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Why, if $X $ and $Y $ are two independent'', continuous random variables, described by probability density functions $f_X $ and $f_Y $, then the distribution of $Z = XY$ is

$$f_Z(z) = \int^{\infty}_{-\infty} f_X \left( x \right) f_Y \left( z/x \right) \frac{1}{|x|}\, dx $$

as stated here

What I tried until now:

$$ P[XY \le z ] = P[XY \in B] \text{ with } B := \{(x,y)|x\in R, y \le z/x\}$$ The part with $y \le z/x$ is obviously wrong, but with that attempt I got this near to the solution till now: $$\int_R \int_{(-\infty, z/x)}f_X(x)f_Y(y) dy dx = \int_R \int_{(-\infty, z)}f_X(x)f_Y(y/x) \frac{1}{x} dy dx = {\int_{(-\infty, z)} \int_Rf_X(x)f_Y(y/x) \frac{1}{x} }dx dy $$

Under the wrong assumption I made, the PDF now would be

$$ f_{XY}(y) = \int_Rf_X(x)f_Y(y/x) \frac{1}{x} dx $$

which already is pretty close to where I want to get to

when using the right $B := \{ (x,y)∈R^2|x>0,y≤z/x\}∪\{(x,y)∈R^2|x<0,y≥z/x \} $ as 'Did' suggested, I come to this solution:

$$ P[XY\in B] = \int_{-\infty}^0 \int_z^{\infty} f_X(x)f_Y(y/x)\frac{1}{x} dy dx + \int_0^{\infty} \int_{-\infty}^z f_X(x)f_Y(y/x)\frac{1}{x} dydx $$

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  • $\begingroup$ This is not a "joint" distribution. How would you compute the PDF of XY if you had to? $\endgroup$ – Did Nov 30 '14 at 19:28
  • $\begingroup$ @Did then you should correct it on wikipedia, I copied the text from there $\endgroup$ – songbook Nov 30 '14 at 19:28
  • $\begingroup$ On which WP page? $\endgroup$ – Did Nov 30 '14 at 19:34
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    $\begingroup$ @Did the one I linked at "as stated here". I am at the moment editing my question to add details of what I have done until now (how I tried to compute the PDF as you asked) $\endgroup$ – songbook Nov 30 '14 at 19:35
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    $\begingroup$ @Did I now added everything I have tried until now and deleted the "joint" to avoid confusion $\endgroup$ – songbook Nov 30 '14 at 19:45

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