I have two random variables $X$ and $Y$ and I know the distribution of their product $Z = X Y$. It is distributed as a power law $$ PDF_Z(z) = A z^{-\alpha} $$ where there naturally is a lower cutoff $z > z_0$.
I'm trying to figure out if based on this I can say anything about the distributions of the variables $X$ and $Y$.
The idea I have so far is to write $CDF_Z$ using the joint distribution $PDF_{X,Y}(x, y)$ by calculating this integral $$ CDF_Z(z) = \int_{x=x_{min}}^{x=x_{max}} \int_{y=\frac{z_0}{x}}^{y=\frac{z}{x}} PDF_{X,Y}(x, y) \, \mathrm{d} y \, \mathrm{d} x $$ where I could for example make an ansatz of the power law form $PDF_{X,Y}(x, y) = B x^{-\beta} y^{-\gamma}$ and see what comes out as the distribution of the product.
The calculation is \begin{align*} CDF_Z(z) &= B \int_{x=x_{min}}^{x=x_{max}} x^{-\beta} \int_{y=\frac{z_0}{x}}^{y=\frac{z}{x}} y^{-\gamma} \, \mathrm{d} y \, \mathrm{d} x \\ &= \frac{B}{1-\gamma} \int_{x=x_{min}}^{x=x_{max}} x^{-\beta} \left[ \left(\frac{z}{x}\right)^{1-\gamma} - \left(\frac{z_0}{x}\right)^{1-\gamma} \right] \, \mathrm{d} x \\ &= \frac{B}{1-\gamma} \left( z^{1-\gamma} - z_0^{1-\gamma} \right) \int_{x=x_{min}}^{x=x_{max}} x^{\gamma-1-\beta} \, \mathrm{d} x \\ &= \frac{B}{1-\gamma} \frac{x_{max}^{\gamma-\beta} - x_{min}^{\gamma-\beta}}{\gamma-\beta} \left( z^{1-\gamma} - z_0^{1-\gamma} \right) = C \left( z^{1-\gamma} - z_0^{1-\gamma} \right) \end{align*} which would directly imply $$ PDF_Z(z) = \frac{\mathrm{d}}{\mathrm{d} z} CDF_Z(z) = C \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{1-\gamma} - z_0^{1-\gamma} \right) = C z^{-\gamma} \mathrm{.}$$
I'm slightly confused about this. It seems to imply that only the exponent $\gamma$ in the power law of the joint distribution has an effect on the exponent of the product distribution (meaning that I should have $\gamma = \alpha$) and that the exponent $\beta$ only has an effect on the normalization factor $C$.
Is there some fundamental flaw in my thinking or is a valid way of arguing that "if the joint distribution of $X$ and $Y$ would look like this it would produce the known distribution for their product $Z = X Y$"?
Since $W:=\ln Z=\ln X +\ln Y$ and $Z$ has pdf $Az^{-\alpha}$ on $z>z_0$ iff $\ln Z$ has pdf $Ae^{-(\alpha -1)w}$ on $w>\ln z_0$, you're trying to express an exponential distribution (in the special case $z_0=1$, which I'll stick to hereafter because the problems it encounters are instructive) as the sum of two independent exponential distributions. (You didn't mention independence, but your PDF Ansatz is of the form $p(x)q(y)$ for functions $p,\,q$, so implies $X,\,Y$ are independent.) This is the wrong choice of Ansatz. In particular $U:=\ln X,\,V:=\ln Y$ each have characteristic functions of the form $(1-\frac{it}{\lambda})^{-1}$, so $U+V$ has a characteristic function of the form $(1-\frac{it}{\lambda_1})^{-1}(1-\frac{it}{\lambda_2})^{-1}$, which isn't of the form $(1-\frac{it}{\lambda})^{-1}$. Obviously, if an inappropriate Ansatz is assumed you'll get nonsensical results.
Although I couldn't find a proof or reference, it's mentioned here that the exponential distribution is infinitely divisible, i.e. $W$ can be represented as the sum of $n$ IIDs for any $n\in\Bbb N$. Thus we can choose $U$ to be the some of $k$ of them and $V$ the sum of $l$ of them, and so one can't be very precise about the distributions of $U,\,V$ or indeed of $X,\,Y$.