Intuition behind product distribution pdf Say we have two distributions $X$ and $Y$. I know that the pdf of the distribution $Z = X + Y$ is given by:
$f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx$
The intuition is that you sum up the probabilities of all possible $x$, $z-x$ pairs. However, now I want to find the pdf of $Z = XY$. Using the same intuition as above, I expect it to be: 
$f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_Y(\frac{z}{x})dx$
However, wikipedia says that the correct pdf is:
$f_Z(z) = \int_{-\infty}^{\infty}\frac{1}{|x|}f_X(x)f_Y(\frac{z}{x})dx$
Where is the extra $\frac{1}{|x|}$ term coming from? What is the intuition? Thanks.
 A: Assuming independence of $X$ and $Y$, we start with deriving the cdf for $Z$: 
$$F_Z(z)=P(XY \leq z).$$
Notice that $(XY \leq z)$ can occur in two ways: 1) $(Y \leq z/X, X >0)$ or 2) $(Y \geq z/X, X <0).$ Since these two possibilities are mutually exclusive events, we can write 
$$F_Z(z) = P(Y \leq z/X, X >0)+ P(Y \geq z/X, X <0).$$
It follows
$$F_Z(z)=\int_{0}^{\infty} \int_{-\infty}^{z/x} f_X(x)f_Y(y)dy dx+\int_{-\infty}^{0} \int_{z/x}^{\infty} f_X(x)f_Y(y)dy dx $$
$$= 
\int_{0}^{\infty} f_X(x) \int_{-\infty}^{z/x} f_Y(y)dy dx+\int_{-\infty}^{0} f_X(x) \int_{z/x}^{\infty} f_Y(y)dy dx$$
$$= 
\int_{0}^{\infty} f_X(x) F_Y(z/x) dx+\int_{-\infty}^{0} f_X(x)(1-F_Y(z/x))dx.$$
Using the fact that $f_Z(z)=F_Z'(z)$, we get 
$$f_Z(z)=\frac{d}{dz}\int_{0}^{\infty} f_X(x) F_Y(z/x) dx+ \frac{d}{dz}\int_{-\infty}^{0} f_X(x)(1-F_Y(z/x))dx$$
$$ =\int_{0}^{\infty} f_X(x) f_Y(z/x)\frac{1}{x} dx+\int_{-\infty}^{0} f_X(x)f_Y(z/x)\frac{-1}{x}dx.$$
In the first term above, $x>0$, so $\frac{1}{x}=\frac{1}{|x|}$. In the second term above, $x<0$, so $\frac{-1}{x}=\frac{1}{|x|}$. Using this, we get
$$ f_Z(z)=\int_{0}^{\infty} f_X(x) f_Y(z/x)\frac{1}{|x|} dx+\int_{-\infty}^{0} f_X(x)f_Y(z/x)\frac{1}{|x|}dx.$$
Combining the two integrals, we get
$$f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z/x)\frac{1}{|x|} dx.$$
