0
$\begingroup$

I have a dataset where I test for correlation and then if the correlation is significant. And basically my results are A correlates with B significantly and B correlates with C significantly, but A doesnt correlate significantly with C.

It is kind of unintuitive for me, I would have thought that A would also significantly correlate with C. Can anyone explain me why this is perfectly reasonable? I read about this question: If A and B are correlated with C, why are A and B not necessarily correlated?

But I dont think thats the exact same scenario here.

$\endgroup$
3
  • $\begingroup$ Also see How to infer correlations from correlations $\endgroup$
    – Glen_b
    Nov 19, 2019 at 11:46
  • $\begingroup$ @Baran The only difference from the question you link seems to be the addition of "significant". But since significant correlations can be small, and (as you should be able to see from whuber's answer at that linked one) the correlation between A and C could then be much smaller in size - even exactly zero, where is there any mystery left? $\endgroup$
    – Glen_b
    Nov 19, 2019 at 11:46
  • $\begingroup$ No you are right Glen. The question is actually the same. I probably had a brain dropout. $\endgroup$ Nov 20, 2019 at 8:12

2 Answers 2

2
$\begingroup$

Let $U, V$ denote uncorrelated random variables with variance one. Suppose that \begin{align*} B &= U + V \\ A & = U \\ C &= V \end{align*} $A$ and $C$ are uncorrelated by definition. However, \begin{align*} cov(A, B) = var(U) = 1 = var(V) = cov(B, C). \end{align*} This shows that $A, B$ and $B,C$ are pairwise correlated.

$\endgroup$
2
$\begingroup$

Happens all the time. Imagine a model:

  • $A \sim N(0,1)$,
  • $C \sim N(0,1)$,
  • $E \sim N(0,1)$,
  • $A$, $C$ and $E$ are independent (and, therefore, uncorrelated),
  • $B = 0.1 * A + 0.2 * C + \sqrt{1 - 0.1^2 - 0.2^2} * E$.

By construction,

  • $B \sim N(0,1)$,
  • ${\rm corr}(A,B) = 0.1$,
  • ${\rm corr}(B,C) = 0.2$.
$\endgroup$

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