1
$\begingroup$

I'm reading this paper and on page 5, it states that the covariance between two elements is positive ($\sigma_{AB} \geq 0$). Why is this true? If $<a_ib_i>$ is just the average over the dot product, can't that be negative?

$\endgroup$
4
  • 1
    $\begingroup$ Certainly covariance in general can be negative. I am working with a very poor screen, cannot find an assertion on page 5 that in this case it is non-negative. $\endgroup$ Apr 19, 2013 at 18:57
  • $\begingroup$ I agree with André Nicolas that there seems to be no assertion that the covariance is nonnegative. But notice that the paper uses the dreadful choice of notation $\sigma_{AB}^2$ for the covariance (presumably by analogy with $\sigma_A^2$ for the variance) which would seem to imply that the covariance is always nonnegative (the paper itself does not make this inference). $\endgroup$ Apr 19, 2013 at 19:16
  • 1
    $\begingroup$ How does $\sigma^2_{AB} \geq 0$ make $\sigma_{AB}$ nonnegative? Or better yet, what is $\sigma_{AB}$? I think I'm just getting confused over the notation... $\endgroup$
    – victor
    Apr 19, 2013 at 19:19
  • $\begingroup$ Also, the author later goes on to say "By definition covariances must be non-negative, thus the minimal covariance is zero" $\endgroup$
    – victor
    Apr 19, 2013 at 19:23

1 Answer 1

2
$\begingroup$

This tutorial on Principal Component Analysis written by Shlens is great! I learned a lot from it. However, those descriptions about covariance $\sigma^2_{AB} \geq 0$, including line 13 on the left side and line 28 on the right side of page 5, were in the previous version 2 of the paper (2005) and were misunderstandingly incorrect. The covariance of $A$ and $B$ equals to $\sigma^2_{AB} = \rho \sigma_A \sigma_B$, where $\rho$ is the correlation coefficient of $A$ and $B$ with $-1 \le \rho \le 1$, $\sigma^2_A$ and $\sigma^2_B$ is the variance of $A$ and $B$, respectively. From the above we know that covariance could be negative.

The above incorrect descriptions were revised on the latest version of the paper; please refer to Shlens's revised paper on 2014.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .