I'm trying to intuitively understand what it means for two random variables to have non-zero covariance.

At the moment, I imagine that the two random variables (which have non-zero covariance) both depend (at least in part) on a common random variable, and this is why they have non-zero covariance.

E.g., if $U$, $V$, $Z$ and $W$ are independent random variables, and $X_1 = \frac{U + V}{Z}$ and $X_2 = W + Z$, then since both $X_1$ and $X_2$ depend on $Z$, the two variables will have non-zero covariance.

Is this a good way of intuitively thinking about covariance? If not, is there a better way which might help?


Here is my intuition on covariance:

  • The covariance of $X$ and $Y$ is positive, if the two have a monotonous correlation. This means that high values of $X$ imply high values of $Y$ and vice versa.
  • It is negative, if they have a negative monotonous correlation. This means that high values of $X$ imply low values of $Y$ and vice versa.
  • If the covariance is zero, then there exists no monotonous correlation between $X$ and $Y$. Other correlations are still possible though.

The covariance provides the direction of a correlation between two random variables, but it does not provide the strength of it. To make the correlation comparable it has to be normalized (compare for example Pearson correlation coefficient).

  • $\begingroup$ Thanks for your reply. I should have been clearer in my question - I understand what covariance is, what I don't understand is what gives random variables a non-zero covariance in the first place. $\endgroup$ – Jack Nov 3 '17 at 12:23
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    $\begingroup$ Your intuition is correct. It is correct that the covariance is only nonzero if the two random variables are not independent, therefore they depend on a common random variable. But one could intuitively think that therefore if the covariance is zero, then the random variables are independent. That is not the case, that's why I provided this answer. Sorry if my point was not clear. $\endgroup$ – RoyPJ Nov 3 '17 at 12:34
  • $\begingroup$ Ah I see. Thanks RoyPJ - this is helpful. $\endgroup$ – Jack Nov 3 '17 at 12:36

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