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I'm interested in obtaining a sample of numbers $x_1,\ldots,x_n$ such that their cumulants approximately match user-provided set of cumulants.

For instance, I can get a sample with first two cumulants 0, 1 by sampling uniformly, then normalizing mean and variance. Is there a recipe that works when user provides first four cumulants?

A more specialized case I'd be interested in is a case of 0 centered symmetric sample, specified through variance + kurtosis

kjetil b halvorsen
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Yaroslav Bulatov
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  • Is $n$ specified or not? Do you want a solution that works for all sufficiently large $n$ or one particular $n$? In what sense to you mean "approximately?" This is unclear because one generally would want to match higher-order cumulants more closely than the lower-order ones, so some quantitative way to measure the approximation would be helpful. Do you need to approximate *any* cumulants, even when they don't correspond to any distribution at all? – whuber Dec 14 '19 at 21:56
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    I need to generate samples (large n) to test how my algorithm performs for various settings of higher order statistics. I've been using predefined distributions (Normal, Uniform), but this limits me in what cumulants I can test, so was looking for a more flexible method – Yaroslav Bulatov Dec 15 '19 at 00:30
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    One solution to the simplified version of the problem is provided by the [Pearson family of distributions](https://en.wikipedia.org/wiki/Pearson_distribution). – whuber Dec 17 '19 at 16:59

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