I know there is already at least one question answer with this, but I think the solution does not apply in my case.

I have a population for which I know the mean, variance and skewness. I saw how to estimate the parameters for the skew normal distribution given this data, but the problem is that when calculating $\alpha$, $\delta > 1$ (by wikipedia notation), and thus $\alpha$ is undetermined?

How can I solve this?

  • $\begingroup$ Could you please include the formula, definition or a reference for $\delta$ , so readers can know what you're referring to and get something from the post? $\endgroup$
    – MSIS
    Aug 23, 2022 at 23:00

1 Answer 1


By looking at the formulae in the wikipedia page, one can see that if we know skewness, we know $\delta$ (note that skewness is monotonic in $\delta$, so the relation is unique). In turn this determines the sign of $\alpha$.

Also, $$|\delta| = \frac{|\alpha|}{\sqrt{1+\alpha^2}} \implies \delta^2(1+\alpha^2)=\alpha^2 \implies (\delta^2-1)\alpha^2 +\delta^2 = 0$$

So if $\delta \geq1$ this is impossible, not undetermined.

And indeed, this is one of the limitations of the skew normal distribution: $\delta$ should be strictly smaller than unity, which in turn imposes limits on the magnitude of the skewness coefficient

So if someone specifies a skew normal distribution with such a theoretical skew coefficient, that $\delta$ ends up equal or above unity, it is simply wrong.

When the sample skew coefficient is higher than theoretically possible, then one can apply maximum likelihood to estimate $\alpha$.

  • $\begingroup$ Forgive my ignorance, but isn't a Skew normal just a t-distribution $\endgroup$
    – MSIS
    Aug 23, 2022 at 23:04
  • 1
    $\begingroup$ @MSIS It is not. It can be represented as the convolution of a Normal plus or minus an independent Half Normal. $\endgroup$ Aug 24, 2022 at 0:16

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