PDF of $Y=g(X)$ when $X$ is absolutely continuous and $g$ is strictly increasing and continuously differentiable

Question

Let $X$ be an absolutely continuous random variable with density function $f_{X}$. Suppose that $g$ is strictly increasing and continuously differentiable. Derive the PDF of the random variable $Y=g(X).$

It is given as hint to find the CDF of $Y$ first and then differentiate it to find the PDF. So I try that for a start \begin{align*} F_{Y}(y) = \mathsf{P}(Y\leq y) &=\mathsf{P}(g(X)\leq y) \\ & = \mathsf P(X \leq g^{-1}(y)) \\ & = F_{X}(g^{-1}(y)). \end{align*}

And differentiating with respect to $y$ on both sides yields \begin{align*} f_{Y}(y)=\frac{d}{dy}F_{Y}(y) &=\frac{d}{dy}F_{X}(g^{-1}(y)) \\ &=f_{X}(g^{-1}(y))\cdot\frac{1}{g'(y)}, \end{align*} where in the final step I used the inverse function theorem for $\mathbb{R}$.

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Is this the right approach to the problem? If it is not, could you please provide a hint to put me in the right direction?

Any feedback is much appreciated. Thank you for your time.

Answer 1

Is this the right approach to the problem?

Yes, that is quite okay.   Your approach is appropriate, and the working correct.

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When $g$ is strictly increasing and continuously differentiable, and $X$ an absolutely continuous random variable with density function $f_X$, then indeed we have the result you discovered.

$$f_{g(X)}(y) =\dfrac 1{g'(y)} f_X(g^{-1}(y))$$

Good work.