Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Variance

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

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

Then, the variance of $X$ is

\[\label{eq:norm-var} \mathrm{Var}(X) = \sigma^2 \; .\]

Proof: The variance is the probability-weighted average of the squared deviation from the mean:

\[\label{eq:var} \mathrm{Var}(X) = \int_{\mathbb{R}} (x - \mathrm{E}(X))^2 \cdot f_\mathrm{X}(x) \, \mathrm{d}x \; .\]

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

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

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

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

Now substituting $z = \sqrt{2} \sigma x$, we have:

\[\label{eq:norm-var-s3} \begin{split} \mathrm{Var}(X) &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{+\infty} (\sqrt{2} \sigma x)^2 \cdot \exp \left[ -\frac{1}{2} \left( \frac{\sqrt{2} \sigma x}{\sigma} \right)^2 \right] \, \mathrm{d}(\sqrt{2} \sigma x) \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot 2 \sigma^2 \cdot \sqrt{2} \sigma \int_{-\infty}^{+\infty} x^2 \cdot \exp \left[ -x^2 \right] \, \mathrm{d}x \\ &= \frac{2 \sigma^2}{\sqrt{\pi}} \int_{-\infty}^{+\infty} x^2 \cdot e^{-x^2} \, \mathrm{d}x \; . \end{split}\]

Since the integrand is symmetric with respect to $x = 0$, we can write:

\[\label{eq:norm-var-s4} \mathrm{Var}(X) = \frac{4 \sigma^2}{\sqrt{\pi}} \int_{0}^{\infty} x^2 \cdot e^{-x^2} \, \mathrm{d}x \; .\]

If we define $z = x^2$, then $x = \sqrt{z}$ and $\mathrm{d}x = 1/2 \, z^{-1/2} \, \mathrm{d}z$. Substituting this into the integral

\[\label{eq:norm-var-s5} \mathrm{Var}(X) = \frac{4 \sigma^2}{\sqrt{\pi}} \int_{0}^{\infty} z \cdot e^{-z} \cdot \frac{1}{2} z^{-\frac{1}{2}} \, \mathrm{d}z = \frac{2 \sigma^2}{\sqrt{\pi}} \int_{0}^{\infty} z^{\frac{3}{2}-1} \cdot e^{-z} \, \mathrm{d}z\]

and using the definition of the gamma function

\[\label{eq:gam-fct} \Gamma(x) = \int_{0}^{\infty} z^{x-1} \cdot e^{-z} \, \mathrm{d}z \; ,\]

we can finally show that

\[\label{eq:norm-var-s6} \mathrm{Var}(X) = \frac{2 \sigma^2}{\sqrt{\pi}} \cdot \Gamma\!\left(\frac{3}{2}\right) = \frac{2 \sigma^2}{\sqrt{\pi}} \cdot \frac{\sqrt{\pi}}{2} = \sigma^2 \; .\]
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Metadata: ID: P18 | shortcut: norm-var | author: JoramSoch | date: 2020-01-09, 22:47.