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Re:Why isn't everything normally distributed? by srean on Thursday July 20, 2017 @11:43PM (#14817907)
> I understand that there are good theoretical motivations for this --- namely, the normal distribution is the distribution that maximizes entropy for a given mean and variance.

That is indeed true, but why should such a property imply that the use of Normal distribution is appropriate ? Just a rhetorical question, of course, because your comment does indicate that Normal is not a good choice unless one has compelling reasons to do so.

Another argument that is used to justify its use is the central limit theorem. That says a sum of (nearly) independent variables with (nearly) identical distributions with finite mean converge to the Normal distribution. If the process under observation is indeed a superposition of such random processes, then yes the choice of the Gaussian can be justified. But it is surprisingly common that one of the 3 requirements are violated. A common violation is that the variance is infinite or it is so high that the process is better modeled as one with an infinite variance. In such situations the family of stable distributions are the more appropriate choice.

Gauss' own use of the Gaussian was motivated by convenience rather than any deep theory. Even at that time it was well known that other distributions, for example, Laplace distribution works better.

Re:Why isn't everything normally distributed? by bonoboTP on Friday July 21, 2017 @06:20AM (#14819299)
As you say, mathematical convenience is another argument.

The negative log of the (unnormalized) Gaussian density is basically x^2 and this fact comes very handy.

It's used extensively in Bayesian regression, modeling noise as additive Gaussian leads to the squared loss function, and if you also model the prior as Gaussian it leads to an L2 regularizer.

It also simplifies many other calculations. Fitting a mixture of Gaussians by expectation-maximization or the Kalman filter come to mind.

Re:Why isn't everything normally distributed? by srean on Friday July 21, 2017 @06:28AM (#14819325)
Absolutely !

For squared loss almost every theorem you want to be to be true are actually true. In a way its dangerous because it sets up bad expectations (* no nerd pun intended).

A result that holds for squared loss but not for others is the SVD as a low rank approximation of a matrix.

*Expectation minimizes the squared loss