4
$\begingroup$

I recently received the following claim from a work colleague with regard to a loose definition of the correlation between two random variables:

“A 60% correlation between two variables approximately implies that in 60% of cases the movement away from the variables' mean position is aligned.”

I have not seen correlations defined in such a way, but I was unable to provide an answer as to a) whether this is correct, b) the alternative official definition.

$\endgroup$
4
  • $\begingroup$ Are these two random variables continuous or discrete? Are they normally or approximately normally distributed? $\endgroup$ Jun 5 '14 at 17:49
  • 4
    $\begingroup$ With a correlation of zero, what would your colleague claim? That in about 0% of cases the movements would be aligned? That's hardly plausible, is it? $\endgroup$
    – whuber
    Jun 5 '14 at 17:57
  • $\begingroup$ What does aligned movement mean here? What's the interpretation of negative correlations? $\endgroup$
    – Nick Cox
    Jun 5 '14 at 18:28
  • $\begingroup$ Leaving aside the merits (or rather the lack of them) in the intended description, even the metric itself is problematic -- tt doesn't really make sense to describe correlation in percentage terms. Correlations are between -1 and 1. With squared correlation, it does, because we can partition the variance in the data into two independent components, that explained by the other variable, and the residual, and discuss that as a percentage of the total variation. In this case trying to do it leads to the bizarre assertion than if the correlation was -0.2, something happens in -20% of cases! $\endgroup$
    – Glen_b
    Jun 6 '14 at 1:32
10
$\begingroup$

There's some merit to the idea, but it's not quantitatively correct.

The standard (Pearson) definition uses the expected product of the standardized values of $X$ and $Y$ to measure the correlation of $(X,Y)$. Specifically, recenter both $X$ and $Y$ (so that we're always discussing how much they differ from their respective means) and let their standard deviations be $\sigma$ and $\tau$, respectively. Then when $F$ is the bivariate distribution of $(X,Y)$,

$$\rho_F = \mathbb{E}_F\left(\frac{X}{\sigma}\frac{Y}{\tau}\right) = \mathbb{E}_F\left(r\left(\frac{X}{\sigma},\frac{Y}{\tau}\right)\right)$$

for the function $r(u,v)=uv.$ This function $r$ is positive when $u$ and $v$ have the same sign (are "aligned") and is negative otherwise. It grows in direct proportion to the values of $u$ and $v$. Thus its average value (the expectation) reflects a value-weighted balance of aligned and unaligned variations from the central point.

We could make $r$ less sensitive to the values of $u$ and $v$. A fairly extreme way to do this would be to give it a unit value when $u$ and $v$ are aligned and negate it when they are not; that is,

$$r^\prime(u,v) = \text{sgn}(u)\text{sgn}(v).$$

One could use this in place of $r$ in the correlation definition to create a robust measure of correlation of empirical distributions. In so doing, we would have the interpretation suggested in the question: the expectation of $r^\prime(X/\sigma,Y/\tau)$ is the chance that $X$ and $Y$ are "aligned" minus the chance they are not aligned. (In practice, the variables would be recentered using robust estimates such as their medians rather than their means and their standard deviations would be replaced by robust estimates of variation such as interquartile ranges or MADs.)

Yes, we could even choose $r^{\prime\prime}$ to be the indicator that $X$ and $Y$ are aligned, so that $\rho_F^{\prime\prime}$ really would be the chance of alignment. In so doing, however, we could only achieve values between $0$ and $1$: that can scarcely be interpreted like the Pearson coefficient, which ranges from $1$ down to $-1$. Since

$$r^{\prime\prime}(u,v) = \frac{1}{2} + \frac{r^\prime(u,v)}{2},$$

there is a clear, simple relationship between the characterization in the question (which appears to refer to $r^{\prime\prime}$) and the more familiar-looking correlation values attained by $r^\prime$. I will therefore continue to discuss $r^\prime$.

To appreciate the distinction between $\rho$ and $\rho^\prime$, consider the archetypical application of correlation: the bivariate normal distribution. When this particular distribution has Pearson correlation $\rho_F$, the alternative measure of correlation is

$$\rho^\prime_F = \frac{1}{2} + \frac{1}{\pi} \arctan\left(\frac{\rho_F}{\sqrt{1-\rho_F^2}}\right).$$

For values of $\rho$ within the range $(-0.6,0.6)$ or so, this function is approximately linear:

$$\rho^\prime_F \approx \frac{1}{2} + \frac{\rho_F}{\pi}.$$

For instance, with $\rho=0.6,$ $\rho^\prime = 0.705.$ This clearly differs (a lot) from the formula quoted in the question, which asserts $\rho^\prime_F \approx \rho_F$. This will only be the case for values of $\rho_F$ close to $\frac{\pi}{2(\pi-1)}\approx 0.733.$

Figure

The solid line graphs $\rho^\prime$ in terms of $\rho$ for bivariate Normal distributions. The dashed line is the linear approximation to $\rho^\prime$ around $\rho=0$. It has slope $\frac{1}{\pi}\approx 0.318$.

This one-to-one correspondence between $\rho$ and $\rho^\prime$ for the bivariate Normal distribution shows we could use either definition equally well for describing correlations where a Normal model applies--but I suspect the sampling distribution of $\rho^\prime$ might be a little more difficult to derive. There is not necessarily such a one-to-one correspondence between $\rho$ and $\rho^\prime$ among all bivariate distributions, though: for a given value of $\rho^\prime$ it is clear we could vary $\rho$ quite a bit by pulling a tiny bit of probability mass around in the $(X,Y)$ plane at large distances from the origin without changing $\rho^\prime$ at all.

$\endgroup$
-1
$\begingroup$

Your colleague is only correct if the multivariate distribution only has two moments (mean,variance). Since each case is weighted both by its distance from the mean, outlier pairs can raise the correlation without changing the most frequent comovement. This is due to using squared residuals as the measure of covariance.

$\endgroup$
1
  • 1
    $\begingroup$ Your argument implies the colleague is not correct. BTW, squared residuals are not used to measure covariance: only the residuals themselves are. $\endgroup$
    – whuber
    Jun 5 '14 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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