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There is a function FindDistribution that finds a distribution from the given data.

But what if I'm given moments and what to find a distribution.

Is there any function for this?

Thank you.

    j[x_] := Moment[JohnsonDistribution["SL", gamma, delta, 0, lambda],x];

    FindRoot[{j[1] == 1.2974425414, j[2] == 5.252267250245984, 
    j[3] == 112.50634898225843576187}, {{delta, 
    1.5}, {gamma, -1.2}, {lambda, 0.4}}, WorkingPrecision -> 250, 
    MaxIterations -> 10000]

It gives Encountered a singular Jacobian at the point {delta,gamma,lambda} = \ {1.5,-1.2,0.4}. Try perturbing the initial point(s)`.

The histograms of the distribution are shown below:enter image description here enter image description here

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    $\begingroup$ "if I'm given moments" All moments or just some moments? With all moments, build the characteristic function, take its inverse Fourier transform to get the PDF. Then use ProbabilityDistribution to define the distribution from the PDF. $\endgroup$
    – Bob Hanlon
    Commented Jan 23 at 18:27
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    $\begingroup$ If you only know a few sample moments AND you have a distribution in mind, then you can equate the sample moments to the theoretical moments and solve for the distribution parameters. This is known as the "Method of Moments". There are several examples on this site. Here's one: mathematica.stackexchange.com/questions/108052/…. $\endgroup$
    – JimB
    Commented Jan 23 at 18:31
  • $\begingroup$ Is this any different from your question at stats.stackexchange.com/questions/635999/… ? $\endgroup$
    – JimB
    Commented Jan 23 at 18:37
  • $\begingroup$ No. I don't have MGF, just a finite number of moments. I tried to use the method of moments, but Mathematica doesn't give a solution. $\endgroup$
    – Paul R
    Commented Jan 23 at 18:41
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    $\begingroup$ I've deleted my answer because I now see that for this particular Johnson distribution, 3 moments are not sufficient to obtain unique estimates of the parameters. (This is typically not the case in that in practice 3 parameters requires just 3 moments.) But for this Johnson distribution the 3rd moment is the cube of the ratio of the second to the first moment. This relationship is probably the reason for the error you're seeing. So very loosely, the first 3 moments only provide 2 of the necessary pieces of information to estimate the 3 parameters. $\endgroup$
    – JimB
    Commented Jan 25 at 5:26

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