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Let $X$ have a Student-t distribution, so that \begin{align*} f_X(x|\nu ,\mu ,\beta) = \frac{\Gamma (\frac{\nu+1}{2})}{\Gamma (\frac{\nu}{2}) \sqrt{\pi \nu} \beta} \left(1+\frac{1}{\nu}\left(\frac{x - \mu}{\beta}\right)^2 \right)^{\text{$-\frac{1+\nu}{2}$}} \end{align*}

I know that Student-t distributions show a power-law in the tail. I also know that Lévy stable distributions ( e.g with the following characteristic function:

\begin{align*} \phi(t|\alpha ,\beta, c ,\mu) = exp[i t \mu - |ct|^\alpha (1-i\beta sgn(t) \Phi)] \end{align*}

where $sgn(t)$ is the sign of $t$ and $\Phi= tan(\frac{\pi \alpha}{2}) \quad \forall \alpha$ except for $\alpha =1$ when $\Phi = -\frac{2}{\pi} log|t|$ ) have a power-law in the tails, so that the asymptotic behaviour for large $x$ of a r.v. $X$ Lévy stable-distributed is:

$$ f_X(x) \propto \frac{1}{|x|^{1+\alpha}}$$

My question is: is the Student-t distribution stable? Or, in other words, does a power-law in the tails imply a Lèvy stable distribution?

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    $\begingroup$ One empirical finding I have found in the past is that for the cut off of around 90% that the t distribution and alpha stable have similar results from the Hill estimator when alpha = 2 - 1/DF, so although they are different there are lots of similarities: - at one extreme alpha =1 and DF =1 both are Cauchy - at other for alpha =2 and DF = infinity both are Normal - in between relationship described above could be used to give rough equivalence between these extremes - shape of tail beyond 90% is different as Cauchy is more extreme in very far tail $\endgroup$ – James65 Feb 14 '17 at 22:37
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One of the characterizing features of a Levy-stable distribution is that linear combinations of independent copies have the same distribution, up to location and scaling. So if this property does not hold, the distribution cannot be Levy stable. Equivalently the characteristic function isn't of the Levy form.

In the case of the student t distribution, it has a characteristic function that looks like:

$$\frac{K_{v/2}(\sqrt{v}|t|)(\sqrt{v}|t|)^{v/2}}{\Gamma(v/2)2^{v/2-1}},$$

which in general will not have the Levy form.

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    $\begingroup$ Since it's probably not immediately obvious to most people that this form of the Student t CF cannot be written in the Levy stable form, it would be nice to see some demonstration of that impossibility. $\endgroup$ – whuber Jan 21 '16 at 19:32
  • $\begingroup$ So a a stable distribution necessarily implies a power law in the tails (apart from the Normal), but a power law in the tails doesn't necessarily imply a stable distribution? $\endgroup$ – Puzzle Jan 21 '16 at 22:03
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    $\begingroup$ From the other side, Student distribution is an infinitely divisible distribution and as such is a distribution of some Levy process. How it can be? $\endgroup$ – zer0hedge Jul 7 '17 at 9:26
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case $\nu>2$

To expand Alex's aswer we can make a different type of argument for $\nu>2$:

  • Lévy-stable distributions have infinite variance for the stability parameter $\alpha < 2$.

  • But the t-distribution has a finite variance for the degrees of freedom parameter $\nu > 2$.

  • And the Gaussian distribution is already the (unique) Lévy-stable distribution with $\alpha=2$.

Thus it must be the case that the generalized t-distribution can not be a Lévy stable distribution.

  • Another way of seeing this is that (due to the finite variance and the CLT) the distribution of a sum of t-distributed variables must converge to the normal distribution. Thus the t-distribution can not be a Lévy stable distribution.

case $1 \leq \nu \leq 2$

In these cases we can not use the argument above. One way to look at is to inspect the characteristic function (as Alex's answer mentions). In the case of the location scale variant this is :

$$\varphi(t) = e^{it\mu} \frac{K_{\frac{\nu}{2}}\left( \sqrt{\nu} \vert \sigma t \vert \right) \left(\sqrt{\nu} \vert \sigma t \vert \right)^{\frac{\nu}{2}} }{\Gamma \left( \frac{\nu}{2}\right) 2^{\frac{\nu}{2}-1} }$$

with $K_{\lambda}(w)$ the modified Bessel function of the second kind.

$$K_{\lambda}(w) = \frac{1}{2} \int_0^\infty x^{\lambda-1} e^{-\frac{1}{2}w\left(x + 1/x \right)} dx$$

See Dae-Kun Song, Hyoung-Jin Park, Hyoung-Moon Kim A Note on the Characteristic Function of Multivariate t Distribution

  • In the case of $\mathbf{\nu = 1}$ the t-distribution is the same as the Cauchy distribution which is known to be Lévy stable.

    In this special case the term with the modified Bessel function is $$K_{\frac{1}{2}}(\vert \sigma t \vert) = \sqrt{\frac{2}{\pi \vert \sigma t \vert }}e^{-\vert \sigma t\vert }$$ and you end up with $$\varphi(t) = e^{it\mu + \vert \sigma t \vert }$$

  • In the case of $\mathbf{1 < \nu \leq 2}$ the t-distribution and the function $K_\nu$ are more difficult to evaluate. But, we can make an argument in reverse direction and suppose that $K_\nu$ must be of some form and then see whether is a solution of Bessel's equation.

    Suppose some t-distribution with $1 < \nu \leq 2$ is Lévy stable, then the characteristic function would need to be of the form $$\varphi(t)=e^{it\mu -ct^\alpha}$$ with $c>0$ and $1 < \alpha < 2$ (in these cases the mean is finite and the variance infinite). Actually the Holtsmark distribution is the only currently known explicit distribution that has this form with these conditions.

    If a t-distribution for a particular $\nu$ is of such form then the modified Bessel function of the third kind can needs to be of the form: $$K_{\lambda = \frac{\nu}{2}}(w) \propto w^{-\nu/2}e^{-w^\alpha }$$ we can check this by plugging it into the modified Bessel equation $$x^2y'' + xy' - (x^2+\lambda^2)y = 0$$ which becomes $$\alpha x^{2\alpha}-\alpha(\alpha - \nu) x^\alpha - x^2 = 0$$ Which has only the solution $\nu = \alpha = 1$ which is the Cauchy distribution case. Thus, there is no other t-distribution that is Lévy stable.

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