> [!NOTE] Theorem > Let $X\sim \mathcal{N}(0,1)$ where $\mathcal{N}(0,1)$ denotes the [[Standard Normal Distribution|standard normal distribution]]. Then $X$ is [[Square-Integrable Continuous Real-Valued Random Variable|square-integrable]] and its [[Variance of a Square-Integrable Continuous Real-Valued Random Variable|variance]] is given by $\text{Var}(X)=1.$ **Proof**: By [[Expectation of Real-Valued Function of Continuous Real-Valued Random Variable]], $\begin{aligned} \mathbb{E}[X^{2}]& =\int_{-\infty}^{+\infty}x^2\frac{e^{-x^2/2}}{\sqrt{2\pi}}\mathrm{d}x \\ &=2\cdot\frac1{\sqrt{2\pi}}\int_0^{+\infty}x\cdot(xe^{-x^2/2})\mathrm{d}x \\ &=\frac2{\sqrt{2\pi}}\lim_{u\to+\infty}\left[-xe^{-x^2/2}+\int_0^ue^{-x^2/2}\mathrm{d}x\right]_0^u\\ &=\frac2{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-x^2/2}\mathrm{d}x=1. \end{aligned}$ By [[Expectation of Standard Normal Distribution]] and [[Alternative Formula for Variance of a Square-Integrable Continuous Real-Valued Random Variable]], $\text{Var}(X)=\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2}=1-0=1.$