Random Variables and Distributions - Key Proof

We present a rigorous derivation of the transformation formula for PDFs, a fundamental result for computing distributions of functions of random variables.

Transformation of PDFs

Theorem

Let $X$ be a continuous random variable with PDF $f_X$ , and let $Y = g(X)$ where $g$ is strictly monotonic (either strictly increasing or strictly decreasing) and differentiable. Then $Y$ has PDF: $f_Y(y) = f_X(g^{-1}(y)) \left|\frac{d}{dy} g^{-1}(y)\right|$

on the range of $Y$ , and $f_Y(y) = 0$ elsewhere.

Proof

We consider two cases based on whether $g$ is increasing or decreasing.

Case 1: $g$ strictly increasing

Since $g$ is strictly increasing and continuous, it has an inverse $g^{-1}$ , and for any $y$ : $P(Y \leq y) = P(g(X) \leq y) = P(X \leq g^{-1}(y))$

The CDF of $Y$ is: $F_Y(y) = P(X \leq g^{-1}(y)) = F_X(g^{-1}(y))$

Differentiating both sides with respect to $y$ using the chain rule: $f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} F_X(g^{-1}(y))$

$= f_X(g^{-1}(y)) \cdot \frac{d}{dy} g^{-1}(y)$

Since $g$ is increasing, $g^{-1}$ is also increasing, so $\frac{d}{dy} g^{-1}(y) > 0$ . Therefore: $f_Y(y) = f_X(g^{-1}(y)) \left|\frac{d}{dy} g^{-1}(y)\right|$

Case 2: $g$ strictly decreasing

For strictly decreasing $g$ : $P(Y \leq y) = P(g(X) \leq y) = P(X \geq g^{-1}(y)) = 1 - F_X(g^{-1}(y))$

The CDF of $Y$ is: $F_Y(y) = 1 - F_X(g^{-1}(y))$

Differentiating: $f_Y(y) = -f_X(g^{-1}(y)) \cdot \frac{d}{dy} g^{-1}(y)$

Since $g$ is decreasing, $g^{-1}$ is also decreasing, so $\frac{d}{dy} g^{-1}(y) < 0$ . Therefore: $f_Y(y) = f_X(g^{-1}(y)) \left|\frac{d}{dy} g^{-1}(y)\right|$

In both cases, we obtain the same formula. □

■

Alternative Form Using Change of Variables

An equivalent formulation uses $x = g^{-1}(y)$ , so $y = g(x)$ :

If we write $h(x) = g^{-1}(x)$ , then by the inverse function theorem: $\frac{dh}{dy} = \frac{1}{g'(h(y))}$

Therefore: $f_Y(y) = f_X(x) \left|\frac{dx}{dy}\right| = \frac{f_X(x)}{|g'(x)|} \bigg|_{x = g^{-1}(y)}$

Example

Linear Transformation: Let $Y = aX + b$ where $a \neq 0$ .

Here $g(x) = ax + b$ , so $g^{-1}(y) = \frac{y-b}{a}$ and $\frac{d}{dy} g^{-1}(y) = \frac{1}{a}$ .

Therefore: $f_Y(y) = f_X\left(\frac{y-b}{a}\right) \cdot \frac{1}{|a|}$

Verification for Normal: If $X \sim \mathcal{N}(\mu, \sigma^2)$ and $Y = \sigma X + \mu$ , then: $f_Y(y) = \frac{1}{\sqrt{2\pi}\sigma} \exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right)$

Indeed, $Y \sim \mathcal{N}(\mu, \sigma^2)$ as expected. ✓

Example

Square Transformation: Let $Y = X^2$ where $X$ has PDF $f_X$ supported on $(0, \infty)$ .

Here $g(x) = x^2$ (increasing on $(0,\infty)$ ), so $g^{-1}(y) = \sqrt{y}$ and: $\frac{d}{dy} g^{-1}(y) = \frac{d}{dy} \sqrt{y} = \frac{1}{2\sqrt{y}}$

Therefore: $f_Y(y) = f_X(\sqrt{y}) \cdot \frac{1}{2\sqrt{y}}, \quad y > 0$

Special Case: If $X \sim \mathcal{N}(0,1)$ , then $Y = X^2 \sim \chi^2_1$ (chi-squared with 1 degree of freedom).

Remark

The transformation formula is fundamental in probability theory and statistics. It generalizes to multivariate settings via the Jacobian determinant and underpins many theoretical results. The absolute value ensures the PDF remains non-negative regardless of whether the transformation is increasing or decreasing.