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 cumulative distribution function of a strictly increasing function as

$\label{eq:cdf-sifct} F_Y(y) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; y < \mathrm{min}(\mathcal{Y}) \\ F_X(g^{-1}(y)) \; , & \text{if} \; y \in \mathcal{Y} \\ 1 \; , & \text{if} \; y > \mathrm{max}(\mathcal{Y}) \; . \end{array} \right.$ $\label{eq:norm-cdf} F_X(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}t \; .$

Applying \eqref{eq:cdf-sifct} to \eqref{eq:norm-cdf}, we have:

$\label{eq:Z-cdf-s1} \begin{split} F_Z(z) &\overset{\eqref{eq:cdf-sifct}}{=} F_X(g^{-1}(z)) \\ &\overset{\eqref{eq:norm-cdf}}{=} \int_{-\infty}^{\sigma z + \mu} \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}t \; . \end{split}$

Substituting $s = (t - \mu)/\sigma$, such that $t = \sigma s + \mu$, we obtain

$\label{eq:Z-cdf-s2} \begin{split} F_Z(z) &= \int_{(-\infty - \mu)/\sigma}^{([\sigma z + \mu] - \mu)/\sigma} \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{(\sigma s + \mu)-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}(\sigma s + \mu) \\ &= \int_{-\infty}^{z} \frac{\sigma}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} s^2 \right] \, \mathrm{d}s \\ &= \int_{-\infty}^{z} \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} s^2 \right] \, \mathrm{d}s \end{split}$

which is the cumulative distribution function of the standard normal distribution.

Sources:

Metadata: ID: P111 | shortcut: norm-snorm | author: JoramSoch | date: 2020-05-26, 23:01.