= Solution
First take integers $1\leq k<n$. For independent copies $X_1,\ldots,X_n$ of $X$, define
$$
A=k^{-1/\alpha}\sum_{j=1}^kX_j,\qquad B=(n-k)^{-1/\alpha}\sum_{j=k+1}^nX_j.
$$
The two blocks involve disjoint families of <independent random variables>, so $A$ and $B$ are independent. The assumed <strictly stable distribution> gives $A\overset d=X$, $B\overset d=X$, and
$$
(k/n)^{1/\alpha}A+(1-k/n)^{1/\alpha}B=n^{-1/\alpha}\sum_{j=1}^nX_j\overset d=X.
$$
Therefore, on any fixed <probability space> carrying $X$ and an independent copy $X'$, their <joint probability distribution> gives
$$
(k/n)^{1/\alpha}X+(1-k/n)^{1/\alpha}X'\overset d=X.
$$
Now 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>,
$$
r_m^{1/\alpha}X+(1-r_m)^{1/\alpha}X'\longrightarrow cX+(1-c^\alpha)^{1/\alpha}X'\quad\text{almost surely}.
$$
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>. Hence
$$
\boxed{R_c=(1-c^\alpha)^{1/\alpha}X',\qquad R_c\text{ independent of }X,\qquad cX+R_c\overset d=X.}
$$
This proves <strict stability implies self-decomposability>. The limit argument uses no <moment-generating function> or finite <moments>, which is essential for general <stable distributions>.
Back to article page
