Index: The Book of Statistical ProofsProbability DistributionsUnivariate continuous distributions ▷ Normal distribution ▷ Cumulative distribution function

Theorem: Let $X$ be a random variable following a normal distribution:

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

Then, the cumulative distribution function of $X$ is

$\label{eq:norm-cdf} F_X(x) = \frac{1}{2} \left[ 1 + \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right) \right]$

where $\mathrm{erf}(x)$ is the error function defined as

$\label{eq:erf} \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} \exp(-t^2) \, \mathrm{d}t \; .$ $\label{eq:norm-pdf} f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \; .$

Thus, the cumulative distribution function is:

$\label{eq:norm-cdf-s1} \begin{split} F_X(x) &= \int_{-\infty}^{x} \mathcal{N}(z; \mu, \sigma^2) \, \mathrm{d}z \\ &= \int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{z-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{x} \exp \left[ -\left( \frac{z-\mu}{\sqrt{2} \sigma} \right)^2 \right] \, \mathrm{d}z \; . \end{split}$

Substituting $t = (z-\mu)/(\sqrt{2} \sigma)$, i.e. $z = \sqrt{2} \sigma t + \mu$, this becomes:

$\label{eq:norm-cdf-s2} \begin{split} F_X(x) &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{(-\infty-\mu)/(\sqrt{2} \sigma)}^{(x-\mu)/(\sqrt{2} \sigma)} \exp(-t^2) \, \mathrm{d}\left( \sqrt{2} \sigma t + \mu \right) \\ &= \frac{\sqrt{2} \sigma}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{\frac{x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\frac{x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{0} \exp(-t^2) \, \mathrm{d}t + \frac{1}{\sqrt{\pi}} \int_{0}^{\frac{x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \\ &= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \exp(-t^2) \, \mathrm{d}t + \frac{1}{\sqrt{\pi}} \int_{0}^{\frac{x-\mu}{\sqrt{2} \sigma}} \exp(-t^2) \, \mathrm{d}t \; . \end{split}$

Applying \eqref{eq:erf} to \eqref{eq:norm-cdf-s2}, we have:

$\label{eq:norm-cdf-s3} \begin{split} F_X(x) &= \frac{1}{2} \lim_{x \to \infty} \mathrm{erf}(x) + \frac{1}{2} \, \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right) \\ &= \frac{1}{2} + \frac{1}{2} \, \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right) \\ &= \frac{1}{2} \left[ 1 + \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right) \right] \; . \end{split}$
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Metadata: ID: P85 | shortcut: norm-cdf | author: JoramSoch | date: 2020-03-20, 01:33.