广西大学：《概率论与数理统计》课程电子教案（PPT课件）第四章 随机变量的数字特征 4.2 方差

4.2方差(2）5利用公式计算D(X)= E(X)-[E(X)I证 D(X)= E{[X -E(X)}"}=E(X? -2XE(X)+[E(X)})= E(X)- 2E(X)E(X)+[E(X)]=E(X)-[E(X)}

(2) 利用公式计算 ( ) ( ) [ ( )] . 2 2 D X = E X − E X 证 D(X) { 2 ( ) [ ( )] } 2 2 E X − XE X + E X 2 2 E(X ) − 2E(X)E(X) + [E(X)] ( ) [ ( )] . 2 2 E X − E X = {[ ( )] } 2 = E X − E X = =

4.2方差(1)例如X21Y0.20.3010.40.1解: E(X) = 1 × 0.2 + 2 × 0.3 + 1 × 0. 4 + 2 × 0. 1= 1. 4E(x2) = 12 × 0. 2 + 22 × 0. 3 + 12 × 0. 4 + 22 × 0. 1= 2. 2D(X) = E(x2) - [E(X)]2 = 2. 2 - 1. 42 = 0.24.K

例如 (1) 解： 𝑬 𝑿 = 𝟏 × 𝟎. 𝟐 + 𝟐 × 𝟎. 𝟑 + 𝟏 × 𝟎. 𝟒 + 𝟐 × 𝟎. 𝟏 = 𝟏. 𝟒 X Y 𝟎 𝟏 𝟏 𝟐 𝟎. 𝟐 𝟎.𝟑 𝟎. 𝟒 𝟎.𝟏 𝑬 𝑿 𝟐 = 𝟏 𝟐 × 𝟎. 𝟐 + 𝟐 𝟐 × 𝟎. 𝟑 + 𝟏 𝟐 × 𝟎. 𝟒 + 𝟐 𝟐 × 𝟎. 𝟏 = 𝟐. 𝟐 𝑫 𝑿 = 𝑬 𝑿 𝟐 − 𝑬 𝑿 𝟐 = 𝟐. 𝟐 − 𝟏. 𝟒 𝟐 = 𝟎. 𝟐𝟒

4.2方差(2)例如[2, (x,y)eG=((x,y)/0≤y≤x≤1))f(x,y) :[0,etc.解: E(X)= [[ xf(x,y)dxdy = [ x · 2dxdy+ 0R2G2I'dxf"2xdy3三E(X°)= [[ x f(x, y)dxdy = J x2 . 2dxdy + 0GR1=J' dxJ’ 2xdy 2D(X) = E(X2) - [E(X)]2 ==- ()12K

例如 (2) 解： 2, ( , ) {( , ) | 0 1}, ( , ) 0, etc. x y G x y y x f x y   =    =   𝑫 𝑿 = 𝑬 𝑿 𝟐 − 𝑬 𝑿 𝟐 = 𝟐 𝟑 − 𝟏 𝟐 𝟐 = 𝟕 𝟏𝟐 . 2 ( ) ( , ) 2 0 R G E X xf x y dxdy x dxdy = =  +   1 0 0 2 2 3 x = = dx xdy   2 2 2 2 ( ) ( , ) 2 0 R G E X x f x y dxdy x dxdy = =  +   1 2 0 0 1 2 2 x = = dx x dy  

4.2方差5.方差的性质1°设C是常数，则D(C)=0证 D(C) = E{[C- E(C)}}= 02°设X是一个随机变量，C是常数，则有D(CX) = CD(X), D(X +C) = D(X)证 D(CX)= E{[CX - E(CX)})=C’E([X - E(X)}"}= C2D(X)

5. 方差的性质 1 设C 是常数，则D(C) = 0. 证 D(C) {[ ( )] } = 0. 2 = E C − E C 2 设 X 是一个随机变量，C是常数， 则有 D(CX) = C 2D(X)， D(X + C) = D(X). D(CX ) {[ ( )] } 2 证 = E CX − E CX {[ ( )] } 2 2 C E X − E X ( ). 2 C D X = =

4.2方差D(X +C) = E(IX +C- E(X +C)})= E([X - E(X)}"})= D(X).3°设X,Y是两个随机变量，则有D(X + Y)= D(X)+ D(Y)+ 2E{[X - E(X)I[Y - E(Y)I)若X，Y相互独立，则有D(X + Y)= D(X) + D(Y)

D(X + C) {[ ( )] } 2 = E X + C − E X + C {[ ( )] } 2 E X − E X D(X). = = 3 设X,Y 是两个随机变量， 则有 D(X + Y ) = D(X) + D(Y ). 若X,Y 相互独立, 则有 D(X + Y ) = D X D Y E X E X Y E Y ( ) ( ) 2 {[ ( )][ ( )]}. + + − −