To understand how the distribution of a random variable changes under a transformation, we use the change of variables formula. Let’s explore this concept with a concrete example involving a Gaussian random variable.

Problem Setup: Suppose $z \sim \mathcal{N}(0, 1)$ is a standard Gaussian random variable, and we apply a transformation $x = z^2$. We want to determine the distribution of $x$, the square of $z$.

Understanding the Transformation: The transformation $x = z^2$ maps $z$ to non-negative values because the square of any real number is always non-negative. This transformation introduces two "folds" since both positive and negative values of $z$ map to the same value of $x$.

Change of Variables Formula: To determine the probability density function (PDF) of $x$, we use the change of variables formula:

$$ p_x(x) = p_z(z) \left| \frac{dz}{dx} \right|, $$

where: $p_z(z)$ is the PDF of the original variable $z$; $\frac{dz}{dx}$ is the derivative (Jacobian matrix) of the transformation $x = z^2$ with respect to $z$. However, since $z$ can take both positive and negative values for the same $x$, we need to account for both branches $z = \pm \sqrt{x}$.

Proof:

Deriving the Distribution: The PDF of $z \sim \mathcal{N}(0, 1)$ is $p_z(z) = \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{z^2}{2} \right)$; For $x = z^2$, the relationship between $x$ and $z$ is $z = \pm \sqrt{x}, \quad \text{and} \quad \frac{dz}{dx} = \frac{1}{2\sqrt{x}}$. Since there are two branches ($z = +\sqrt{x}$ and $z = -\sqrt{x}$), the total density $p_x(x)$ is the sum of contributions from both branches.

Substituting $z$ into the formula, we get

$$ p_x(x) = p_z(\sqrt{x}) \left| \frac{d(\sqrt{x})}{dx} \right| + p_z(-\sqrt{x}) \left| \frac{d(-\sqrt{x})}{dx} \right|. $$

Simplifying, since $p_z(\sqrt{x}) = p_z(-\sqrt{x})$ (symmetry of the Gaussian), and $\frac{d\sqrt{x}}{dx} = \frac{1}{2\sqrt{x}}$, we get:

$$ p_x(x) = 2 \cdot p_z(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}, \quad x \geq 0. $$