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I know the from the basic rule of the covariance we have: $$\text{Cov(aX,Y)=aCov(X,Y)}$$ however now i'm looking at a case that is creating me some doubt: Looking at the covariance of the same random variable:

$1)$ $\text{Cov(aX,X)=aCov(X,X)=aVar(X)}$

$2)$ $\text{Cov(aX,X)=Var(aX)=}a^2\text{Var(X)}$

which one is the correct solution?

Thank you in advance

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  • $\begingroup$ The first one is correct. $\endgroup$ Commented Feb 20, 2020 at 15:24
  • $\begingroup$ The first is correct. The second is wrong. We do not have in general $\mathsf{Cov}(aX,X)=\mathsf{Var}(aX)$ but $\mathsf{Cov}(aX,aX)=\mathsf{Var}(aX)=a^2\mathsf{Var}(X)$. $\endgroup$
    – drhab
    Commented Feb 20, 2020 at 15:25

1 Answer 1

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The first is correct.

The second is not: By definition $\operatorname{Cov}(X,X)=\operatorname{Var}(X)$.

So $$\operatorname{Var}(aX)= \operatorname{Cov}(aX,aX)= a\operatorname{Cov}(X,aX)= a^2\operatorname{Cov}(X,X)=a^2 \operatorname{Var}(X)$$

So the last $=$ of 2 is correct, the first is not.

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