# Is the covariance of standardized variables the correlation?

I have a basic question. Say I have two random variables, $X$ and $Y$. I can standardize them by subtracting the mean and dividing by the standard deviation, i.e., $X_{standardized} = \frac{(X - E(X))}{(SD(X))}$.

Is the correlation of $X$ and $Y$, $Cor(X, Y)$, the same as the covariance of the standardized versions of $X$ and $Y$? That is, is $Cor(X, Y) = Cov(X_{standardized}, Y_{standardized})$?

• Yes. ${}{}{}{}{}$ – Dilip Sarwate Apr 26 '15 at 3:34

\begin{align} \operatorname{corr}(X,Y)&=\frac{E\Big((X-E(X))\times(Y-E(Y))\Big)}{SD(X)\times SD(Y)}\\ \operatorname{Cov}(X_{\text{standardized}}, Y_{\text{standardized}}) &=E\Bigg[\Bigg(\frac{(X - E(X))}{(SD(X))}-0\Bigg)\times\Bigg(\frac{(Y - E(Y))}{(SD(Y))}-0\Bigg)\Bigg]\\ &= \frac{E\Big((X-E(X))\times(Y-E(Y))\Big)}{SD(X)\times SD(Y)} \end{align}So, Yes!
• Errr no. The question is about correlation and covariance of random variables whereas your answer is about sample correlation and covariance. For example, the result asked about holds for continuous random variables whereas at best what you have applies only to discrete random variables taking on values $(X_1,Y_1), \ldots, (X_n,Y_n)$ with equal probability $\frac 1n$. – Dilip Sarwate Apr 26 '15 at 14:35
• Not quite. You don't need the subscripts $i$ at all, so I have gone ahead and deleted them, and improved the presentation a little bit. Feel free to roll back if you don't like the changes. – Dilip Sarwate Apr 26 '15 at 16:00