一元高斯分布

最新推荐文章于 2025-03-02 00:16:57 发布

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本文详细探讨了一元高斯分布的性质，包括概率密度函数、期望与方差，以及与标准正态分布的关系。通过解析公式展示了如何计算分布函数，并讨论了其对称性和标准化性质。

$f\left (x; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} \sigma} e ^ { - \dfrac {\left (x - \mu\right ) ^2} {2 \sigma ^2} }$
$f\left (\mu; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} \sigma} , f\left (\mu \pm \sigma; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} \sigma} e ^ { - \dfrac {1} {2} }$
$f\left (\mu + x; \mu, \sigma \right ) = f\left (\mu - x; \mu, \sigma \right )$
$f'\left (x; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} \sigma} e ^ { - \dfrac {\left (x - \mu\right ) ^2} {2 \sigma ^2} } \left [- \dfrac {2 \left (x - \mu\right )} {2 \sigma ^2} \right ] = - \dfrac {x - \mu}{\sigma ^2} f\left (x; \mu, \sigma \right )$
$\dfrac {\partial} {\partial {\mu}} f\left (x; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} \sigma} e ^ { - \dfrac {\left (x - \mu\right ) ^2} {2 \sigma ^2} } \left [ - \dfrac {1} {2 \sigma ^2} \right ] 2 \left (\mu - x \right ) = f\left (x; \mu, \sigma \right ) \dfrac {x - \mu} {\sigma ^2}$
$\dfrac {\partial} {\partial {\sigma}} f\left (x; \mu, \sigma \right ) = \dfrac {1} {\sqrt {2 \pi} } \left \{ \left (- \dfrac {1} {{\sigma} ^2} \right ) e ^ { - \dfrac {\left (x - \mu\right ) ^2} {2 \sigma ^2} } + \dfrac {1} {\sigma} e ^ { - \dfrac {\left (x - \mu\right ) ^2} {2 \sigma ^2} } \left [- \dfrac {\left (x - \mu\right ) ^2} {2} \right ] \left ( -2 \right ) \dfrac {1} { \sigma ^3} \right \}$
$= f\left (x; \mu, \sigma \right ) \left [ - \dfrac {1} { \sigma} + \dfrac {\left (x - \mu\right ) ^2} { \sigma ^3} \right ]$
$= f\left (x; \mu, \sigma \right ) \dfrac {1} { \sigma ^ 3} \left [ \left (x - \mu\right ) ^2 - \sigma ^ 2 \right ]$

$\phi \left (x\right ) = \dfrac {1} {\sqrt {2 \pi}} e ^ { - \dfrac {x ^2} {2} }$
$f\left (x; \mu, \sigma \right ) = \dfrac {1}{\sigma} \phi \left (\dfrac {x - \mu}{\sigma}\right )$
$\phi '\left (x\right ) = - x \phi \left (x\right )$
$f''\left (x; \mu, \sigma \right ) = - \dfrac {1}{\sigma ^2} \left [f\left (x; \mu, \sigma \right ) + \left (x - \mu\right ) f'\left (x; \mu, \sigma \right )\right ]$
$= - \dfrac {1}{\sigma ^2} \left [f\left (x; \mu, \sigma \right ) - \left (x - \mu\right ) \dfrac {x - \mu}{\sigma ^2} f\left (x; \mu, \sigma \right )\right ]$
$=\dfrac {1}{\sigma ^2} f\left (x; \mu, \sigma \right ) \left [ \dfrac {\left (x - \mu\right ) ^2}{\sigma ^2} - 1\right ]$
$=\dfrac {1}{\sigma ^4} f\left (x; \mu, \sigma \right ) \left [ \left (x - \mu\right ) ^2 - \sigma ^2\right ]$
$\phi ''\left (x\right ) = \phi \left (x\right ) \left (x ^2 - 1\right )$
$\Phi \left (x\right ) = \int _{- \infty} ^{x} \phi \left (t \right ) \operatorname {d} t$
$\int _{a} ^{b} f\left (x; \mu, \sigma \right ) \operatorname {d} x = \int _{a} ^{b} \dfrac {1}{\sigma} \phi \left (\dfrac {x - \mu}{\sigma}\right ) \operatorname {d} x$
$= \int _{a} ^{b} \phi \left (\dfrac {x - \mu}{\sigma}\right ) \operatorname {d} \dfrac {x - \mu}{\sigma} = \Phi \left (\dfrac {b - \mu}{\sigma}\right ) - \Phi \left (\dfrac {a - \mu}{\sigma}\right )$
$F \left (x; \mu, \sigma \right ) = \int _{- \infty} ^{x} f\left (t; \mu, \sigma \right ) \operatorname {d} t = \Phi \left (\dfrac {x - \mu}{\sigma}\right )$

