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 补充例题

4.2方差补充1设随机变量X具有概率密度[1+x,-1≤x<0,f(x)== 3 1-x, 0≤x<1,其他.0,求 D(X)解E(X) =I,x(1+x)dx+,x(1-x)dx= 0

补充1 设随机变量X 具有概率密度 求 D(X). 0, 其他 . 1 − x, 0  x  1, 1 + x, − 1  x  0,      f (x) = 解 E(X) = 0,   = + + − − 1 0 0 1 x(1 x)d x x(1 x)d x

4.2方差E(X")-f' x*(1+x)dx+ f'x(1-x)dx=6于是D(X) = E(X")-[E(X)}’ = =-02

( ) 2 E X , 6 1 = 于是 D(X) 2 0 6 1 = − . 6 1 =   = + + − − 1 0 2 0 1 2 x (1 x)d x x (1 x)d x 2 2 = E(X ) − [E(X)]

4.2方差补充2设连续型随机变量X的概率密度为元-2cosx, 0≤x≤f(x) =-0,其他。求随机变量Y=X2的方差D(Y)解E(X") =[~x"f(x)dx2元-Jix° cos xdx =2A

设连续型随机变量X 的概率密度为 解 ( ) 2 E X  = 2 π 0 2 x cos xd x ( ). 2 求随机变量Y = X 的方差 D Y 0, 其他. 2 π cos x, 0  x     f (x) = x f (x)d x 2   − = 2, 4 π 2 = − 补充2

4.2方差元2AE(X*) = [°x*f(x)dx=xcosxdx0公3元2 + 24,16因为 D(X)= E(X°)-[E(X)}°,所以 D(X°)= E(X*)-[E(X°)1代元3元+ 2416=20-2元2

( ) 4 E X  = 2 π 0 4 x f (x)d x x cos xd x 4   − = 因为 2 2 2 4 2 4 3 24 16         −  −  + −  = 20 2 . 2 = −  3 24, 16 2 4 −  +  = 所以 ( ) ( ) [ ( )] , 2 2 D X = E X − E X 2 4 2 2 D(X ) = E(X ) − [E(X )]

4.2方差21补充3 设 X~1'求 D(2X3 + 5)231212解 D(2X3 +5) = D(2X3)+ D(5)= 4D(X3)= 4[E(X)-(E(X)11493+ 0°E(X°)= (-2)°×= +312126福

解 (2 5) 3 D X + 4 ( ) 3 = D X 4[ ( ) ( ( )) ] 6 3 2 = E X − E X ( ) 6 E X , 6 493 = , 12 1 12 1 2 1 3 1 2 0 1 3 ~           − 设 X (2 5). 3 求 D X + (2 ) (5) 3 = D X + D 12 1 3 12 1 1 2 1 0 3 1 ( 2) 6 6 6 6 = −  +  +  +  补充3

4.2方差[E(X3)" =(-2) ×=+0°×= ++13二+31121故 D(2X3 +5) = 4[E(X)-(E(X3)29549

3 2 [E(X )] (2 5) 3 故 D X + , 9 1 = 4[ ( ) ( ( )) ] 6 3 2 = E X − E X . 9 2954 = 2 3 3 3 3 12 1 3 12 1 1 2 1 0 3 1 ( 2)       = −  +  +  + 