Suppose $X$ is symmetric and stable with index $\alpha$ and $Y$ is stable and nonnegative with index $\beta$. Show that $XY^{1/\alpha}$ is stable with index $\alpha \beta$.
The text also gives the hint that the Laplace transform of $Y$ is $L_Y(s) = \exp(-s^{\beta})$. I have no idea how this is helpful.
My first (only) idea is to use characteristic functions. We know that the characteristic function of $X$ is $\exp(-c|t|^{\alpha})$. However, that's where I get stuck. There is no independence assumption in the problem, and I have absolutely no idea how to proceed then.
False without independence assumption. If $X$ and $Y$ are independent then $Ee^{itXY^{1 /\alpha}}=E(Ee^{itXY^{1 /\alpha}}|Y)=Ee^{-c^{\alpha}|t|^{\alpha} Y}=e^{-c^{\beta \alpha}|t|^{\beta \alpha}}$ so $XY^{1 /\alpha}$ is symmetric stable with index $\beta \alpha$.
Counterexample: if $Z$ has standard normal distribution that $Y=\frac 1 {Z^{2}}$ is a positive stable random variable with index $\beta =\frac 1 2$. Take $X=Z$ so that $X$ is stable with index $2$. But $XY^{\frac 1 2}=\frac Z {|Z|}$ is not stable with index $1$ so the statement is false without independence.