指数分布

定义

指数分布的期望

\[EX = \frac{1}{\lambda} \]

证明

\[EX = \int_{-\infty}^{+\infty}xf(x)dx = \int_{0}^{+\infty}x\lambda e^{-\lambda x}dx = -\int_{-0}^{+\infty}xde^{-\lambda x} = \int_{0}^{+\infty}e^{-\lambda x}dx = \frac{1}{\lambda } \]

指数分布的方差

\[DX = \frac{1}{\lambda ^ 2} \]

证明

\[EX^{2} = \int_{-\infty }^{+\infty }g(x)f(x)dx = \int_{0}^{+\infty }x^{2}\lambda e^{-\lambda x}dx = -\int_{0}^{+\infty }x^{2}de^{-\lambda x} \]

\[= -x^2e^{-\lambda x}|_0^{+\infty} + 2\int_{0}^{+\infty}xe^{-\lambda x}dx = \frac{2}{\lambda }\int_{0}^{+\infty}e^{-\lambda x}dx = \frac{2}{\lambda ^{2}}, \]

所以，

\[DX = EX^{2} - (EX)^{2} = \frac{2}{\lambda ^{2}} - (\frac{1}{\lambda})^{2} = \frac{1}{\lambda ^{2}}. \]

标签：infty,frac,int,EX,dx,指数分布,lambda
来源： https://www.cnblogs.com/fanlumaster/p/14008895.html