正态分布

正态分布的性质 Properties of Normal Distributions

正态分布的定义 Definition

Definition and p.d.f. A random X has the normal distribution with mean $\mu$ and variance $\sigma^2$ ($-\infty<\mu<\infty$ and $\sigma > 0$) if X has a contimuous distribution with the following p.d.f.
$$f(x|\mu,\sigma^2)=\frac{1}{(2\pi)^{\frac{1}{2}}\sigma}e^{-\frac{1}{2}(\frac{(x-\mu)}{\sigma})^2}\text{for} -\infty<x<\infty$$

Theorem $f(x|\mu,\sigma^2)=\frac{1}{(2\pi)^{\frac{1}{2}}\sigma}e^{-\frac{1}{2}(\frac{(x-\mu)}{\sigma})^2}\text{for} -\infty < x< \infty$ is a p.d.f.

$$\int^{\infty}_{-\infty}f(x|\mu,\sigma^2)dx=\int^{\infty}_{-\infty}\frac{1}{(2\pi)^{1/2}}e^{-\frac{1}{2}y^2}dy\\ \text{we shall now let:}\\ I=\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}dy$$

\begin {aligned} I^2&=I\times I=\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}dy \cdot \int^{\infty}_{-\infty}e^{-\frac{1}{2}z^2}dz\\ &=\int^{\infty}_{-\infty} \int^{\infty}_{-\infty}e^{-\frac{1}{2}(y^2+z^2)}dydz\\ \text{to the polar coordinates } r \text{ and } \theta :\\ I^2&=\int^{2\pi}_{0} \int^{\infty}_{0}e^{-\frac{1}{2}(r^2)}rdrd\theta \\ \text{substitute }v=r^2/2\\ &\int^{\infty}_{0}e^{-v}dv=1 \end{aligned}

正态分布的距生成函数 m.g.f.

m.g.f. 一旦得到相应的均值和方差就非常简单了。

Theorem Moment Generating Function.The m.g.f. of the distribution with p.d.f. given by upside is
\begin{aligned} \psi(t)&=e^{\mu t+\frac{1}{2}\sigma^2t^2}&\text{ for }-\infty<t<\infty \end{aligned}

\begin{aligned} \psi(t)&=E(e^{tX})=\int^{\infty}_{-\infty}\frac{1}{(2\pi)^{1/2}}e^{tx-\frac{(x-\mu)^2}{2\sigma^2}}dx\\ \text{square inside the brackets:}\\ tx-\frac{(x-\mu)^2}{2\sigma^2}&=\mu t+\frac{1}{2}\sigma^2t^2-\frac{[x-(\mu+\sigma^2t)]^2}{2\sigma^2}\\ \text{Therefore:}\\ \psi(t)&=Ce^{\mu t+\frac{1}{2}\sigma^2t^2}\\ \text{where: }\\ C&=\int^{\infty}_{-\infty}\frac{1}{(2\pi)^{1/2}\sigma}e^{-\frac{[x-(\mu+\sigma^2t)]^2}{2\sigma^2}}dx \end{aligned}

正态分布的均值和方差 Mean and Variance

Theorem Mean and Variance.The mean and variance of the distribution with p.d.f. given by definition upside are $\mu$ and $\sigma^2$ ,repectively.

\begin{aligned} \psi'(t)&=(\mu+\sigma^2t)e^{\mu t+\frac{1}{2}\sigma^2t^2}\\ \psi”(t)&=([\mu+\sigma^2t]^2+\sigma^2)e^{\mu t +\frac{1}{2}\sigma^2t^2}\\ \text{Plugging } t=0\\ E(X)&=\psi'(0)=\mu \\ Var(X)&=\psi”(0)-[\psi'(0)]^2=\sigma^2 \end{aligned}

正态分布的形状 the Shapes of Normal Distribution

1. 均值和中值都是 $\mu$
2. $f(x|\mu,\sigma^2)$ 在 $x=\mu$ 时得到最大值。
3. 二次求导后在 $\mu \pm \sigma$ 处为0，为曲线拐点。

线性变换 Linear Transformations

Theorem If $X$ has the normal distribution with mean $\mu$ and variance $\sigma^2$ and if $Y =aX+b$ ,where $a$ and $b$ are given constants and $a \neq 0$ ,then $Y$ has the normal distribution with mean $a\mu+b$ and variance $a^2\sigma^2$ .

$$\psi_Y(t)=e^{bt}\psi(at)=e^{(a\mu+b)t+\frac{1}{2}a^2\sigma^2t^2} \text{ for }-\infty <t<\infty$$

\begin{aligned} \psi(t)&=e^{\mu t+\frac{1}{2}\sigma^2t^2}&\text{ for }-\infty<t<\infty \end{aligned}

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