Why a normal variance is \sigma^2

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Let X be standard normal. Show that \operatorname{Var}(X)=1, and deduce \operatorname{Var}(Y)=\sigma^2 for Y\sim\operatorname{N}(\mu,\sigma^2).

Worked answer (click to show)

The standard normal density satisfies \varphi'(x)=-x\varphi(x), and \mathbb{E} X=0, so \operatorname{Var}(X)=\mathbb{E} X^2. Integrate by parts with u=x and \,\mathrm{d} v=x\varphi(x)\,\,\mathrm{d} x, so v=-\varphi(x). The boundary term x\varphi(x) vanishes at both infinities, and

\mathbb{E} X^2=\int_{-\infty}^{\infty}\varphi(x)\,\,\mathrm{d} x=1.

If Y=\mu+\sigma X with the same X, then \operatorname{Var}(Y)=\sigma^2\operatorname{Var}(X)=\sigma^2.