定义

20201120011035

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,

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

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