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I have a data set that includes positive and negative values and I performed a Pearson correlation on it, getting quite a high value: +0.77. Admittedly it is quite a small sample (n=50) on which to perform a correlation.

The analysis is on pairs of questionnaire items - each pair has one item that is fairly socially desirable or undesirable, and one corresponding item that is more neutral in social desirability (we intentionally neutralised this item). Each item-pair is considered one case, with two values associated with it: 1. difference between the two items in social desirability; 2. difference between the two items in self-rating. These differences can be positive or negative, depending on which item is more socially desirable/undesirable. Scatterplot of Social Desirability versus Self-Rating

My question is whether having values in the negative x negative space, in addition to the positive x positive space, artificially increases the Pearson correlation? It's such a high correlation and interesting result that I'm suspicious of it.

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Having negative values and/or positive values of the variables doesn't affect the Pearson correlation coefficient because adding a constant to a variable does not change the correlation of that variable with other variables.
If $X,Y$ are random variables and $a,b$ are constants then $$\rho(X+a,Y+b)=\rho(X,Y)$$

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  • $\begingroup$ Thanks Kourabi for your answer. That is also very helpful. $\endgroup$
    – Bruegel
    Aug 16, 2019 at 1:59
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No. You can translate the points anywhere on the plane (together!) and not change the correlation.

$$\rho_{XY} = \dfrac{cov(XY)}{\sigma_X\sigma_Y}$$

We know that changing the mean of a variable $Z$ will not change its standard deviation: $\sigma_Z = \sigma_{Z+k}$. That's why normalizing a variable divides by the original standard deviation after subtracting the mean.

Likewise, covariance will remain the same no matter how you shift the data around the plane.

The key is that, for both variance (so standard deviation) and covariance, the equation subtracts out the mean: $cov(X,Y) = \mathbb{E}[(X-\mu_X)(Y-\mu_Y)]$ and $var(X) = \mathbb{E}[(X-\mu_X)^2]$. If you shift $X$ or $Y$, you also shift $\mu_X$ or $\mu_Y$ by the same amount which you then subtract out from $X$ or $Y$.

Together, those mean that correlation is translation invariant, which it should be.

The scatterplot looks like it has correlation around 0.75: tight but not too tight. I find your result totally plausible.

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  • $\begingroup$ Brilliant, thanks Dave. Much appreciated! $\endgroup$
    – Bruegel
    Aug 16, 2019 at 1:59

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