Distribution of a stable and exponential r.v

Question

I am given an absolutely continuous positive r.v $X$, such that $\mathbb{E}[e^{-sX}]= e^{{-s}^\alpha}, \ s>0.$ The goal is to prove that $(Y/X)^\alpha$ is an Exponential r.v with intensity $1$, provided that $Y$ is independent of $X$ and also Exp$(1)$.

My initial attempt was to prove that $$\mathbb{E}[e^{t(Y/X)^\alpha}]= \int_0^{\infty}f_X(x)dx\int_0^{\infty}e^{t(y/x)^\alpha}e^{-y}dy=\frac{1}{1-t}=\psi_Y(t),$$ which is the mgf of $Y$, but I can't evaluate that integral, so there must be some other approach. Or am I not doing it right? Any insight would be helpful..

Answer 1

Here is a solution from first principles.

$Z = (Y/X)^a$.

$P(Z \leq z) = P(Y/X \leq z^{1/a})$

$ = \int_0^\infty P(Y \leq xz^{1/a})f_X(x)dx$

$ = \int_0^\infty (1-e^{-xz^{1/a}})f_X(x)dx$

$ = 1 - \int_0^{\infty}e^{-xz^{1/a}}f_X(x)dx$

$ = 1 - \mathbb{E}[e^{-z^{1/a}X}]$

$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).

This shows that $Z \sim \exp(1)$.