> [!NOTE] Theorem > Let $a<b.$ Let $X\sim \text{Exp}(\lambda)$ where $\text{Exp}(\lambda)$ denotes the [[Exponential Distribution|exponential distribution]] with parameter $\lambda.$ 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 $\text{Var}(X)=\frac{1}{\lambda^{2}}.$ **Proof**: By [[Expectation of Real-Valued Function of Continuous Real-Valued Random Variable]], $\mathbb{E}[X^{2}] = \int_{0}^{\infty} x^{2} \lambda e^{-\lambda x} \, dx= \lim_{z\to+\infty}\left[-x^2e^{-\lambda x}-\frac{2x}\lambda e^{-\lambda x}-\frac2{\lambda^2}e^{-\lambda x}\right]_0^z=\frac2{\lambda^2}. $ Thus by [[Expectation of Exponential Distribution]] and [[Alternative Formula for Variance of a Square-Integrable Continuous Real-Valued Random Variable]], $\text{Var}(X)=\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2} = \frac{2}{\lambda^{2}}-\frac{1}{\lambda^{2}}= \frac{1}{\lambda^{2}}.$