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 ∣ = 2 Π 计算 : E ∣ X ∣ = ∫ − ∞ + ∞ ∣ X ∣ f ( x ) d x E Z = E ∣ x − μ ∣ E|X|={\sqrt{\frac{2}{\Pi}} }\\ 计算:E|X|=\displaystyle \int^{+\infty}_{-\infty}{|X|f(x)dx}\\ EZ=E|x-\mu| E∣X∣=Π2计算:E∣X∣=∫−∞+∞∣X∣f(x)dxEZ=E∣x−μ∣
二、习题
X ∼ N ( 0 , 1 ) , E ( ∣ X ∣ ) = 2 Π X\sim N(0,1),E(|X|)=\sqrt{\frac{2}{\Pi}} X∼N(0,1),E(∣X∣)=Π2
X ∼ N ( 3 , 9 ) , E ( ∣ X − 3 ∣ ) = X\sim N(3,9),E(|X-3|)= X∼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 ∣ X − 3 3 ∣ = 2 π 等式左右两边同时乘以 3 E ∣ X − 3 ∣ = 3 2 π 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}}\ E∣3X−3∣=π2等式左右两边同时乘以3E∣X−3∣=π32
X ∼ N ( μ , σ 2 ) , Z = ∣ X − μ ∣ , E ( Z ) = X\sim N(\mu,\sigma^2),Z=|X-\mu|,E(Z)= X∼N(μ,σ2),Z=∣X−μ∣,E(Z)=
标准化 : E ∣ X − μ σ ∣ = 2 π 默认 σ 大于 0 ，即在等式两边同时乘以 σ E ∣ X − μ ∣ = 2 μ π 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}}\\ 标准化:E∣σX−μ∣=π2默认σ大于0，即在等式两边同时乘以σE∣X−μ∣=π2μ
设（ x , y ) ∼ N ( 0 , 0 ; 1 2 , 1 2 ; 0 ) , ϕ ( x ) 为标准正态函数，则 P { X − Y < E ∣ X − Y ∣ } = S E T ( x , y ) ∼ N ( 0 , 0 ; 1 2 , 1 2 ; 0 ) , ϕ ( x ) i s a s t a n d a r d n o r m a l f u n c t i o n , t h e n P { X − Y < E ∣ X − Y ∣ } = 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= 设（x,y)∼N(0,0;21,21;0),ϕ(x)为标准正态函数，则P{X−Y<E∣X−Y∣}=SET (x,y)∼N(0,0;21,21;0),ϕ(x) is a standard normal function, then P{X−Y<E∣X−Y∣}=
X ∼ N ( 0 , 1 2 ) Y ∼ N ( 0 , 1 2 ) ( 正态分布的线性组合服从一维正态 ) L i n e a r c o m b i n a t i o n s o f n o r m a l d i s t r i b u t i o n s o b e y o n e − d i m e n s i o n a l n o r m a l X − Y ∼ N ( 0 , 1 ) ( 期望相减，方差相加 ) E x p e c t a t i o n s a r e s u b t r a c t e d , a n d v a r i a n c e s a r e a d d e d E ( X − Y ) = 2 π p l e a s e l e t Z = x − y S o P Z < E ∣ X − Y ∣ = Φ ( 2 π ) 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}})\\ X∼N(0,21)Y∼N(0,21)(正态分布的线性组合服从一维正态)Linear combinations of normal distributions obey one−dimensional normalX−Y∼N(0,1) (期望相减，方差相加)Expectations are subtracted,and variances are addedE(X−Y)=π2please let Z=x−ySo PZ<E∣X−Y∣=Φ(π2)
设 X 1 ， X 2 为来自总体 N ( μ , σ 2 ) 的简单随机样本，记 σ ^ = a ∣ X 1 − X 2 ∣ , 若 E ( σ ) = σ , 则 a = L e t X 1 a n d X 2 b e s i m p l e r a n d o m s a m p l e s f r o m t h e p o p u l a t i o n N ( μ , σ 2 ) a n d d e n o t e σ ^ = a ∣ X 1 − X 2 ∣ , i f E ( σ ) = σ , t h e n a = 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= 设X1，X2为来自总体N(μ,σ2)的简单随机样本，记σ^=a∣X1−X2∣,若E(σ)=σ,则a=Let X1 and X2 be simple random samples from the population N(μ,σ2) and denote σ^=a∣X1−X2∣,if E(σ)=σ,then a=
