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I have some troubles with understanding of product distribution calculation.

Consider a simple example: $f_X(x)$=$\frac{1}{2}$ for $1\le x \le 3$ and $f_Y(y)$=$\frac{1}{4}$ for $2 \le y \le 6$. Find $f_Z(z)$ if $Z=XY$.

To find pdf of $Z=XY$, I'm using the equation:

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

Then, I suppose that $f_X(x)=\frac{1}{2}$ and $f_Y(\frac{z}{x})=\frac{1}{4}$, and getting $$f_Z(z)=\frac{1}{8}\int_1^3 \frac{1}{|x|} dx=\frac{1}{8} ln(3)$$

So, I'm losing $z$ in the right part (guess, this is wrong). In addition, I'm not sure if my integration limits are correct.

May you please show how to solve this particular case? Thanks.

user474290
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2 Answers2

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Your integration needs to reflect the support of the integrand. Since the integrand includes two pdfs, the support of both must be reflected in the integration limits.

As you said, $f_X(x)$ has support for $1<x<3.$ But we need to use the fact that $f_z(z/x)$ has support for $2<z/x < 6.$ This inequality can be arranged to give a pair of conditions on $x$:$$x<z/2$$ and $$x>z/6.$$

So the integral should be $$ \frac{1}{8}\int_{\max(1,z/6)}^{\min(3,z/2)}\frac{dx}{|x|}.$$

Notice that this effectively splits up into two cases: $$2<z <6$$ and $$6 < z < 18.$$

spaceisdarkgreen
  • 58,508
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Look to the supports of your functions.

$\begin{align}f_Z(z) ~&=~ \int_\Bbb R \lvert x^{-1}\rvert~f_X(x)f_Y(z/x)\operatorname d x \\ &=~ \tfrac 18\int_\Bbb R\lvert{ x^{-1}}\rvert\mathbf 1_{1\leq x\leq 3}\mathbf 1_{2\leq z/x\leq 6}\operatorname d x \\ &=~ \tfrac 18\mathbf 1_{2\leq z\leq 18} \int_\Bbb R\lvert{ x^{-1}}\rvert\mathbf 1_{1\leq x\leq 3}\mathbf 1_{z/6\leq x\leq z/2}\operatorname d x \\ &=~ \tfrac 18\mathbf 1_{2\leq z\leq 18} \int_\Bbb R\lvert{ x^{-1}}\rvert\mathbf 1_{\max(1,z/6)\leq x\leq \min (3,z/2)}\operatorname d x \\ &=~ \tfrac 18\mathbf 1_{2\leq z\lt 6} \int_{1}^{z/2} x^{-1}\operatorname d x + \tfrac 18\mathbf 1_{6\leq z\leq 18} \int_{z/6}^3{ x^{-1}}\operatorname d x \end{align}$

Graham Kemp
  • 129,094