概率论习题之标准正态绝对值的期望-Toy模板网

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概率论习题之标准正态绝对值的期望

一、主要注意的点

$E|X|={\sqrt{\frac{2}{\Pi}} }\\ 计算:E|X|=\displaystyle \int^{+\infty}_{-\infty}{|X|f(x)dx}\\ EZ=E|x-\mu|$

二、习题

$X\sim N(0,1),E(|X|)=\sqrt{\frac{2}{\Pi}}$

$X\sim N(3,9),E(|X-3|)=$
处理一般正态的基本操作：1.标准化 2.标准正态
$E|\frac{X-3}{3}|=\sqrt{\frac{2}{\pi}}\\ 等式左右两边同时乘以3\\ E|X-3|=\frac{3\sqrt 2}{\sqrt{\pi}}\$

$X\sim N(\mu,\sigma^2),Z=|X-\mu|,E(Z)=$
$标准化:E|\frac{X-\mu}{\sigma}|=\frac{2}{\pi}\\ 默认\sigma大于0，即在等式两边同时乘以\sigma\\ E|X-\mu|=\frac{\sqrt{2}\mu}{\sqrt{\pi}}\\$

$设（x,y)\sim N(0,0;\frac{1}{2},\frac{1}{2};0),\phi(x)为标准正态函数，则P \lbrace X-Y<E|X-Y|\rbrace=\\ SET ~(x,y)\sim N(0,0; \frac{1}{2},\frac{1}{2}; 0), \phi(x) ~is ~a ~standard~ normal ~function, ~then ~P \lbrace X-Y<E|X-Y|\rbrace=$
$X\sim N(0,\frac{1}{2})\\ Y\sim N(0,\frac{1}{2})\\ (正态分布的线性组合服从一维正态)\\ Linear~combinations~\\ of ~normal~ distributions ~obey~ one-dimensional~normal\\ X-Y\sim N(0,1) ~~(期望相减，方差相加)\\ Expectations ~are ~subtracted, and ~variances~ are~added\\ E(X-Y)=\frac{\sqrt{2}}{\sqrt{\pi}}\\ please~let~~Z=x-y\\ So~~P{Z<E|X-Y|}=\Phi(\sqrt{\frac{2}{\pi}})\\$

$设X1，X2为来自总体N(\mu,\sigma^2)的简单随机样本，记\hat{\sigma}=a|X1-X2|,若E(\sigma)=\sigma,则a=\\ Let ~X1 ~and~ X2 ~be ~simple ~random ~samples~ from ~the~ population ~N(\mu,\sigma^2) ~and ~denote ~\hat{\sigma}=a|X1-X2|, \\ if ~E(\sigma)=\sigma, then ~a=$
$X1\sim N(\mu,\sigma^2)\\ X2\sim N(\mu,\sigma^2)\\ E(\hat{\sigma})=E(a|X1-X2|)=aE|X1-X2|\\ X1-X2\sim N(0,2\sigma^2)\\ \frac{X1-X2}{\sqrt{2}\mu} \sim N(0,1)\\ E|\frac{X1-X2}{\sqrt{2}\mu}=\sqrt{\frac{2}{\pi}}\\ Both~sides~of~the~equation~are~multiplied~by~ \sqrt{2}\mu~at~the~same~time\\ E|X1-X2|=\frac{2\sigma}{\sqrt{\pi}}\\ So~~E(\hat{\sigma})=a \cdot \frac{2\sigma}{\sqrt{\pi}}=\sigma\\ \frac{2}{\sqrt{\pi}} \cdot a=1\\ a = \frac{\sqrt{\pi}}{2}\\$

$设二维随机变量(X1,X2)\sim N(0,0;1,1;0),记X=\max\lbrace X1,X2 \rbrace,Y=\min\lbrace X1,X2 \rbrace,Z=X-Y,求E(Z)\\ Let ~the~ two-dimensional ~random ~variable(X1,X2)\sim N(0,0; 1,1; 0),\\ ~ write ~X=max\lbrace X1,X2 \rbrace,Y=min\lbrace X1,X2 \rbrace,Z=X-Y, and ~find E(Z)$
$X1\sim N(0,1)\\ X2\sim N(0,1)\\ E(Z)=E(X-Y)\\ if~~X1>X2,Z=X-Y=X1-X2\\ else~~Z=X-Y=X2-X1\\ So~~Z=|X1-X2|\\ E(Z)=E(|X-Y|)\\ standardization~:\frac{X1-X2}{\sqrt{2}}\sim N(0,1)\\ E|\frac{X1-X2}{\sqrt{2}}|=\sqrt{\frac{2}{\pi}}\\ Both~sides~of~the~equation~are~multiplied~by~ \sqrt{2}~at~the~same~time\\ E(X1-X2)=\frac{2}{\sqrt{\pi}}\\$

$设X1与X2相互独立，且服从正态分布N(\mu,\sigma^2),求E\lbrace (X1,X2\rbrace)\\ Let ~X1 ~and~ X2 ~be ~independent ~of ~each ~other ~and ~obey~ the ~normal ~distribution N(\mu,\sigma^2), and ~find ~E\lbrace(X1,X2\rbrace)$
$X1\sim N(\mu,\sigma^2)\\ X2\sim N(\mu,\sigma^2)\\ E\lbrace\max(x1,x2)=\frac{1}{2}(x1+x2+|x1-x2|))\rbrace\\ =\frac{1}{2}(E(X1)+E(X2)+E|X1-X2|)\\ X1-X2\sim N(0,2\sigma^2)\\ \frac{X1-X2}{\sqrt{2}\sigma}\sim N(0,1)\\ E|\frac{X1-X2}{\sqrt{2}\sigma}|=\sqrt{\frac{2}{\pi}}\\ E(X1-X2)=\frac{2\sigma}{\sqrt{\pi}}\\ E\lbrace\max(x1,x2)=\frac{1}{2}(x1+x2+|x1-x2|))\rbrace\\ =\frac{1}{2}(E(X1)+E(X2)+E|X1-X2|)\\ =\frac{1}{2}(\mu+\mu+\frac{2\sigma}{\sqrt{\pi}})\\ =\mu+\frac{\sigma}{\sqrt{\pi}}\\$ 文章来源地址https://www.toymoban.com/news/detail-839890.html