已知随机变量X1和X2的概率分布为-|||-X1 -1 0 1 X2 0 1-|||-pi dfrac (1)(4) .dfrac (1)(2) dfrac (1)(4) pi .dfrac (1)(2) dfrac (1)(2)-|||-且 {X)_(1)(X)_(2)=0} =1 。-|||-(1)求X1和X2的联合分布律;-|||-(2)问X1和X2是否独立?

设随机变量X的概率密度为(x)=dfrac (1)(2sqrt {pi )}(e)^-dfrac ({(x-3)^2)(4)}((x)=dfrac (1)(2s

设随机变量X的分布函数为(X)=dfrac (1)(2)Phi (x)+dfrac (1)(2)Phi (dfrac (x-4)(2))(X)=dfrac (1

某随机变量 X 的概率密度函数为(x)=dfrac (2)(pi )dfrac (1)({e)^x+(e)^-x}，则分布函数为(x)=dfrac (2)(pi

设随机变量X的分布律为X -1 0 1-|||-P dfrac (1)(5) dfrac (3)(5) dfrac (1)(5)设X -1 0 1-|||-P

(13)设随机变量(X,Y)的联合概率分布如下.-|||-X-|||-①-|||-0 dfrac (1)(8) a dfrac (1)(4)-|||-1 dfr

5、已知随机变量X的密度函数为 (x)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac ({(x-1)^2)(8)}, 则 =dfrac (

设X1,X2是来自任意总体UND的一个容量为2的-|||-样本, =dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) =dfrac

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

[例2]设随机变量X与Y相互独立,其概率分布为-|||-x 0 1-|||-P .dfrac (1)(3) .dfrac (2)(3)-|||-Y 0 1-||

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}