> [!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 [[Integrable Continuous Real-Valued Random Variable|integrable]] and its [[Expectation of Integrable Continuous Real-Valued Random Variable|expectation]] is given by $\mathbb{E}[X]=0.$ **Proof**: We have $\int_{0}^{\infty} \frac{1}{\sqrt{2\pi}}x e^{-\frac{x^{2}}2} \, dx = \lim_{ z \to \infty } \left[ \frac{1}{\sqrt{2\pi}}\cdot -e^{-\frac{x^2}2} \right]_{0}^{\infty}=\frac{1}{\sqrt{ 2\pi }} $By symmetry, $\int_{-\infty}^{0} \frac{1}{\sqrt{2\pi}}x e^{-\frac{x^{2}}2} \, dx =-\frac{1}{\sqrt{ 2\pi }}$Thus $\mathbb{E}[X]=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}x e^{-\frac{x^{2}}2} \, dx =0.$ # Applications By [[Variance of Standard Normal Distribution]], if $X\sim\mathcal{N}(0,1)$ then $\text{Var}(X)=\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2}=1.$