(1)当z<1时,F2(=)=0 (2)当1≤z≤2时,有 F()=P{z≤+}=PX+Y≤=f(x,y)dt=2 dxdy 其中D={(x,y)|0≤x≤1,0≤y≤1,1≤x+y≤z},DcG 故F2(z)=2D的面积= (3)当x>2时,F2(=)=1 于是F2()={1-(2-),1≤=≤2,从而f(c) 1, 其它 E(X+Y)=E(2)=2()d=1=2(2-) (x+12)(k=:2)k D(X+Y)=D(Z)=E(2)-E(Z)2= 18 方法二E(x+)=(x+y)(x)b EICX +)]=f(x+y)2/(x, y)drdy=ar[2(x+y)dy 所以D(x+》)=E(X+y)1-[E(x+)=18 方差的性质 例7(E06)设f(x)=E(X-x)2,x∈R,证明:当x=E(X)时,f(x)达到最小值 证依题f(x)=E(X-x)2=E(X2)-2xE(X) 两边对x求导数,有出(x) a=2x-2E(X),显然当x=E(们时,(x) /,p图入 =2>0,所以当x=E(X)时,∫(x)达到最小值,最小值为 f(E(XD=E(X-E(D)则 (1) 当 z 1 时, F (z) = 0; Z (2) 当 1 z  2 时, 有 F (z) P{Z z} Z =  = P{X +Y  z}  +  = x y z f (x, y)dxdy 2 ,  = D dxdy 其中 D ={(x, y)| 0  x 1,0  y 1, 1 x + y  z},D  G, 故 FZ (z) = 2  D 的面积       = − − 2 (2 ) 2 1 2 1 2 z 1 (2 ) ; 2 = − − z (3) 当 z  2 时, F (z) =1. Z 于是 , 1, 2 1 (2 ) , 1 2 0, 1 ( ) 2       − −    = z z z z F z Z 从而 , 0, 2(2 ), 1 2 ( )    −   = 其它 z z f z z E(X +Y) = E(Z)  + − = zf z dz Z ( )  =  − 2 1 z 2(2 z)dz 3 4 = [( ) ] ( ) 2 2 E X +Y = E Z  + − = z f z dz Z ( ) 2  =  − 2 1 2 z 2(2 z)dz 6 11 = D(X +Y) = D(Z) 2 2 = E(Z ) −[E(Z)] . 18 1 = 方法二   + − + − E(X +Y) = (x + y) f (x, y)dxdy   − = + 1 1 1 0 2( ) x dx x y dy  = + 1 0 2 (x 2x)dx . 3 4 3 1 0 2 3 =         = + x x   + − + − E[(X + Y) ] = (x + y) f (x, y)dxdy 2 2   − = + 1 1 2 1 0 2( ) x dx x y dy  = + + 1 0 3 2 ( 3 3 ) 3 2 x x x dx 6 11 = 所以 2 2 D(X +Y) = E[(X +Y) ]−[E(X +Y)] . 18 1 = 方差的性质 例 7 (E06) 设 ( ) ( ) , , 2 f x = E X − x xR 证明: 当 x = E(X) 时, f (x) 达到最小值. 证 依题 2 f (x) = E(X − x) ( ) 2 ( ) , 2 2 = E X − xE X + x 两边对 x 求导数, 有 2 2 ( ), ( ) x E X dx df x = − 显然当 x = E(X ) 时, 0. ( ) = dx df x 又因 2 0, ( ) 2 2 =  dx d f x 所以当 x = E(X ) 时, f (x) 达到最小值, 最小值为 ( ( )) ( ( )) ( ). 2 f E X = E X − E X = D X