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If I make some experiment, which I repeat $N$ times, and look at the distribution of the experiment. I would usually expect (and hope) for a Gaussian or normal distribution.

If however my data is T-distributed, what is this saying about my data? What can I interpret from the T-distributed.

I realise this may be a slightly open ended question, but if I may illustrate what I am after by means of example. If I make an experiment and my data comes out as a Rayeligh distribution, I can already say that my data is comes from the absolute magnitude of two components which are both Gaussian distributed.

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  • $\begingroup$ Do your data come from the ratio between normally distributed data and the sum of squares of normally distributed data? en.wikipedia.org/wiki/Student%27s_t-distribution $\endgroup$ Mar 4, 2020 at 14:25
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    $\begingroup$ Your logic is inverted: when your data are the norms of Gaussian vectors, it's reasonable to use a Rayleigh distribution to model them; but when a Rayleigh model fits your data, that does not imply your data were generated by norms of some (hidden) Gaussian vectors. Similarly, there's very little you can say about the process by which $t$-distributed data were generated. $\endgroup$
    – whuber
    Mar 4, 2020 at 15:01
  • $\begingroup$ @whuber I did a little more reading, and I now interpret a t-distribution as data drawn from a Gaussian distribution but where there are not enough points in the sample to fully represent the parent distribution, from which the sample is generated. But I do take your point about my logic being inverted. $\endgroup$
    – user27119
    Mar 4, 2020 at 15:11
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    $\begingroup$ The link is this: the particular Student $t$ in question has "one degree of freedom:" it is the distribution of the $t$ statistic for a sample of size $2.$ In this case the standard error is a multiple of the size of the difference between the two numbers in the sample. Because the difference follows a Gaussian distribution, its size follows a "half-Gaussian." Thus, the Student $t(1)$ arises by dividing one Gaussian by an (independent) half-Gaussian. However, because the numerator is equally likely to be negative as positive, you get the same distribution as dividing by the full Gaussian. $\endgroup$
    – whuber
    Mar 4, 2020 at 17:39
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    $\begingroup$ Historically, the first proof would have appeared in a paper by R. Fisher around the first World War. Any theoretical or sufficiently advanced stats textbook includes one. The result in a disguised form is proven at stats.stackexchange.com/questions/85916. I presented the argument in my previous comment, with more details, at stats.stackexchange.com/a/437424/919, and the same thread contains other demonstrations, including the one at mathworld.wolfram.com/NormalRatioDistribution.html. $\endgroup$
    – whuber
    Mar 5, 2020 at 21:19

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If your data resembles as if sampled from a t-distribution rather than from a normal distribution, that means it has fatter tails, while it is still symmetric. You cannot say from this how the data was generated, there must be many possibilities. But it must be some process that for some reason tends to produce many atypical values outliers.

One way to learn about this is using simulation, in R try x <- rt(100, df=5) and experiment, plot, ...

Sometimes a t-distribution is assumed to make a more robust model, see Why should we use t errors instead of normal errors? and Fitting t-distribution in R: scaling parameter

Your conclusion in the last paragraph is wrong, as said in comment by whuber:

Your logic is inverted: when your data are the norms of Gaussian vectors, it's reasonable to use a Rayleigh distribution to model them; but when a Rayleigh model fits your data, that does not imply your data were generated by norms of some (hidden) Gaussian vectors.

Similarly, there's very little you can say about the process by which 𝑡-distributed data were generated.

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