Correlation and Covariance on Standardized X

Question

I am stuck on the following problem:

Let $Z_X$ be the standardized $X$, $Z_X=(X-\mu_X)/\sigma_X$, and let $Z_Y$ be the standardized $Y$, $Z_Y=(Y-\mu_Y)/\sigma_Y$. Show that $Corr(X,Y)=Cov(Z_X,Z_Y)=E(Z_XZ_Y)$.

I have tried just plugging in the definition into what we are trying to prove but I don't know how $ Corr(X,Y)=Cov(\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y})=E(\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y}) $ helps. I think I have to use the facts that $Cov(aX+b,cY+d)=acCov(X,Y)$ and $Corr(aX+b,cY+d)=Corr(X,Y)$ when $ a $ and $ c $ the same sign, but I am not sure where exactly. Any suggestions?

Answer 1

From the formula for covariance, we have $$\operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$ Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so $$\operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$

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Using the facts mentioned at the end of your question, we have $$\operatorname{Cov}(Z_X, Z_Y) = \operatorname{Cov}(\frac{X-\mu_X}{\sigma_X}, \frac{Y-\mu_Y}{\sigma_Y}) = \frac{\operatorname{Cov}(X,Y)}{\sigma_X \sigma_Y} = \operatorname{Corr}(X,Y).$$