Variance of the normal distribution

Statement

Let $\mu \in \mathbb{R}$ , let $\sigma^2 > 0$ and let $X$ have the normal distribution $\mathcal{N}(\mu, \sigma^2)$ . Then $\operatorname{Var}(X) = \sigma^2.$

Remarks

Proof: $Z = (X - \mu)/\sigma$ has the law $\mathcal{N}(0, 1)$ , where $\sigma = \sqrt{\sigma^2}$ . Integration by parts gives $E[Z^2] = \int z^2 \varphi(z) \, dz = \int \varphi(z) \, dz = 1$ . Since $E[Z] = 0$ by the mean of the normal distribution, $\operatorname{Var}(Z) = 1$ . The scaling and translation rules then give $\operatorname{Var}(X) = \sigma^2 \operatorname{Var}(Z) = \sigma^2$ . So the two parameters of $\mathcal{N}(\mu, \sigma^2)$ are its mean and its variance.

Tag	`03EESZDB9A`
URI	`m2://03EESZDB9A`
Kind	statement, proposition
Writing	W0 stub
MSC	60E05
Status	This copy of the site has no build data. Run `m2 status 03EESZDB9A`.
Uses	Normal distribution, Variance, Mean of the normal distribution, Variance under scaling, Variance under translation

Tag

03EESZDB9A

URI

m2://03EESZDB9A

Kind

statement, proposition

Writing

W0 stub

MSC

60E05

Status

This copy of the site has no build data. Run m2 status 03EESZDB9A.

Uses

Normal distribution, Variance, Mean of the normal distribution, Variance under scaling, Variance under translation

Variance of the normal distribution

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Remarks

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