let $X_1,X_2,X$ be i.i.d (idenpendant, identically distributed) with stable probability density function $l_\alpha(x)$.
For $\forall A>0,B>0$ find $C$ such that
$AX_1 + BX_2 \overset{d}{=}CX \qquad (1)$
where $\overset{d}{=}$ means equal in distribution.
$\textrm{My Idea}$:
In the lecture we showed that the characteristic function $\hat{l}_\alpha$ is given by $\hat{l}_\alpha(k) = <e^{ikx}>:= \int_{-\infty}^\infty e^{ikx}l_\alpha(x)dx=e^{-\sigma^\alpha|k|^\alpha} \quad \sigma>0$.
Using this result i get the following relation:
$(1) \quad \Rightarrow \quad <e^{i(AX_1 +BX_2)k}>=<e^{ikAX_1}><e^{ikBX_2}>=<e^{ikCX}>$
which should be equivalent to
$e^{-\sigma^\alpha|k|^\alpha A^\alpha} e^{-\sigma^\alpha|k|^\alpha B^\alpha} = e^{-\sigma^\alpha|k|^\alpha C^\alpha} \Rightarrow A+B=C$
Is this the correct solution?