> [!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 [[Integrable Continuous Real-Valued Random Variable|integrable]] and its [[Expectation of Integrable Continuous Real-Valued Random Variable|expectation]] is given by $\mathbb{E}[X]=\frac{1}{\lambda}.$ **Proof**: By definition, $\begin{align} \mathbb{E}[X] &= \int_{0}^{\infty} \lambda e^{-\lambda x} x \, dx \\ &= \lim_{ n \to \infty } \left[ -xe^{-\lambda x} - \frac{1}{\lambda} e^{-\lambda x} \right]_{0}^{n} = \frac{1}{\lambda} \end{align}$ # Applications By [[Variance of Exponential Distribution]], if $X\sim \text{Exp}(\lambda)$ then $\text{Var}(X)= \frac{1}{\lambda^{2}}.$