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If I take $X$ to be a degenerate random variable, i.e. $X=1$ WP1 and $Y=X$ defined over the singleton sample space $\Omega=\{1\}$. Then

$$\mathbb{P}(X=1|Y=1)=1=\mathbb{P}(X=1)$$

i.e. I'd assume they're independent. But, we have

$$\rho_{XY}=\frac{\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)}{\sigma_X\sigma_Y}=\frac{0}{0}$$

which is undefined, not zero. Where am I being imprecise here/what's the misunderstanding? Thanks!

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    $\begingroup$ Interesting question and a +1 from me. Notice that you get into similar trouble if you try to calculate the correlation between independent Cauchy-distributed random variables (not $0/0$, but a similar situation of not even being able to define the division). $\endgroup$
    – Dave
    Commented Mar 4, 2022 at 1:54
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    $\begingroup$ You have not misunderstood. If either of the centered second moments are zero, then the Pearson correlation coefficient is indefinite. Now consider again if you had computed the reflective correlation coefficient instead. $\endgroup$
    – Galen
    Commented Mar 4, 2022 at 3:10

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Independence implies the correlation can't be any number other than zero, but as this example and @Dave's example of independent Cauchy variables shows, the correlation might not be a number at all. So the usual statement is slightly inexact.

The point-mass example is worse, in a way. In the Cauchy example it's still true that the correlation is zero for, eg, all bounded functions $f(X)$, $g(Y)$ whereas in the point-mass example there's no way to get a well-defined correlation.

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    $\begingroup$ The second moments have to exist to compute a Pearson product-moment correlation coefficient (PPMCC). But for Cauchy variables they do not exist, so there is not a defined PPMCC between them. $\endgroup$
    – Galen
    Commented Mar 4, 2022 at 3:04
  • $\begingroup$ @DifferentialPleiometry That's why Thomas specified bounded functions: this ensures the correlations do exist, provided we understand that almost surely constant functions are excluded from that statement. $\endgroup$
    – whuber ♦
    Commented Mar 4, 2022 at 14:40
  • $\begingroup$ So Pearson's correlation implicitly requires the variance of the random variables to be bounded and nonzero $\endgroup$
    – pSrIoGcNeAsLs - bye stackGPT
    Commented Mar 4, 2022 at 22:17

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