# Proving covariance using (Pearson) correlation coefficient

Could anyone guide or show me how to prove the covariance

$$2\cdot Cov\left(\frac{X}{\sigma_X},\frac{Y}{\sigma_Y}\right)=\frac{2}{\sigma_X\sigma_Y}Cov(X,Y)$$

with

$$Cov(X,Y) = E(X − E(X))(Y − E(Y ))$$

• Is this self-study? Please read the tag description and add it if you think it fits, otherwise ignore. Nov 25, 2016 at 10:39
• The result is immediate from the explanation of covariance in terms of signed areas of rectangles, because the divisions by $\sigma_X$ and $\sigma_Y$ obviously factor through the calculation of area.
– whuber
Nov 25, 2016 at 17:32

Knowing that: $$E[X\cdot c] = c\cdot E[X]$$ Where c is constant. $$\text{Cov}\left(\frac{X}{\sigma_X},\frac{Y}{\sigma_Y}\right)= E\left[ \left( \frac{X}{\sigma_X}−E\left[\frac{X}{\sigma_X}\right] \right) \left( \frac{Y}{\sigma_Y}−E\left[\frac{Y}{\sigma_Y}\right] \right) \right]= E\left[\frac{1}{\sigma_X}(X - E[X])\frac{1}{\sigma_Y}(Y-E[Y])\right] =\frac{1}{\sigma_X\sigma_Y}E[(X - E[X])(Y-E[Y])]=\frac{1}{\sigma_X\sigma_Y}\text{Cov}(X,Y)$$
Using the fact that the mean is a linear function (ie. $E(aX)=aE(X)$) several times, you get:
The first equality is just the definition. At the second equality we the use fact that the mean is linear $E(\frac{X}{\sigma_{X}})=E(\frac{1}{\sigma_{X}}X)=\frac{1}{\sigma_X}E(X)$. Now, factoring out $\frac{1}{\sigma_X}$ gives you the third equality. The fourth equality also follows from the fact that the mean is linear. The fifth equality is obviously true, and the sixth equality is just the definition again.
• Thank you, Jens! About the $\frac{1}{\sigma_X}$ and $\frac{1}{\sigma_Y}$ Why can they be taken out of the brackets since they are in two different brackets and multiplied with each other? I hope you understand my question. Nov 25, 2016 at 9:46