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Is it possible to design a probability function P(x) such that the distribution will have a specified kurtosis, K, and skewness, S? I am not otherwise interested in any other properties of this distribution.

P(x) = f(K,S)

...f is some function of kurtosis, skewness.

Nick Cox
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user3068957
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    Just for the record, note that any such distribution will not be unique, except possibly within its own family. – Nick Cox Dec 11 '13 at 18:46

2 Answers2

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Is it possible to design a probability function P(x) such that the distribution will have a specified kurtosis, K, and skewness, S?

Certainly.

Consider, for example, the Pearson system of distributions.

This is a system of distributions (a collection of distribution families) indexed by the standardized third and fourth central moments (which is usually what people mean when they say 'skewness' and 'kurtosis'), or more usually, the square of the standardized third moment and the standardized fourth moment.

These families - twelve in all, if I remember correctly (but 4 main ones, the remaining ones are special cases) - divide up the skewness-kurtosis plane into regions where each family applies.

To my recollection these families include the normal, t, gamma, inverse gamma, F, and beta distribution families.

There's a fairly detailed discussion of the Pearson type IV distribution family, including the relationship between its parameters and the mean, variance, skewness and kurtosis.

Heinrich, J. (2004)
A Guide to the Pearson Type IV Distribution,
Univ. Pennsylvania, Philadelphia, Tech. Rep. CDF/Memo/Statistics/Public/6820

[Incidentally, the R package PearsonDS (available on CRAN) offers the usual set of functions for pdf, cdf, quantiles and random numbers, as well as fitting (via maximum likelihood or method of moments). There's a function, pearsonFitM to fit a given mean, variance, skewness and kurtosis.]

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But you can do just fine, simply by taking a few discrete points, say 5-6, and holding the mean constant (at 0), move points around until the desired values are obtained. Such an approach can be automatic via use of optimization routines, though in specific cases it doesn't take long by trial and error (as long as you don't try to realize unattainable values!).

For example, let's say I want standardized 3rd central moment to be 1 and standardized 4th central moment 5.

Consider these 6 values, arrived at by trial and error:

-2.9935,-1.51988,0,0,.51988,3.9935

they give (for the standardized 3rd and 4th central moments):

1.000026
4.999994

which are accurate to 5 significant figures.

Edit: see also this related question

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Glen_b
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    Discussions of the Pearson system of distributions usually don't reference an excellent, characteristically concise and original, exposition within Jeffreys, H. 1939/1948/1961. Theory of Probability. Oxford: Oxford University Press. I just checked and the Wikipedia entry is another example of oversight. (Jeffreys' book is not so much probability theory, but more data analysis with a Bayesian twist.) – Nick Cox Dec 11 '13 at 18:50
  • +1 Thanks for the reference, @NickCox - I considered mentioning the discussion in Kendall and Stuart (Stuart and Ord in more recent versions), but in the end decided to hold off while I pondered what else to say more generally. I shall endeavour to take a look. – Glen_b Dec 11 '13 at 21:54
  • Jeffreys is full of odd surprises e.g. a nice little discussion of Spearman correlation; comments on "too good to be true", i.e. remarkably good fits; wishful thinking as a pitfall in data analysis. – Nick Cox Dec 11 '13 at 21:57
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The maximum entropy distribution with your constraints is the exponential family

\begin{align} f(x) &\propto e^{ax^4 + bx^3} \end{align}

You'll have to convert a given kurtosis and skewness to natural parameters $a, b$.

Neil G
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