Distribution function technique

Suppose $X_{1}, \dots, X_{n}$ is a set of continuous random variables with a known joint pdf $f_{X_{1}, \dots X_{n}} (x_{1}, \dots, x_{n})$ and

Y = h (X_{1}, \dots X_{n})

Given that $X_{1}, \dots X_{n}$ are continuous, $Y$ is continuous, we can find the pdf of $Y$ by first getting its cdf

F_{Y} (y) = P (Y \leq y) = P (h (X_{1}, \dots, X_{n}) \leq y)

and then derivating it to get the pdf

f_{Y} (y) = \frac{\partial F_{Y} (y)}{\partial y}

When working with discrete variables, we usually work on the pmf directly

Examples

Example with $U = 1 - e^{- x / β}$ , $X \sim E x p o n e n t i a l$

x \in (0, \infty), e^{- x / β} \in (0, 1), 1 - e^{- x / β} \in (0, 1)

\begin{aligned} F_{U} (u) & = P (U \leq u) = P (1 - e^{- x / β} \leq u) \\ = P (e^{- x / β} \geq 1 - u) \\ = P (\frac{- x}{β} \geq \ln (1 - u)) \\ = P (x \leq - β \ln (1 - u)) \\ = \int_{0}^{- β \ln (1 - u)} \frac{1}{β} e^{- x / β} d x \\ = {[- e^{- x / β} |}_{0}^{- β \ln (1 - u)} \\ = 1 - e^{- - β \ln (1 - u) / β} \\ = 1 - e^{\ln (1 - u)} \\ = 1 - (1 - u) \\ = u \\ f_{U} (u) = \frac{\partial F_{U}}{\partial u} & = \frac{\partial}{\partial u} u \\ f_{U} (u) & = 1, 0 < u < 1 \end{aligned}

The random variable $U$ follows a uniform distribution, in the interval $(0, 1)$