The above seems intuitively true, (where $|a|$ refers to the absolute value of $a$), but I'm struggling to prove it - would be very grateful for either a proof or a reference.


2 Answers 2


Unfortunately, it's not true, so you won't get far trying to prove it.

Counterexamples are easy to come up with.

Imagine $X\sim U(-1,1)$ and $Y=X+1$

$\text{Cov}(X,Y) = \text{Var}(X)=1/3$

Note that $|Y|=Y$

So $\text{Cov}(|X|,|Y|)= \text{Cov}(|X|,Y)$

but by the symmetry of $X$ about $0$, and hence the fact that the relationship between $|X|$ and $Y-E(Y)$ is an even function, we can see that $\text{Cov}(|X|,Y) = 0$.

So as stated, it isn't so; let's generate some data which illustrate the general sense of what I'm getting at there:

enter image description here

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    $\begingroup$ As much as it makes me feel a bit silly this is a great answer! Thanks!! $\endgroup$ Jun 20, 2015 at 10:42
  • $\begingroup$ You desperately need to improve your intuition in probability. Per my answer in stats.stackexchange.com/questions/33776/… , I recommend you buy and read at least the first 6 chapters (first 218 pages) of William J. Feller "An Introduction to Probability Theory and Its Applications, Vol. 2" amazon.com/dp/0471257095/ref=rdr_ext_tmb . At least read all of the Problems for Solution, and preferably try solving as many as you can. You don't need to have read Vol 1, which in my opinion is not particularly meritorious. $\endgroup$ Jun 20, 2015 at 14:07
  • $\begingroup$ Certainly not my finest hour! Thanks for the reference, looks like a read will do me good - I'll take a look. $\endgroup$ Jun 20, 2015 at 19:51
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    $\begingroup$ You should not feel bad about posting what is a fine question for the site. It's the sort of thing many people would share your intuition about, and it's useful to know some counterexamples as they can help generate better intuition. Untrained intuition about these things is quite rare, and even experienced practitioners can get surprises; indeed, basic texts in stats often have surprising errors about things like skewness or the central limit theorem. That said, the suggested reference is a good one. Counterexamples in Probability And Statistics, Romano and Siegel (1986) may help later on $\endgroup$
    – Glen_b
    Jun 20, 2015 at 22:49
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    $\begingroup$ I like to collect neat counterexamples (and also to generate my own); not only does it help clarify where our intuitions fail, collecting them can serve as a whispered Memento homo in ones ear-- a useful reminder when we apply intuition not to trust the ideas it leads us to. We all have our intuitions, but it helps to try to break the ideas we get (find counterexamples to them) before we try to prove them. Practice at finding counterexamples also seems to generate a degree of intuition. $\endgroup$
    – Glen_b
    Jun 21, 2015 at 0:20

A somewhat more extreme example, let $a \equiv b$ be distributed as $$\mathbb{P}(a = 1) = \mathbb{P}(a = -1) = 1/2.$$

Then $\text{Cov}(a, b) = \text{Cov}(a, a) = \text{var}(a) = 1$, while, $\text{Cov}(\left|a\right|, \left|b\right|) = \text{Cov}(1, 1) = 0$. Hence the conjecture is not true.


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