Assume that we have two variables, $a_{i}$ and $b_{i},$measured across individuals $i.$ They have a strong, positive correlation, say $corr(a_{i},b_{i})\approx0.9$. Now, there is a another variable, $c_{i}$ , which varies across individuals. Note also that $$ 0<a_{i}\leq b_{i}\leq c_{i} $$

We compute, then, for each individual: $$ A_{i}=\frac{a_{i}}{c_{i}} $$ and $$ B_{i}=\frac{b_{i}}{c_{i}} $$

We notice that $corr(A_{i},B_{i})$ becomes strongly negative! This is a bit perplexing, but the only reasonable explanation I can come up with is this- across individuals, when $b_{i}$ increases, although $a_{i}$ increases, $c_{i}$ increases by more than $a_{i}$ increases on average, rendering this correlation negative. Is this reasoning correct? Is there perhaps a formal condition linking $a,b$ and $c$ which can capture the essence of this reverse in correlation?

  • $\begingroup$ The nature of the answers may vary depending on what you assume about the signs of these variables. The context hints that all of them are non-negative: is this the case or not? $\endgroup$
    – whuber
    Jun 28, 2021 at 14:35
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    $\begingroup$ Yes indeed, they are! Well observed.. $\endgroup$ Jun 28, 2021 at 14:47
  • $\begingroup$ I have included the requisite information in the body of the question. $\endgroup$ Jun 28, 2021 at 14:50

1 Answer 1


This can happen when the $c_i$ are positively correlated with $a_i+b_i.$ The geometry of the $(a_i,b_i)$ scatterplot reveals that such behavior should not be surprising.

Without any loss of generality we may scale all the $a_i$ by some positive value, all the $b_i$ by some other positive value, and all the $c_i$ by yet a third positive value, without changing any of the conditions.

Thus, it's no restriction to imagine the $(a_i,b_i)$ scatterplot takes on a classic "cigar" shape, with its axis oriented around 45 degrees.

The rescaling by the $c_i$ moves each point to another one on the line connecting it to the origin. Thus, we need to envision a situation in which the points in the original cigar can be moved in this way to fall along a cigar-shaped cloud that is oriented negatively. This is easily done: pick any nonzero number $\mu$ and for each point $i$ set

$$c_i \approx (a_i + b_i)/\mu.$$

The projected point

$$\left(\frac{a_i}{c_i},\frac{b_i}{c_i}\right) = \mu\left(\frac{a_i}{a_i+b_i},\frac{b_i}{a_i+b_i}\right)$$

will lie close to the line $a_i+b_i=\mu,$ which has negative slope. Thus, you can achieve any negative correlation, even $-1.0$ exactly, by varying the closeness of the approximation. The value $\mu\sqrt{2}$ will be the typical distance between the new points and the line $b = -a.$

Here is an example to illustrate. The $n=30$ original points with correlation coefficient $0.9$ are plotted as open circles. Arrows project them to new solid red points with correlation coefficient $-0.8.$ Those points all lie near the line $a+b \approx 5.57$ (the median of all the $a$ and $b$ coordinates).


  • $\begingroup$ What an answer!! I'll be sure to go over it carefully. You, sir, should write a textbook on Statistics. $\endgroup$ Jun 28, 2021 at 16:12
  • $\begingroup$ Just to clarify- a silly follow up I had- given that a_i and b_i are both scaled by the same amount, c_i, why does c_i not cancel out on both sides? I would imagine it would in the case where a_i and b_I are perfectly correlated... $\endgroup$ Jun 28, 2021 at 17:04
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    $\begingroup$ It does cancel: the slope of the line of projection (that is, of any arrow) is $b_i/a_i = (b_i/c_i)/(a_i/c_i).$ $\endgroup$
    – whuber
    Jun 28, 2021 at 17:10
  • $\begingroup$ Indeed- however, it seems it is not true when there is a constant. I am posting another question to that effect, and would appreciate your insight! $\endgroup$ Jun 28, 2021 at 18:55
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    $\begingroup$ A basic property of correlation is that constant scaling of either variable doesn't change the correlation: see the initial remarks in this answer. $\endgroup$
    – whuber
    Jun 28, 2021 at 19:37

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