13
$\begingroup$

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

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$
    – Dilip Sarwate
    Apr 26, 2015 at 3:34

1 Answer 1

Reset to default
12
$\begingroup$

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

Improve this answer
$\endgroup$
7
  • 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, 2015 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, 2015 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, 2015 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, 2018 at 11:34
  • 1
    $\begingroup$ @Erdogan constants can be taken outside of Expected() function without change. $\endgroup$
    – Hemant Rupani
    Jul 2, 2018 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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