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}xlambda 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} + 2int_{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}}.
]