Tsallis $q$-Gaussian and applications

Question

Why is not $q$-Gaussian distribution merely the substitution of q exponential into the gaussian function?, i.e. substitution of equ.2 in equ.1. Where would there be three cases as below. When to use each, the direct substitution and the defined $q$-Gaussian function? different applications?

statistical mechanics thermodynamics

Brick · Accepted Answer

There are a couple of things going on here.
First, the $q$-Gaussian is defined as closely as possible to a direct swap of exponential for $q$-exponential as you could hope.  To the extent it looks like that isn't true, it is only because in the $q$-Gaussian case, the shape factor ("covariance") and the normalization have been written differently.  For the Gaussian, the normalization $A$ was written in the numerator and for the $q$-Gaussian, the normalization $C_q$ was written in the denominator.  Likewise, for the Gaussian, you have factors of $w$ (in the notation of the question) in the denominators for the Gaussian and corresponding factors of $sqrt{beta}$ in the numerators.
Now this notation does hide some things, some of which are related to the normalization choices:

The $q$-exponential is only defined over a bounded subset of the real line for $q<1$, and so the distribution there is fundamentally different than in the unbounded cases.  This is enforced by the innocuous looking $+$ subscript in the definition of the $q$-exponential.
In the "fat-tail" cases of $1 < q < 3$, the distribution goes to 0 at infinity with power-law tails rather than the exponential decay in the normal Gaussian.
Points #1 & #2 give rise to the completely different forms of the normalization factors in the different cases.
A point never well-emphasized with these distributions is that the transition between the cases is not defined by a properly smooth limit in $q$.  This is especially true in the limit $q rightarrow 1^-$ where $q$ approaches 1 from below because the truncation to the finite domain is not a smooth operation.  It's also true, though more subtly so, in the limit $q rightarrow 1^+$ because there's infinite Fisher distance between the fat-tails and the exponential tails.

Kurt G. · Answer

This doesn't work, supposing that by substitution of equ.2 into equ.1 you mean $e_q(g(x))$ where
$$
g(x)=frac{A}{wsqrt{pi/2}}e^{-2(x-xi_C)^2/w^2}
$$
and $e_q(x)=[(1+(1-q)x)]_+^{1/(1-q)},:$ Since $g(pminfty)=0$ we get
$e_q(g(pminfty))=1$ regardless what $q$ is. For a function to be a distribution function on $(-infty,+infty)$ it is however necessary that its limits at $pminfty$ are zero.
For further applications of q-Gaussian distributions, see this thesis.

Tsallis $q$-Gaussian and applications

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