Question

G with the definition of covariance, prove cov[(ay − b),(cy − d)] = accov(x, y ), where x, y are random variables and a, b, c, d are constants.

Answer 1

By definition of covariance,

`\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]`

`\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]`

We have

`\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]`

`=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd`

`\mathbb E[aX-b]=a\mathbb E[X]-b`

`\mathbb E[cY-d]=c\mathbb E[Y]-d`

`\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd`

Putting everything together, we find the covariance reduces to

`\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)`

as desired.