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 $\mathbb{R}$.
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.
Best Answer
Yes, that is quite okay. Your approach is appropriate, and the working correct.
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.