X 1 ∼ N ( μ , σ 2 ) X 2 ∼ N ( μ , σ 2 ) E ( σ ^ ) = E ( a ∣ X 1 − X 2 ∣ ) = a E ∣ X 1 − X 2 ∣ X 1 − X 2 ∼ N ( 0 , 2 σ 2 ) X 1 − X 2 2 μ ∼ N ( 0 , 1 ) E ∣ X 1 − X 2 2 μ = 2 π B o t h s i d e s o f t h e e q u a t i o n a r e m u l t i p l i e d b y 2 μ a t t h e s a m e t i m e E ∣ X 1 − X 2 ∣ = 2 σ π S o E ( σ ^ ) = a ⋅ 2 σ π = σ 2 π ⋅ a = 1 a = π 2 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}\\ X1∼N(μ,σ2)X2∼N(μ,σ2)E(σ^)=E(a∣X1−X2∣)=aE∣X1−X2∣X1−X2∼N(0,2σ2)2μX1−X2∼N(0,1)E∣2μX1−X2=π2Both sides of the equation are multiplied by 2μ at the same timeE∣X1−X2∣=π2σSo E(σ^)=a⋅π2σ=σπ2⋅a=1a=2π
设二维随机变量 ( X 1 , X 2 ) ∼ N ( 0 , 0 ; 1 , 1 ; 0 ) , 记 X = max { X 1 , X 2 } , Y = min { X 1 , X 2 } , Z = X − Y , 求 E ( Z ) L e t t h e t w o − d i m e n s i o n a l r a n d o m v a r i a b l e ( X 1 , X 2 ) ∼ N ( 0 , 0 ; 1 , 1 ; 0 ) , w r i t e X = m a x { X 1 , X 2 } , Y = m i n { X 1 , X 2 } , Z = X − Y , a n d f i n d E ( Z ) 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) 设二维随机变量(X1,X2)∼N(0,0;1,1;0),记X=max{X1,X2},Y=min{X1,X2},Z=X−Y,求E(Z)Let the two−dimensional random variable(X1,X2)∼N(0,0;1,1;0), write X=max{X1,X2},Y=min{X1,X2},Z=X−Y,and findE(Z)
X 1 ∼ N ( 0 , 1 ) X 2 ∼ N ( 0 , 1 ) E ( Z ) = E ( X − Y ) i f X 1 > X 2 , Z = X − Y = X 1 − X 2 e l s e Z = X − Y = X 2 − X 1 S o Z = ∣ X 1 − X 2 ∣ E ( Z ) = E ( ∣ X − Y ∣ ) s t a n d a r d i z a t i o n : X 1 − X 2 2 ∼ N ( 0 , 1 ) E ∣ X 1 − X 2 2 ∣ = 2 π B o t h s i d e s o f t h e e q u a t i o n a r e m u l t i p l i e d b y 2 a t t h e s a m e t i m e E ( X 1 − X 2 ) = 2 π 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}}\\ X1∼N(0,1)X2∼N(0,1)E(Z)=E(X−Y)if X1>X2,Z=X−Y=X1−X2else Z=X−Y=X2−X1So Z=∣X1−X2∣E(Z)=E(∣X−Y∣)standardization :2X1−X2∼N(0,1)E∣2X1−X2∣=π2Both sides of the equation are multiplied by 2 at the same timeE(X1−X2)=π2
设 X 1 与 X 2 相互独立，且服从正态分布 N ( μ , σ 2 ) , 求 E { ( X 1 , X 2 } ) L e t X 1 a n d X 2 b e i n d e p e n d e n t o f e a c h o t h e r a n d o b e y t h e n o r m a l d i s t r i b u t i o n N ( μ , σ 2 ) , a n d f i n d E { ( X 1 , X 2 } ) 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) 设X1与X2相互独立，且服从正态分布N(μ,σ2),求E{(X1,X2})Let X1 and X2 be independent of each other and obey the normal distributionN(μ,σ2),and find E{(X1,X2})
X 1 ∼ N ( μ , σ 2 ) X 2 ∼ N ( μ , σ 2 ) E { max ( x 1 , x 2 ) = 1 2 ( x 1 + x 2 + ∣ x 1 − x 2 ∣ ) ) } = 1 2 ( E ( X 1 ) + E ( X 2 ) + E ∣ X 1 − X 2 ∣ ) X 1 − X 2 ∼ N ( 0 , 2 σ 2 ) X 1 − X 2 2 σ ∼ N ( 0 , 1 ) E ∣ X 1 − X 2 2 σ ∣ = 2 π E ( X 1 − X 2 ) = 2 σ π E { max ( x 1 , x 2 ) = 1 2 ( x 1 + x 2 + ∣ x 1 − x 2 ∣ ) ) } = 1 2 ( E ( X 1 ) + E ( X 2 ) + E ∣ X 1 − X 2 ∣ ) = 1 2 ( μ + μ + 2 σ π ) = μ + σ π 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}}\\ X1∼N(μ,σ2)X2∼N(μ,σ2)E{max(x1,x2)=21(x1+x2+∣x1−x2∣))}=21(E(X1)+E(X2)+E∣X1−X2∣)X1−X2∼N(0,2σ2)2σX1−X2∼N(0,1)E∣2σX1−X2∣=π2E(X1−X2)=π2σE{max(x1,x2)=21(x1+x2+∣x1−x2∣))}=21(E(X1)+E(X2)+E∣X1−X2∣)=21(μ+μ+π2σ)=μ+πσ