex]Cov(X,Y) = E[(x – M_x)(Y – M_y)][/tex]where Mx and My are means of X and Y respectivelya) [tex]E[(x – M_x)(Y – M_y)]\\= E(x,y) - E(x,My)-E(y,Mx)+M_x M_y\\=E(x,y) -E(x) M_y -E(y) M_x +M_x M_y\\=E(x,y)-M_x M_y-M_x M_y+M_x M_y\\=E(x,y)-M_x M_y[/tex]b) Since right side inside expectation is commutative, we get cov(x,y) = cov (y,x)c) [tex]cov (x,x) = E(x-M_x)(x-M_x) = E(x-M_x)^2\\= Var(X)[/tex]d) [tex]Cov(X + Z,Y)=E(x+z-M_{x+z} )(Z-M_z)\\=E{(x-M_x)+(Y-M_y)}(Z-M_z)\\\\=E{(x-M_x)}(Z-M_z)+E{+(Y-M_y)}(Z-M_z)= cov (x,y)+cov (z,y)[/tex]f) Var(x+y) = [tex]E{(x-M_x)+(Y-M_y)}^2[/tex]Expanding inside we get=[tex]Var(x)+var(y)+2 cov (x,y)[/tex]