How to calculate the density function of $g(X)$?

Question

Suppose $X$ is a random variable on probability space $(\Omega,\mathcal{F},P)$ and has continuous density $f$. $P(\alpha\le X\le \beta)=1$ and $g$ is a function that is strictly increasing and differentiable on $(\alpha,\beta)$. Then $g(X)$ has density $f(g^{-1}(y))/g'(g^{-1}(y))$ for $y\in(g(\alpha),g(\beta))$ and $0$ otherwise.

How to prove this?

My thought is taking $y\in(g(\alpha),g(\beta))$ then $P(g(x)\le y)=P(x\le g^{-1}(y))= F(g^{-1}(y))$. Here $F(x)$ is the distribution function of $X$, $F(x)=\int_{-\infty}^x f(t) dt$, where $f(t)$ is the density function of $X$. Now I need to get the density function of $g(X)$. How to calculate it from $F(g^{-1}(y))$ ?

Anyone know the answer are welcome to comment!:)

Answer 1

So you now have that $Y = g(X)$ has distribution function $F(g^{-1}(y))$.

Taking derivatives via the chain rule, we have $F^{\prime}(g^{-1}(y))\dfrac{\text{d}}{\text{d}y}[g^{-1}(y)]$ as the density function.

First, $F^{\prime} = f$.

Next, recall that the derivative of an inverse function is $\dfrac{\text{d}}{\text{d}y}[g^{-1}(y)] = \dfrac{1}{g^{\prime}(g^{-1}(y))}$.

Then you're done.