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Suppose that I have $n$ independent samples from a probability distribution. I believe that the distribution is Gaussian ($G$) but I don't know; then my question is: does there exist a test such that I can interpret the result of the test to be something like "the probability that my samples are the output of a non-Gaussian distribution T with $d(G,T)>\epsilon$ is smaller than $x$%", where $d$ is the variational distance.

I have Googled and found the Kolmogorov-Smirnov test and some others, but I'm not sure if I can use the results of those tests to make a strong statement like the one above.

Nick Cox
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Al Learner
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  • Thanks for the tip, but in my case I want the statement as part of a proof for something completely different. I'm not trying to process data in any form. – Al Learner Dec 11 '13 at 19:43
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    If you read the answers in the link carefully it should be clear that a major problem with what you want to be able to say is that there's no restriction on how close a process can be to a Gaussian one yet still be non-Gaussian. – Scortchi - Reinstate Monica Dec 11 '13 at 20:03
  • You are completely right. In fact, I'd be happy with a statement that includes a distance parameter, I'll modify the question. – Al Learner Dec 11 '13 at 20:16
  • What do you mean by "process" ? Do you have time-indexed observations ? – Stéphane Laurent Dec 11 '13 at 20:18
  • Just a distribution, if it is clearer, I can change process to probability distribution. – Al Learner Dec 11 '13 at 20:20
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    If you really want a probability, you might be able to formulate a Bayesian approach (though not exactly a 'test'). In a frequentist hypothesis-testing framework, you instead obtain a conditional probability of a somewhat different kind. – Glen_b Dec 11 '13 at 22:59
  • However, you may be able to get somewhere with that approach. – Glen_b Dec 11 '13 at 23:06
  • @Glen_b I'd be really interested in an expanded version of the comment. Just with a couple of pointers and a clearer idea of what I can or cannot ask I'd be definitely satisfied. – Al Learner Dec 12 '13 at 08:55

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