# Solved – Does this Q-Q plot indicates leptokurtic or platykurtic distribution

The points have shown a "hump" pattern but I can't really identify that is it a narrow hump or wide hump?

Contents

The normal $$q$$$$q$$ plot is a graph of $$(z_{0(i)}, z_{(i)})$$.

Excess kurtosis is $$kappa_e = sum z_{(i)}^4/n – sum z_{0(i)}^4/n ,$$ where $$z_{(i)} = (x_{(i)} – bar{x})/s_x$$ is the $$i$$th ordered standardized value and $$z_{0(i)}$$ is a corresponding theoretical quantile, which can be chosen so that $$sum z_{0(i)}^4/n = 3.$$

Rewriting, $$kappa_e = (1/n)sum (z_{(i)}^2 – z_{0(i)}^2)(z_{(i)}^2 + z_{0(i)}^2) .$$

This expression shows how you can relate excess kurtosis directly to the visual appearance of the normal $$q$$$$q$$ plot. Assuming the line is the 45 degree $$(z_{0(i)}, z_{0(i)})$$ line, you have positive excess kurtosis when the average difference between squared vertical deviations from zero of the data and the squared vertical deviation from the line to zero, weighted by squared Euclidean distance from the plotted point $$(z_{0(i)}, z_{(i)})$$ to 0, is positive. You have negative excess kurtosis when it is negative.

Hence, when there are large positive values of $$z_{(i)}^2 – z_{0(i)}^2$$ in the tails, you have positive excess kurtosis. This result confirms the well-known result that kurtosis measures tail extremity relative to the normal distribution.

If the line in the graph of the OP corresponds to the $$(z_{0(i)}, z_{0(i)})$$ line, then the graph shows positive excess kurtosis, because the tail extremities are more extreme than the normal distribution predicts.

Another correspondence between the normal quantile-quantile plot with the excess kurtosis $$kappa_e$$ concerns the detrended plot, which can be constructed as a plot of $$(z_{0(i)}, z_{(i)} – z_{0(i)})$$.

Rewriting again, $$kappa_e = (1/n)sum (z_{(i)} – z_{0(i)})(z_{(i)} + z_{0(i)})(z_{(i)}^2 + z_{0(i)}^2) .$$

Thus, $$kappa_e$$ is an average of the vertical deviations (the $$z_{(i)} – z_{0(i)}$$ values) from zero in the detrended $$q$$$$q$$ plot, each weighted by a signed measure of distance from the data to the mean, $$w_{(i)} = (z_{(i)} + z_{0(i)})(z_{(i)}^2 + z_{0(i)}^2)$$.

For example, a symmetric heavy-tailed distribution has $$z_{(i)} – z_{0(i)} <0$$ for small $$i$$ ($$i$$ near 1) and $$z_{(i)} – z_{0(i)} >0$$ for large $$i$$ ($$i$$ near $$n$$). The weights $$w_{(i)}$$ are negative and large in magnitude for small $$i$$, and positive and large in magnitude for large $$i$$; after multiplying and averaging, this will tend to give a positive excess kurtosis.

For a symmetric light-tailed distribution, $$z_{(i)} – z_{0(i)} >0$$ for small $$i$$ ($$i$$ near 1) and $$z_{(i)} – z_{0(i)} <0$$ for large $$i$$ ($$i$$ near $$n$$); after multiplying these deviations by the signed weights (the $$w_{(i)}$$) and averaging, this case will tend to give a negative excess kurtosis.

Thus, the same conclusion is obtained for either way of corresponding the excess kurtosis with the normal $$q$$$$q$$ plot: Whether excess kurtosis is positive or negative depends mainly on the comparison of the ordered standardized data quantiles $$z_{(i)}$$ with the corresponding theoretical normal quantiles $$z_{0(i)}$$ in the tails.

I completely disagree with the recommendation to use density plots as regards diagnosing kurtosis. Since kurtosis measures only tails, and since the tails are very close to zero even for heavy-tailed distributions, it is very difficult to discern tail behavior from density plots. On the other hand, tail behavior relative to the normal distribution is obvious in the normal $$q$$$$q$$ plot. That's what the plots are for. Further, the math I just showed gives a direct connection between the normal $$q$$$$q$$ plot and the excess kurtosis statistic.

I also disagree that kurtosis is confusing, bizarre and acts counterintuitively. Sure, if your intuition states that kurtosis measures peakedness, then it behaves counterintuitively. But as long as you understand that it measures tail extremity relative to the normal distribution (as the math above shows), then it is perfectly understandable.

Rate this post