9
$\begingroup$

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 $Cor(X, Y) = Cov(X_{standardized}, Y_{standardized})$?

$\endgroup$
  • 1
    $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – Dilip Sarwate Apr 26 '15 at 3:34
10
$\begingroup$

$$\begin{align} \operatorname{corr}(X,Y)&=\frac{E\Big((X-E(X))\times(Y-E(Y))\Big)}{SD(X)\times SD(Y)}\\ \operatorname{Cov}(X_{\text{standardized}}, Y_{\text{standardized}}) &=E\Bigg[\Bigg(\frac{(X - E(X))}{(SD(X))}-0\Bigg)\times\Bigg(\frac{(Y - E(Y))}{(SD(Y))}-0\Bigg)\Bigg]\\ &= \frac{E\Big((X-E(X))\times(Y-E(Y))\Big)}{SD(X)\times SD(Y)} \end{align}$$So, Yes!

$\endgroup$
  • 1
    $\begingroup$ What???? The right side of your first equation is a random variable while the left side is a constant. $\endgroup$ – Dilip Sarwate Apr 26 '15 at 14:22
  • 2
    $\begingroup$ 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 $\frac 1n$. $\endgroup$ – Dilip Sarwate Apr 26 '15 at 14:35
  • 2
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Apr 26 '15 at 16:00
  • 1
    $\begingroup$ You are taking SD(X) and SD(Y) out of expectation. Explain the reasoning of this step a little bit more please. $\endgroup$ – Erdogan CEVHER Jun 27 '18 at 11:34
  • 1
    $\begingroup$ @Erdogan constants can be taken outside of Expected() function without change. $\endgroup$ – Hemant Rupani Jul 2 '18 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.