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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.

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    $\begingroup$ 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. $\endgroup$ Mar 23, 2019 at 22:21
  • $\begingroup$ 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. $\endgroup$
    – knrumsey
    Mar 24, 2019 at 0:11
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Mar 24, 2019 at 14:26

1 Answer 1

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

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