习题4 1.设离散型随机变量X具有概率分布律: 0102020301 试求E(1),E(2+5,E(1) 解E(X=(-2)×0.1+(-1)×0.2+0×0.2+1×0.3+2×0.1+3×0.1 =0.4, E(H2+5)=E(2)+5 =(-2)2×0.1+(-1)2×0.2+02×0.2 +12×0.3+22×0.1+32×0.1 2+5=7.2. E(X1)=-21×0.1+-11×0.2+10×02+11×0.3+2×0.1+3×0.1 5.设随机变量X具有概率密度 x<0 f(x)=x (1)求常数A; 解由1=厂(kx+!Ach=号+4C,得4=号 (2)求X的数学期望 A E(X)= f()dx=Lx2dx+eroxo-xd.4 6.设随机变量X的概率密度为 f(r)=rexx>0 x<0 求E(),E(-2x+5),E(e-3) 解因为E)x(x)k=x2cb=2,所以 E(3X)=3E(X=3×2=6, E(-2X+5)=-2E(X)+5=-2×2+5=1 E(c3)=c3f(x)= ex.xedx= 17.(1)求第1题中X的方差D( 解E(2)=(-2)2×0.1+(-1)2×0.2+02×02+12×0.3+22×0.1+32×0.1习题 4 1．设离散型随机变量 X 具有概率分布律 X −2 −1 0 1 2 3 pk 01 02 02 03 01 01 试求 E(X) E(X 2+5) E(|X|) 解 E(X)=(−2)0.1+(−1)0.2+00.2+10.3+20.1+30.1 =0.4 E(X 2+5)=E(X 2 )+5 =(−2)20.1+(−1)20.2+0 20.2 +1 20.3+2 20.1+3 20.1 =2.2+5=7.2 E(|X|)=|−2|0.1+|−1|0.2+|0|0.2+|1|0.3+|2|0.1+|3|0.1 =1.2. 5．设随机变量 X 具有概率密度         = − 1 0 1 0 0 ( ) Ae x x x x f x x  (1)求常数 A 解 由 1 1 1 0 2 1 1 ( ) − + − + − = = + = +    f x dx xdx Ae dx Ae x  得 2 e A=  (2)求 X 的数学期望 解 3 4 2 ( ) ( ) 1 1 0 2 = = + =    + − + − xe dx e E X xf x dx x dx x  6．设随机变量 X 的概率密度为      = − 0 0 0 ( ) x xe x f x x  求 E(3X) E(−2X+5) E(e −3X) 解 因为 ( ) ( ) 2 0 2 = = =   + − + − E X xf x dx x e dx x  所以 E(3X)=3E(X)=32=6 E(−2X+5)=−2E(X)+5=−22+5=1. 16 1 ( ) ( ) 0 4 0 3 3 3 = =  = =    + − + − − + − − − E e e f x dx e x e dx x e dx X x x x x . 17．(1)求第 1 题中 X 的方差 D(X) 解 E(X 2 )=(−2)20.1+(−1)20.2+0 20.2+1 20.3+2 20.1+3 20.1 =2.2