I was informed in my last question that the Inverse function theorem:
$$(f^{-1})^{\prime}(f(a))=\cfrac{1}{f^{\prime}(a)}\tag{I*}$$
was needed to show that
$$\rho_x (x)=\rho_\alpha(\alpha)\left|\frac{\mathrm{d}x}{\mathrm{d}\alpha}\right|^{-1}\tag{A}$$
is the same formula as
$$\rho_y(y)=\rho_x(\phi^{-1}(y))\left|\frac{\mathrm{d}\phi^{-1}}{\mathrm{d}y}\right|\tag{B}$$
Where $(\mathrm{A})$ and $(\mathrm{B})$ are formulae that relate the probability densities for random variables $x$, $\alpha$ in $(\mathrm{A})$ and $y$ and $x$ in $(\mathrm{B})$.
It's probably pretty simple to do but I'm not seeing it thus far; So how can I use $(\mathrm{I^{*}})$ to prove equivalence of $(\mathrm{A})$ and $(\mathrm{B})$?
Thank you.
Rewrite B as $$\rho_y(y)=\rho_z(\phi^{-1}(y))\left|\frac{\mathrm{d}\phi^{-1}}{\mathrm{d}y}\right|$$ I just renamed $x$ with $z$ to not mix it with the $x$ in A.
Next, let Let $z=\phi^{-1}(y)$ then $$\frac{\mathrm{d}y}{\mathrm{d}z} =\left( \frac{\mathrm{d}z}{\mathrm{d}y}\right)^{-1}=\left( \frac{\mathrm{d}\phi^{-1}}{\mathrm{d}y} \right)^{-1}$$ subsequently $ \left( \frac{\mathrm{d}y}{\mathrm{d}z} \right)^{-1}=\frac{\mathrm{d}\phi^{-1}}{\mathrm{d}y} $. Note also that $|a^{-1}|=\left|\frac{1}{a}\right|=\frac{1}{|a|}=|a|^{-1}$.
Now substitute this result in A with $x=y$ and $\alpha=z$ to get
$$\rho_y (y)=\rho_x (x)=\rho_\alpha(\alpha)\left|\frac{\mathrm{d}x}{\mathrm{d}\alpha}\right|^{-1}= \rho_z(z)\left|\frac{\mathrm{d}y}{\mathrm{d}z}\right|^{-1} =\rho_z(\phi^{-1}(y))\left|\frac{\mathrm{d}\phi^{-1}\color{blue}{(y)}}{\mathrm{d}y} \right|\tag{1} $$
Note
Due to a question in comment about where is inverse function theorem where used, I add the following
$$\frac{\mathrm{d}z}{\mathrm{d}y}=\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y}\Bigg|_{y=f(a)}= (f^{-1})^{\prime}(f(a))=\cfrac{1}{f^{\prime}(a)} =\cfrac{1}{ \frac{\mathrm{d}f(a)}{\mathrm{d}a} }=\left(\frac {\mathrm{d}f(a)}{\mathrm{d}\color{red}{a}}\right)^{-1}=\left(\frac {\mathrm{d}y}{\mathrm{d}\color{red}{z}}\right)^{-1}\tag{2}$$