Proving covariance using (Pearson) correlation coefficient

Question

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

$$Cov(X,Y) = E(X − E(X))(Y − E(Y ))$$

Answer 1

Knowing that: $$ E[X\cdot c] = c\cdot E[X] $$ Where c is constant. $$ \text{Cov}\left(\frac{X}{\sigma_X},\frac{Y}{\sigma_Y}\right)= E\left[ \left( \frac{X}{\sigma_X}−E\left[\frac{X}{\sigma_X}\right] \right) \left( \frac{Y}{\sigma_Y}−E\left[\frac{Y}{\sigma_Y}\right] \right) \right]= E\left[\frac{1}{\sigma_X}(X - E[X])\frac{1}{\sigma_Y}(Y-E[Y])\right] =\frac{1}{\sigma_X\sigma_Y}E[(X - E[X])(Y-E[Y])]=\frac{1}{\sigma_X\sigma_Y}\text{Cov}(X,Y) $$

Answer 2

Using the fact that the mean is a linear function (ie. $E(aX)=aE(X)$) several times, you get:

\begin{eqnarray} Cov\left(\frac{X}{\sigma_{X}},\frac{Y}{\sigma_{Y}}\right) &=& E\left(\frac{X}{\sigma_{X}} - E\left(\frac{X}{\sigma_{X}}\right)\right)E\left(\frac{Y}{\sigma_{Y}} - E\left(\frac{Y}{\sigma_{Y}}\right)\right) \\ &=&E\left(\frac{1}{\sigma_X}X-\frac{1}{\sigma_X}E(X)\right)E\left(\frac{1}{\sigma_Y}Y-\frac{1}{\sigma_Y}E(Y)\right) \\ &=&E\left(\frac{1}{\sigma_{X}}(X-E(X))\right)E\left(\frac{1}{\sigma_{Y}}(Y-E(Y))\right) \\ &=& \frac{1}{\sigma_{X}}E((X-E(X))\frac{1}{\sigma_{Y}}E((Y-E(Y))\\ &=&\frac{1}{\sigma_{X}\sigma_{Y}}E(X-E(X))E(Y-E(Y)) \\ &=&\frac{1}{\sigma_{X}\sigma_{Y}}Cov(X,Y). \end{eqnarray}

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.

Multiplying both sides by 2 gives you your answer