Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Mean

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

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

Then, the mean or expected value of $X$ is

$\label{eq:norm-mean} \mathrm{E}(X) = \mu \; .$

Proof: The expected value is the probability-weighted average over all possible values:

$\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .$

With the probability density function of the normal distribution, this reads:

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

Substituting $z = x -\mu$, we have:

$\label{eq:norm-mean-s2} \begin{split} \mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty-\mu}^{+\infty-\mu} (z + \mu) \cdot \exp \left[ -\frac{1}{2} \left( \frac{z}{\sigma} \right)^2 \right] \, \mathrm{d}(z + \mu) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{+\infty} (z + \mu) \cdot \exp \left[ -\frac{1}{2} \left( \frac{z}{\sigma} \right)^2 \right] \, \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \left( \int_{-\infty}^{+\infty} z \cdot \exp \left[ -\frac{1}{2} \left( \frac{z}{\sigma} \right)^2 \right] \, \mathrm{d}z + \mu \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( \frac{z}{\sigma} \right)^2 \right] \, \mathrm{d}z \right) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \left( \int_{-\infty}^{+\infty} z \cdot \exp \left[ -\frac{1}{2 \sigma^2} \cdot z^2 \right] \, \mathrm{d}z + \mu \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2 \sigma^2} \cdot z^2 \right] \, \mathrm{d}z \right) \; . \end{split}$

The general antiderivatives are

$\label{eq:exp-erf-anti-der} \begin{split} \int x \cdot \exp \left[ -a x^2 \right] \mathrm{d}x &= -\frac{1}{2a} \cdot \exp \left[ -a x^2 \right] \\ \int \exp \left[ -a x^2 \right] \mathrm{d}x &= \frac{1}{2} \sqrt{\frac{\pi}{a}} \cdot \mathrm{erf} \left[ \sqrt{a} x \right] \end{split}$

where $\mathrm{erf}(x)$ is the error function. Using this, the integrals can be calculated as:

$\label{eq:norm-mean-s3} \begin{split} \mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi} \sigma} \left( \left[ -\sigma^2 \cdot \exp \left[ -\frac{1}{2 \sigma^2} \cdot z^2 \right] \right]_{-\infty}^{+\infty} + \mu \left[ \sqrt{\frac{\pi}{2}} \sigma \cdot \mathrm{erf} \left[ \frac{1}{\sqrt{2} \sigma} z \right] \right]_{-\infty}^{+\infty} \right) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \left( \left[ \lim_{z \to \infty} \left( -\sigma^2 \cdot \exp \left[ -\frac{1}{2 \sigma^2} \cdot z^2 \right] \right) - \lim_{z \to -\infty} \left( -\sigma^2 \cdot \exp \left[ -\frac{1}{2 \sigma^2} \cdot z^2 \right] \right) \right] \right. \\ &\hphantom{\sqrt{2 \pi}\sigma \;} + \mu \left. \left[ \lim_{z \to \infty} \left( \sqrt{\frac{\pi}{2}} \sigma \cdot \mathrm{erf} \left[ \frac{1}{\sqrt{2} \sigma} z \right] \right) - \lim_{z \to -\infty} \left( \sqrt{\frac{\pi}{2}} \sigma \cdot \mathrm{erf} \left[ \frac{1}{\sqrt{2} \sigma} z \right] \right) \right] \right) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \left( [0 - 0] + \mu \left[ \sqrt{\frac{\pi}{2}} \sigma - \left(- \sqrt{\frac{\pi}{2}} \sigma \right) \right] \right) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \mu \cdot 2 \sqrt{\frac{\pi}{2}} \sigma \\ &= \mu \; . \end{split}$
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Metadata: ID: P15 | shortcut: norm-mean | author: JoramSoch | date: 2020-01-09, 15:04.