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I have this empirical sample, which I drew on the scatter plot. It's got to be unimodal according to theory, and it surely does look like one. A simple way to model it is with bivariate normal. However, the marginals appear to be fat-tailed and slightly skewed, e.g. skewed t-distribution has a similar shape.

Can you suggest candidate distributions for this kind of data?

I fit the Gaussian kernel density estimator for a contour plot just for fun, I don't think this particular kernel would be a good fit because it can't produce fat tails.

enter image description here

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  • $\begingroup$ What is the purpose of your fitting? What do you want to do with the fitted distribution? $\endgroup$ Dec 17, 2015 at 14:42
  • $\begingroup$ I'm thinking about simulating the tails $\endgroup$
    – Aksakal
    Dec 17, 2015 at 15:12
  • $\begingroup$ And what is the purpose of simulating the tails? $\endgroup$ Dec 17, 2015 at 15:23
  • $\begingroup$ If you just want to simulate the tails, and they are fat, you could model the tails by fitting a power law to them. $\endgroup$
    – Wintermute
    Dec 17, 2015 at 15:49
  • $\begingroup$ @MarkL.Stone, to create stress scenarios. $\endgroup$
    – Aksakal
    Dec 17, 2015 at 16:51

1 Answer 1

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It sort of looks somewhere between

  1. a bivariate $t$ (correlated normals divided by the square root of the same chi-squared variate, and then scaled and/or shifted as needed)

  2. correlated normals divided by the square root of the independent chi-squared variates, and then scaled and/or shifted as needed

enter image description here

 # code:
 x=rnorm(100000)
 y=.8*x+.6*rnorm(100000)
 z=sqrt(rchisq(100000,8)/8)
 z2=sqrt(rchisq(100000,8)/8)
 plot(-x/z,y/z,pch=16,cex=0.1,col=4)
 plot(-x/z,y/z2,pch=16,cex=0.1,col=4)

as such it's possible that you might get a reasonable approximation by dividing correlated normals by the square root of somewhat correlated chi-squared variates, perhaps something like this:

![enter image description here

z=rchisq(10000,3)
z1=sqrt((rchisq(10000,5)+z)/8)
z2=sqrt((rchisq(10000,5)+z)/8)
plot(-x/z1,y/z2,pch=16,cex=0.1,col=4)

z=rchisq(10000,6)
z1=sqrt((rchisq(10000,2)+z)/8)
z2=sqrt((rchisq(10000,2)+z)/8)
plot(-x/z1,y/z2,pch=16,cex=0.1,col=4)
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  • $\begingroup$ This is interesting $\endgroup$
    – Aksakal
    Dec 21, 2015 at 4:08

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