I have a data set where some x input has a 0.59 Pearson correlation with variable y (sample size is about 300). After performing simple linear regression I get the following standardized residuals scatter plot and qq plot:

Residuals scatter plot

Residuals QQ plot

Regression result and scatter plot relating x to y:

Regression result and scatter plot relating x to y

The residuals scatter plot seems fairly random with no discernible pattern to me, but the QQ plot shows a slight curvature and Shapiro-Wilk test rejects the null hypothesis so I assumed there might be some non linear relationship between x and y. I tried then applying a few transformations like sqrt(x), log(x) or x^2 to the input vector and to my surprise the Pearson correlation barely changes.

With x^2 it goes down to 0.58, for log(x) and sqrt(x) the correlation increases by about 0.002 and 0.003 compared to just x. Performing regression on the transformed variable also yields very similar results to the non transformed variable and the QQ plot still shows the same slight curvature.

At first I thought there was a mistake in my code, but I checked it several times and I'm fairly confident that's not the case, so now I'm wondering what kind of conclusions can I take away from this. Maybe y is related to x through a combination of linear and non linear terms? Maybe some other unknown variable that is collinear to x but has non linear relation to y is the cause of the curvature in the QQ plot? Maybe there's some other variable transformation I haven't tried that could explain it? What other method could I use to further explore the relationship between the two variables?

  • $\begingroup$ You have one predictor x and one outcome y. Have you looked at the plot of y against x. It will be esp. interesting to see this plot since the question is whether there is a non-linear relationship between x and y. $\endgroup$
    – dipetkov
    May 13 at 16:44
  • $\begingroup$ I edited the post to include the scatter plot for x and y as well as the line obtained by the regression. I have more variables but they mostly have either significantly lower correlation to y or high collinearity to x, so I've tried PCR and PLS as alternative methods and they yielded slight improvements but not to a huge degree (r^2 went from 0.43 to 0.47), while also showing the same pattern on the QQ plot. $\endgroup$
    – Janilson
    May 13 at 17:01
  • $\begingroup$ What are the actual scales/ranges of y and x? x must be strictly positive as you tried a log transform. $\endgroup$
    – EdM
    May 13 at 17:04
  • 2
    $\begingroup$ Thanks for adding the plot. There seems to be an influential data point. You can see it to the right of the plot, perfectly interpolated by the regression line. If you exclude this one point, you might get better results without any transformations. Visually, there is nothing nonlinear. $\endgroup$
    – dipetkov
    May 13 at 17:04
  • 1
    $\begingroup$ I understand that your analysis might be more complex. But for the step that you show in this question, you end up trying complex transformation of x just because of one influential point. It biases the regression, judging from the scatterplot of y against x. $\endgroup$
    – dipetkov
    May 13 at 17:26

1 Answer 1


Don't be surprised that Pearson correlation coefficients aren't greatly affected by log or square-root or square or similar simple monotonic transformations of $X$ in your case. The rank orders aren't affected by the transformation, meaning that non-parametric correlations aren't affected. In terms of the Pearson correlation:

$$\rho_{X,Y}=\frac{\operatorname{E}[(X-\mu_X)(Y-\mu_Y)]}{\sigma_X\sigma_Y} $$

when $X$ is transformed and $Y$ isn't, it's a matter of how much the transformation of $X$ moves values above and below the mean in the transformed scale, relative to the corresponding change in $\sigma_X$. The net effect can be very small. The sample Pearson correlation is biased and not robust, further complicating attempts at intuition in practice.

Consider the following bivariate normal data in R; you need to have the mvtnorm package available:

norm2 <- mvtnorm::rmvnorm(300,mean=c(1,1),sigma=matrix(c(.1,-.059,-.059,.1),byrow=TRUE,nrow=2))
norm2 <- data.frame(norm2)
names(norm2) <- c("x","y")
norm2$y <- 35 * norm2$y ## get roughly into reported ranges of values
plot(y~x,data=norm2) ## not shown, similar to cloud in the plot from the question, no outlier
# [1] -0.5559027
# [1] -0.5400946
# [1] -0.5426008
# [1] -0.5546897

If you don't have a solid theoretical reason for a particular transformation, it's often good to let the data suggest the functional form of the relationship by modeling the continuous predictor variable with a regression spline.

Also, see the discussion on: Is normality testing 'essentially useless'? I would worry more about the potential high-leverage point noted by @dipetkov, but you need to apply your knowledge of the subject matter.

  • $\begingroup$ Could you elaborate why it's a matter of how the transformation affects values above vs below the mean? That's not super intuitive to me. $\endgroup$ May 13 at 20:59
  • $\begingroup$ @thomaskeefe that's a poorly worded version of Wikipedia's statement on interpretation of a correlation coefficient: "the correlation coefficient is positive if $X_i$ and $Y_i$ tend to be simultaneously greater than, or simultaneously less than, their respective means. The correlation coefficient is negative (anti-correlation) if $X_i$ and $Y_i$ tend to lie on opposite sides of their respective means. Moreover, the stronger is either tendency, the larger is the absolute value of the correlation coefficient." $\endgroup$
    – EdM
    May 14 at 13:40
  • $\begingroup$ @thomaskeefe so, crudely, if a transformation restricted to $X$ tends to put more $X_i$ values on the "correct" side of their transformed mean, with respect to corresponding $Y_i$ values, then it can increase the absolute value of the correlation coefficient. That needs to be balanced off against any corresponding increase in the transformed $\sigma_X$ in the denominator of the formula. $\endgroup$
    – EdM
    May 14 at 14:02

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