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The Wiki article on PDFs describes the transformations of random variables.

In deriving the formula (one-dimensional), the article states "This follows from the fact that the probability contained in a differential area must be invariant under change of variables."

$$|f_Y(y)dy|=|f_X(x)dx|$$

I do not understand what "probability contained in a differential area" means and why it must be invariant under transformation.

Wikipedia: Change of Variable

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It means that if transformation is $Y=g(X)$, we should have $$P(X=x)=P(g(X)=g(x))=P(Y=g(x))=P(Y=y)$$ where, $g(x)=y$, where $g$ is monotonic.

Since these are continuous random variables, $P(X=x)$ is normally $0$, because it's the limit $$P(X=x) = \lim_{dx\rightarrow 0} f_X(x)|dx|$$

i.e. the integral under the curve while the limit goes to $0$. However, at infinitesimal scale, the probabilities, i.e. $P(x<X<x+dx)$ and $P(y<Y<y+dy)$ must be equal. These are $f_X(x)|dx|$ and $f_Y(y)|dy|$ respectively.

A better way to think in this scale is to have the approximation $P(X=x)\approx f_X(x)|dx|$.

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  • $\begingroup$ Thanks for the reply. Why do you take the absolute over dx, |dx|? $\endgroup$ Dec 15, 2020 at 22:33
  • $\begingroup$ Because we're interested in the area/length of the infinitesimal region. For example, if $Y=-X$, then $dy=-dx$. Probabilities can't be negative, we need to remove the sign. $\endgroup$
    – gunes
    Dec 16, 2020 at 5:21
  • $\begingroup$ @CarolEisen is the answer ok for you? $\endgroup$
    – gunes
    Dec 18, 2020 at 20:07

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