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I would like to know if my interpretation is correct with respect to the Q-Q plot I made (see the first image below). The second image is a plot of a chi-square distribution.

enter image description here

enter image description here

At the first half, it seems more or less similar to what I would expect for a chi-square distribution. However, because my data (the dots) are a bit below the straight line, it means it doesn't increase as fast (so relative to the distribution of the 2nd image, my distribution is a little shifted to the right). More or less at the second half of the Q-Q plot, my data is above the line. This means it descends faster than I would expect for a chi-square distribution. So, my right tail isn't as thick as a regular chi-square distribution.

Is this interpretation correct?

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    $\begingroup$ Your right tail is heavier than for a chi-square (more right skew). The interpretation (in terms of relative skewness) is along the same lines as the discussion here $\endgroup$
    – Glen_b
    Jul 22 '18 at 11:34
  • $\begingroup$ Do you infer the right skewness from the Q-Q plot because my data (the dots) are more spread out in the right part of the Q-Q plot (i.e. D² is larger than expected relative to the straight line)? But it also has something to do with the number of 'dots' / observations that are above it, right? Thank you for posting the discussion, I have already seen your answer there and have already used it multiple times. But for some reason I keep having difficulty interpreting Q-Q plots $\endgroup$
    – Amonet
    Jul 22 '18 at 14:20
  • $\begingroup$ I would also be interested in your advice as to how this right skewness affects multivariate normality. As I understand, if the data were perfectly on the line, I would (not yet) have an indication that my data departs from a multivariate normal distribution, as the D² are chi-squared distributed. However, as you indicated: the right tail is heavier here, so there is evidence against multivariate normality. But to what extent is this departure extreme, or reason to really worry about multivariate normality? $\endgroup$
    – Amonet
    Jul 22 '18 at 15:01
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    $\begingroup$ That's a different question, but one we cannot address on present evidence. The sensitivity of whatever you're doing to whatever kind of departure from multivariate normality we have** cannot be determined from this. $\:$ **(which we cannot really guess at from the present information other than to know that it produces a long tail of Mahalanobis distance) $\quad$ To figure out how much each kind of deviation from MVN affects whatever you're trying to do would require some study (which may have been looked at for some cases or it may never have been studies). ... ctd $\endgroup$
    – Glen_b
    Jul 23 '18 at 1:19
  • $\begingroup$ ctd... It would require knowing a lot more about the circumstances and is beyond the extent of a comment on this question. $\endgroup$
    – Glen_b
    Jul 23 '18 at 1:21
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The interpretation is not correct.

Looking at the top diagram here we see:

plot showing denser and less dense QQ points compared against average or expected trend

Since your plot is the same way around as the QQ plots discussed there (with the theoretical quantiles on the x-axis), judging it is simply a matter of identifying corresponding patterns in your plot.

Now let us examine your plot. The pattern at the top right above corresponds to the region marked in blue at the bottom left below:

QQ plot from question, which shows a convex curve-shape, with lower left and upper right regions marked, corresponding to the left and right tails of the distribution of Mahalanobis' distance

which is to say the points are more dense than for a $\chi^2_9$ at the far left end. Correspondingly, the pattern at the top left in the first plot can be seen at the upper right in your plot -- the largest values are more spread out; i.e. the tail is longer/heavier. The two together give a clear indication that the Mahalanobis' distances in the sample are more right skew than you would expect to see with a mutivariate normal.

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  • $\begingroup$ Thank you for visualising it like this. This really helped! $\endgroup$
    – Amonet
    Jul 23 '18 at 19:22

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