Prove that though covariance is independent of the choice of origin, it depends upon the scale. If u=ax+b, v=cy+d, show that cov(u,v)=a.c. cov(x,y).

Dear student

Cov(x,y) = E((x-m(x))(y-m(y))), where m() is the mean.

Let u = ax + b and v = cy + d

So, Cov(u, v) = E((u-M(u))(v-M(v)))

= E((ax + b - (am(x) + b))(cy + d - (cm(y) + d))

=E((ax - am(x))(cy - cm(y)))

=acE(x - m(x))(y - m(y)))

=acCov(x,y)

Regards