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I tried to make a comparison among various candidate distributions fitting for my data. These data are daily returns of S&P500 US equity Index. Among others I tried with t-location-scale (https://en.wikipedia.org/wiki/Student%27s_t-distribution#Generalized_Student's_t-distribution) and t-skewed (https://en.wikipedia.org/wiki/Skewed_generalized_t_distribution). The first can be viewed as a restricted version of the second. More precisely, at least in the specifications I consider, t-skewed have one parameter more for asymmetry. If this parameter added is zero the two distributions coincide. Therefore I can use the LR test (Likelihood Ratio) for comparison.

In particolar the log-likelihood function (LLF) are: LLF(t-loc-scale)= 16.909,2 and LLF(t-skewed)= 16.914,0. The p-value of related LR test is close to 0,002 therefore the evidence in favour of t-skewed is quite strong. I do not have problem until here, if I miss something please let me know.

Moreover I tried with Stable distribution (https://en.wikipedia.org/wiki/Stable_distribution). Unfortunately Stable is not a general and/or particular case of t-skewed. So it remain a comparison between non nested models.

Now my guess start considering that LLF(t-loc-scale) is higher than LLF(Stable)=16.865. Now if we consider that these two distribution is quite similar (at least in this case, see the graph below), become notable that Stable distribution have one parameter more and reach a lower level of LLF in the same data. This sound me as evidence in favour of t-loc-scale. If this reasoning holds we can deduct that, a fortiori, t-skewed is better than Stable.

Is this reasoning correct? If it is not, I can conclude something about the comparison between Stable and t-skewed distribution with the output below?

Moreover, in both cases shown below standard error for estimated parameters are reported. This sound me as suggestion for using hypothesis testing through usual z-test. However stable distribution deal with unfinite variance. This is not a problem? Limiting distribution for estimated parameters converge to Normal yet?

Here the Matlab output: enter image description here enter image description here

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  • $\begingroup$ Is there a reason why you need to fit a parametric distribution as opposed to a nonparametric density estimate if you otherwise just need to describe the distribution of the values with a figure? Also, such fitting procedures assume independence of observations. Is that a reasonable assumption for daily returns of the S&P500 US equity Index? $\endgroup$
    – JimB
    Nov 17 '21 at 18:25
  • $\begingroup$ I have to compare the distributions above and pick the best among them. I know that independence among observations probably is violated. However there is no reason why some dependance should favor one distribution over another among the above. Actually this point is frequently ignored, sometimes asymptotic argument is used for consider not so relevant dependance issues in sample. In any case my question is only about comparison between not nested distributions. $\endgroup$
    – markowitz
    Nov 18 '21 at 10:00
  • $\begingroup$ I understand "I have to compare the distributions". But my question is about what do you do with the distribution once you've settled on a "best" distribution. If just a picture of the distribution is the end result, then it would seem that a nonparametric density would be adequate. Or maybe you're going to look at change in the estimated parameters over time. Or maybe you intend to take random samples from that distribution. In short, stating the purpose as to why you want to compare distributions would be helpful. (Also, do you have a definition of "nested distributions"?) $\endgroup$
    – JimB
    Nov 18 '21 at 16:42
  • $\begingroup$ My question is legitimate and well posed. I do not have to justify it. About nested or not nested distribution, the reply is in my question: t-dist is nested in t-skew; stable and t-skew are not nested. $\endgroup$
    – markowitz
    Nov 18 '21 at 17:24
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    $\begingroup$ Your question is fine and I wasn't intending to imply otherwise. It's just that knowing how you intend to use the result will get you better and more targeted answers. $\endgroup$
    – JimB
    Nov 18 '21 at 17:52
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I achieved some conclusions.

Is this reasoning correct? If it is not, I can conclude something about the comparison between Stable and t-skewed distribution with the output below?

Yes it is. Indeed BIC criterion can be applied among non nested models and if the number of parameters is the same, BIC boil down in a comparison among LLF. Therefore, a fortiori, t-location-scale distribution is better than stable one.

Moreover, in both cases shown below standard error for estimated parameters are reported. This sound me as suggestion for using hypothesis testing through usual z-test. However stable distribution deal with unfinite variance. This is not a problem? Limiting distribution for estimated parameters converge to Normal yet?

About that I have doubts yet.

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