$X$ is r.v. with Exponential with $\lambda=1$ and Y has a characteristic function $e^{-|t|^a}$ for $a\in (0,2)$. What is the characteristic function of $Z=X^{\frac{1}{a}}Y$ if $X$ and $Y$ are independent?
Here's what I have found out so far:
$$
\varphi_{X^{\frac{1}{a}}Y}(t) = \mathbb{E}(e^{itX^{\frac{1}{a}}Y}) = \mathbb{E}(\mathbb{E}(e^{itX^{\frac{1}{a}}Y}|X)) = \dots
$$
Now
$$
\mathbb{E}(e^{itX^{\frac{1}{a}}Y}) = \mathbb{E}(e^{itx^{\frac{1}{a}}Y})|_{x=X} = \varphi_Y(tx^{\frac{1}{a}})|_{x=X} = e^{-|x^{\frac{1}{a}}t|^a}|_{x=X} = e^{-|t|^aX}
$$
so
$$
\dots = \mathbb{E}e^{-|t|^aX}
$$
Here I got stuck. I haven't been formally introduced to Laplace's transform and I don't know how to proceed.
Best Answer
Since $X$ has an exponential distribution of parameter one, it follows that $$ \mathbb E\left[f(X)\right]=\int_0^\infty f(x)e^{-x}dx $$ for each continuous bounded function $f$. Apply this for a fixed $t$ to $f(x)=\exp\left(-\lvert t\rvert^\alpha x\right)$; you will be reduced to compute an integral of the form $\int_0^\infty e^{-Ax}dx$ for some $A$.