Recently I've came across "Stable distributions" and have seen ways of generating them. I am interested in the so-called "p-stable" distributions (e.g. see here), i.e. those such that if $x_1$ and $x_2$ are copies of a random variable $\zeta,$ then for any constants $a_1,a_2 > 0$ ones has that $a_1x_1 + a_2x_2$ has the distribution of $(a_1^p+a_2^p)^{1/p}\zeta.$ Here, $p > 0$ is a certain constant. Would anyone enlighten me as to whether there exists a random variable $\zeta$ whose distribution is stable and whose values always lie in $(0,1)$?
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If $X_1$ and $X_2$ are independent with support $(0,1)$, then $X_1+X_2$ have support $(0,2)$ while $(1+1)^{1/p}X_1$ have support $(0,2^{1/p})$. This means that the only suitable candidate for $p$ is $p=1$.
However if we suppose (towards a contradiction), that there exists a $1$-stable distribution on $(0,1)$ and consider two independent variables $X_1$ and $X_2$ with this distribution, then we have that $$Var(X_1 + X_2) = 2 Var(X_1)$$ while $$Var(2X_1) = 4 Var(X_1)$$ implying that $2Var(X_1)=4Var(X_1)$, which is of course a contradiction unless $Var(X_1)=0$, which would imply that the distribution is degenerate.
Actually one can show that for stable distributions with $p\in (0,2)$ the variance is undefined and that the mean is also undefined if $p\in (0,1]$. But such distributions are necessarily unbounded.
