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For standard Brownian motion and $a>0$, $T_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely andThe Brownian reflection principle gives this distribution function. The Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ and the Strong Markov property show its probability distribution is a strictly stable distribution with index $1/2$.

Taking $x=y=a$ in (c) and using that $B_t$ has no atom at $a$ givesThis is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ for $c>0$.

For an integer $n\ge 1$, set $T_0=0$ and successively hit the levels $a,2a,\dots ,na$. The Strong Markov property at these finite stopping times and spatial translation imply thatare independent random variables, each with the probability distribution of $T_a$. Thus, for independent copies $T_a^{(j)}$,ThereforeThe paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

First take integers $1\le k<n$. For independent copies $X_1,\dots ,X_n$ of $X$, defineThe two blocks involve disjoint families of independent random variables, so $A$ and $B$ are independent. The assumed strictly stable distribution gives $A\stackrel{d}{=}X$, $B\stackrel{d}{=}X$, andTherefore, on any fixed probability space carrying $X$ and an independent copy $X^{\prime }$, their joint probability distribution givesNow fix an arbitrary $0<c<1$ and choose rationals $r_m=k_m/n_m\in (0,1)$ tending to $c^{\alpha }$. Continuity of the real powers gives, on that fixed probability space,The left sides all have the probability distribution of $X$. Since almost sure convergence implies convergence in distribution, their limit has that same probability distribution. HenceThis proves strict stability implies self-decomposability. The limit argument uses no moment-generating function or finite moments, which is essential for general stable distributions.

For independent and identically distributed random variables $X_j\sim \mathcal{N}(0,{\sigma }^2)$, the linear combination of independent normal random variables $n^{-1/2}{\sum }_{j=1}^n{X}_j$ has expected value zero and variance $n^{-1}n{\sigma }^2={\sigma }^2$. ThusThis is exactly the requested $2$-stability. The paper's definition is a strictly stable distribution, since it allows no centering term.

Let $X$ have a strictly stable distribution of index $\alpha$, and let $X^{\prime }$ be an independent copy. For integers $1\le k<n$, split a sum of $n$ independent copies into blocks of sizes $k$ and $n-k$. Strict stability of each block and of the full sum givesFor any $0<c<1$, choose rationals $k_m/n_m\to c^{\alpha }$. On the fixed probability space carrying $X,X^{\prime }$, the left side converges almost surely to $cX+(1-c^{\alpha }{)}^{1/\alpha }{X}^{\prime }$. Thus it converges in distribution; since its probability distribution is constant, the limit also has the law of $X$. This proves a self-decomposable distribution with the displayed remainder. The proof does not require any finite moments.

A symmetric stable distribution has this characteristic function, with $c>0$ and location $\mu$. At location zero it is a strictly stable distribution: weighted sums of independent copies have scale factor $({\sum }_j|a_j{|}^{\alpha }{)}^{1/\alpha }$. The index $\alpha$ need not permit a finite mean or variance. The scale in this convention is $c^{1/\alpha }$, so the coefficient of $|t{|}^{\alpha }$ and the scale parameter should not be confused.