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lapo4ka [179]
3 years ago
9

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.
Mathematics
1 answer:
____ [38]3 years ago
5 0

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.

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