$\left [ \Phi \left (+ \infty \right ) \right ] ^2 = \left [ \int _{- \infty} ^{+ \infty} \dfrac {1} {\sqrt {2 \pi}} e ^ { - \dfrac {x ^2} {2} } \operatorname {d} x \right] ^2$
$= \int _{- \infty} ^{+ \infty}\int _{- \infty} ^{+ \infty} \dfrac {1} {2 \pi} e ^ { - \dfrac {x ^2 + y ^2} {2} } \operatorname {d} x \operatorname {d} y$
$= \int _{0} ^{2 \pi} \operatorname {d} \theta \int _{0} ^{+ \infty} \dfrac {1} {2 \pi} e ^ { - \dfrac {r ^2} {2} } r \operatorname {d} r = 1$
$\Rightarrow \Phi \left (+ \infty \right ) = 1$
$\Phi \left (- \infty \right ) = \lim \limits_ {a \to - \infty} \Phi \left (a \right ) = \Phi \left (0 \right ) - \lim \limits_ {a \to - \infty} \int _{a} ^{0} \phi \left (x \right ) \operatorname {d} x = \Phi \left (0 \right ) - \Phi \left (0 \right ) = 0$
$\Phi \left (-x \right ) = \int _{- \infty} ^{-x} \phi \left (t \right ) \operatorname {d} t$
$= - \int _{+ \infty} ^{x} \phi \left (-t \right ) \operatorname {d} t$
$= \int _{x} ^{+ \infty} \phi \left (t \right ) \operatorname {d} t$
$= 1 - \Phi \left (x \right )$
$\Rightarrow \Phi \left (x \right ) + \Phi \left (-x \right ) = 1$
$\Phi \left (0 \right ) = \dfrac {1} {2}$
$E \left (X \right ) = \int _{- \infty} ^{+ \infty} x f\left (x; \mu, \sigma \right ) \operatorname {d} x = \int _{- \infty} ^{+ \infty} \left (x - \mu \right ) f\left (x; \mu, \sigma \right ) \operatorname {d} x + \mu = \mu$
$\operatorname {Var} (X) = E \left (X - E X \right ) ^2$
$= \int _{- \infty} ^{+ \infty} \left (x - \mu \right ) ^2 f\left (x; \mu, \sigma \right ) \operatorname {d} x$
$= \int _{- \infty} ^{+ \infty} \left (x - \mu \right ) ^2 \dfrac {1}{\sigma} \phi \left (\dfrac {x - \mu}{\sigma}\right ) \operatorname {d} x$
$= \sigma ^2 \int _{- \infty} ^{+ \infty} x ^2 \phi \left (x \right ) \operatorname {d} x$
$= \sigma ^2$
其中 $\int _{- \infty} ^{+ \infty} x ^2 \phi \left (x \right ) \operatorname {d} x = \int _{- \infty} ^{+ \infty} x ^2 \dfrac {1} {\sqrt {2 \pi}} e ^ { - \dfrac {x ^2} {2} } \operatorname {d} x$
$= - \dfrac {1} {\sqrt {2 \pi}} \left [ x e ^ { - \dfrac {x ^2} {2} } \Bigg \vert _{- \infty} ^{+ \infty} - \int _{- \infty} ^{+ \infty} e ^ { - \dfrac {x ^2} {2} } \operatorname {d} x \right ] = 1$

$\forall 0 \le a \le 1, \begin{cases} 1 - \Phi \left (z_{a} \right )= a, \\ 1 - \Phi \left (z_{1- a} \right )= 1- a, \\ \Phi \left (z_{a} \right )= 1 - a , \\ \Phi \left (z_{1- a} \right ) = a, \\ \Phi \left (- z_{a} \right )= a , \\ \Phi \left (- z_{1- a} \right ) = 1 - a, \end{cases}$
$z_{1- a} = - z_{a} , 0 \le a \le 1$
$z_{\dfrac {1} {2}} = 0$
$z_{0} = + \infty$
$z_{1} = - \infty$
$z_{1- a} = - z_{a}, 0 \le a \le 1$