Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Relationship to standard normal distribution

Theorem: Let $X$ be a random variable following a normal distribution with mean $\mu$ and variance $\sigma^2$:

$\label{eq:X-norm} X \sim \mathcal{N}(\mu, \sigma^2) \; .$

Then, the quantity $Z = (X-\mu)/\sigma$ will have a standard normal distribution with mean $0$ and variance $1$:

$\label{eq:Z-snorm} Z = \frac{X-\mu}{\sigma} \sim \mathcal{N}(0, 1) \; .$

Proof: Note that $Z$ is a function of $X$

$\label{eq:Z-X} Z = g(X) = \frac{X-\mu}{\sigma}$

with the inverse function

$\label{eq:X-Z} X = g^{-1}(Z) = \sigma Z + \mu \; .$

Because $\sigma$ is positive, $g(X)$ is strictly increasing and we can calculate the probability density function of a strictly increasing function as

$\label{eq:pdf-sifct} f_Y(y) = \left\{ \begin{array}{rl} f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; , & \text{if} \; y \in \mathcal{Y} \\ 0 \; , & \text{if} \; y \notin \mathcal{Y} \end{array} \right.$

where $\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace$. With the probability density function of the normal distribution, we have

$\label{eq:pdf-Z} \begin{split} f_Z(z) &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{g^{-1}(z)-\mu}{\sigma} \right)^2 \right] \cdot \frac{\mathrm{d}g^{-1}(z)}{\mathrm{d}z} \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{(\sigma z + \mu)-\mu}{\sigma} \right)^2 \right] \cdot \frac{\mathrm{d}(\sigma z + \mu)}{\mathrm{d}z} \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} z^2 \right] \cdot \sigma \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} z^2 \right] \end{split}$

which is the probability density function of the standard normal distribution.

Sources:

Metadata: ID: P176 | shortcut: norm-snorm2 | author: JoramSoch | date: 2020-10-15, 11:42.