The Distribution Function Method With Many-to-One Transformations

The distribution function method still works when $g$ is not one-to-one — we just have to identify every value of $X$ that produces $Y \le y.$

For a transformation $Y = g(X),$ we compute

$$F_Y(y) = P(g(X) \le y) = P\!\left(X \in \{x : g(x) \le y\}\right),$$

then differentiate to obtain $f_Y(y) = F_Y'(y).$ When $g$ is many-to-one, the preimage $\{x : g(x) \le y\}$ is a union of intervals, not a single one.

The most common many-to-one example is $Y = X^2.$ For $y \ge 0,$

$$\{x : x^2 \le y\} = \left[-\sqrt{y},\, \sqrt{y}\right],$$

so

$$F_Y(y) = P\!\left(-\sqrt{y} \le X \le \sqrt{y}\right) = F_X(\sqrt{y}) - F_X(-\sqrt{y}).$$

For $y < 0,$ we have $F_Y(y) = 0$ since $X^2 \ge 0.$

To illustrate, suppose $X \sim \textrm{Uniform}(-3,3),$ so $F_X(x) = \dfrac{x+3}{6}$ for $x \in [-3,3].$ Then for $y \in [0,9],$

$$F_Y(y) = \frac{\sqrt{y}+3}{6} - \frac{-\sqrt{y}+3}{6} = \frac{\sqrt{y}}{3},$$

and differentiating gives $f_Y(y) = \dfrac{1}{6\sqrt{y}}$ on $(0,9].$