Question:

Given r.v. $X$ with cdf $F(X)$ with pdf $f_X(x)$, $Y = g(X)$, derve the pdf for $Y$. Here, we assume $g$ is monotonically increasing, and both $g^{-1}(y)$ and its derivative wrt. $x$ exists. Since $g$ is monotonically increasing, so $dg^{-1}/dy \ge 0$

\[\begin{align} F_Y(y) &= \mathbb{P}(Y \le y) \\ &= \mathbb{P}(Y \le g(x)) \\ &= \mathbb{P}(X \le g^{-1}(y)) \\ \label{eq:Y_to_X_v1} &= F_X(g^{-1}(y)) \\ \frac{d F_Y}{dy} &= \frac{F_X(g^{-1}(y))}{dx} \frac{dx}{dy} \\ f_Y(y) &= f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy} \end{align}\]

Note,

  • Eq. $\eqref{eq:Y_to_X_v1}$ relies on $g$ being monotonically increasing, so $g^{-1}(y)$ is also monotonically increasing.

In the case when $g(x)$ is monotonically decreasing, then $g^{-1}(x)$ is also monotonically decreasing, and $dg^{-1}/dy \le 0$, which is quite similar

\[\begin{align*} F_Y(y) &= \mathbb{P}(Y \le y) \\ &= \mathbb{P}(Y \le g(x)) \\ &= \mathbb{P}(X \ge g^{-1}(y)) \\ \label{eq:Y_to_X_v2} &= 1 - F_X(g^{-1}(y)) \\ \frac{d F_Y}{dy} &= - \frac{F_X(g^{-1}(y))}{dx} \frac{dx}{dy} \\ f_Y(y) &= - f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy} \end{align*}\]

Merging the two cases together, we have

\[\begin{align*} f_Y(y) &= f_X(g^{-1}(y)) \left| \frac{dg^{-1}(y)}{dy} \right| \end{align*}\]