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It is well known that being $X$ and $Y$ two independent random variables with distributions $f_X(x)$ and $f_Y(y)$, respectively, then the probability distribution of the multiplicative function $z = xy$ is given by $$ f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z/x)\frac{1}{|x|}dx, $$ where $\frac{1}{|x|}$ is the Jacobian of the transformation.

What if $X$ and $Y$ are not independent, in such a way we cannot write the probability distributions by separate functions? We need to use the joint distribution probability $f_{X,Y}(x,y)$. So, will the probability distribution of $z=xy$ be $$ f_Z(z)=\int_{-\infty}^{\infty}f_{X,Y}(x,z/x)\frac{1}{|x|}dx, $$ that is, the only difference is that we cannot separate the pdfs of $x$ and $y$ in the integral?

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Yes, this is the only difference - though sometimes you can simplify the calculation if you can phrase the dependency as $f_{X,Y}=f_X(x)\cdot f_Y(y|x)$, using the conditional probability density function, yielding: $$ f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z/x | x)\frac{1}{|x|}dx, $$ (In the case of independence, $f_Y(z/x | x) = f_Y(z/x)$, returning to your first equation).

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