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If $Y$ is a transformation of $X$, then why is it that $|f_X(x) dx| = |f_Y(y) dy|$ instead of simply $f_X(x) dx = f_Y(y) dy$? This should be an intuitively easy question but I'm trying to think of an example for when this would be the case. Is it because $dy$ could "intuitively" be negative? For example, the mapping from $X$ to $-X.$

I just need some clarification here. Thanks!

Reference: http://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables

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Sorry but the question has no meaning whatsoever. You could begin by defining the quite odd notion of $|dx|$. And, since $|\ |$ is usually applied to real numbers or to complex numbers, by defining the object $dx$... Seriously, you might google change of variables, don't you think? –  Did May 10 '11 at 17:08
    
Change of variables in 1D (i.e., single variables) are sometimes done with a negative derivative (and thereby integrating the "wrong"-way round, i.e., from large to low values in the integration region). Multivariable change of variable necessarily involves the absolute value of the Jacobian. –  Fabian May 10 '11 at 17:21
    
The $| |$ notation was used in a wikipedia article on the topic --- I just wanted to make sure I was understanding it properly. The $|dx|$ thrown in the article threw me off a little bit as well. And also it's a lot more efficient to ask here than to google several different sources. Thanks Fabian! –  Glassjawed May 10 '11 at 17:50
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