Transformations of random variable: How does $|g’(g^{-1}(y))|$ become $1 \over |{dy \over dx}| $?

Question

In transformation of random variable: $dy \over dx$ = $g’(x)$. Why?

In a homework I found in an older website of my university, I see that they mention that:

For r.v. $Y=g(X)$:

$f_Y(y)={\Sigma f_X(g^{-1}(y)) \over |g’(g^{-1}(y))|}$

Now on this website, where the transformation is monotonous and $Y=g(X)$ has always one solution, they write it as:

${f_X|det({dx \over dy})|}$

My question is: According to what theorem, or material that I’m yet not familiar with, did $|g’(g^{-1}(y))|$ become $1 \over |{dy \over dx}| $ ?

That looks like the derivative of the inverse function's theorem... — DonAntonio, Sep 30 '20 at 22:38
I’ll need someone to elaborate if possible, I see that the theorem talk about multiple variables. And I’m pretty weak here. — Vitali Pom, Sep 30 '20 at 22:45
I see this for example: https://youtu.be/gS0TYC78lnw But what does dx/dy even mean? — Vitali Pom, Sep 30 '20 at 22:58

score 1 · Accepted Answer · answered Sep 30 '20 at 23:10

The inverse function theorem (in one variable) says something like: if $f(x) = y$ and $g = f^{-1}$ then

$$ g'(y) = \frac{1}{f'(x)}. $$

You can check this by implicitly differentiating $g(f(x)) = x$ to get $g'(f(x))f'(x) = 1$ and we said that $f(x) = y$ so this is $g'(y)f'(x) = 1$.

To put it in a different notation: if where we had $y$ as a function of $x$, we now are writing $x = g(y)$ as a function of $y$, then

$$ \frac{dx}{dy} = \frac{dg(y)}{dy} = \frac{1}{\frac{df(x)}{dx}} = \frac{1}{\frac{dy}{dx}}. $$

So if you like:

$$ g'(g^{-1}(x)) = g'(f(x)) = g'(y) = \frac{1}{f'(x)} = \frac{1}{\frac{dy}{dx}}. $$

Oh so $dy \over dx=f’(x)$ cool! Many many thanks!!!! – Vitali Pom Oct 01 '20 at 06:51 — Vitali Pom, Oct 01 '20 at 06:51

1 Answers1