# 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.

• You have assumed, without explicitly saying so, that $X$ and $Y$ are independent random variables. The intuition is that when we do a change of variables, there is something called a Jacobian that is involved and that $\frac{1}{|x|}$ is in essence the Jacobian. – Dilip Sarwate Mar 23 '19 at 22:21
• Dilips comment pretty much sums it up imo. I'll add that a Jacobian is also required for the $Z=X+Y$ case, it's just that the Jacobian happens to be constant (i.e. $1$) for linear transformations. – knrumsey Mar 24 '19 at 0:11
• A deeper, and therefore perhaps more satisfying reason, than any calculation of a Jacobian is an appeal to symmetry as described at stats.stackexchange.com/a/185709/919. – whuber Mar 24 '19 at 14:26

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 $$f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z/x)\frac{1}{|x|} dx.$$