Let X be a continuous random variable with cumulative distribution function $F_X(x)=P(X<x)$ and density function $f_x(x)=F_X'(x)$. Let g(x) be another real function, then Y = g(x) is also a continuous random variable.
I'm trying to understand the following statement:
If g is monotonous with inverse function $g^{-1}$ and both g and $g^{-1}$ are differentiable, then:
$f_Y(x)=f_X(g^{-1}(x))|(g^{-1}(x))'|=\frac{f_X(g^{-1}(x))}{|g'(x)|}$
Isn't this just the chain rule, why is $f_X(g^{-1}(x))$ divided by $|g'(x)|$? Shouldn't it be multipled by the derivative of $g^{-1}(x)$?
Note that $$ |g^{-1} | \ \neq \frac{1}{|g|}. $$ Generally, if $Y = g(X)$, and $g$ is monotone increasing function then $$ F_Y(y) = P(Y\le y) = P(g(X)\le y)=P(X\le g^{-1}(y)) = F_X(g^{-1}(y)), $$ hence using the chain rule you can get the density, $$ f_Y(y) =F'_Y(y)= \frac{d}{dy}F_X(g^{-1}(y)) = f_X(g^{-1}(y))\frac{d}{dy}g^{-1}(y)). $$