概率论 设随机变量X的分布函数为 0 ,x<=-1 1/2+1/πarcsinX,-1<x<=1 1,x>1 求E(x)和D(x)

如题所述

概率密度f(x)= F'(x).
故:|x|<=1时, f(x)=(1/π)*{1/[(1-x^2)^0.5]}
其它, f(x)=0.
E(X)=在[-1, 1]积分[xf(x)]dx =0 (奇函数在对称区间的积分).
E(X^2)=在[-1, 1]积分[x^2f(x)]dx =2*{在[0, 1]积分[x^2f(x)]dx} (偶函数在对称区间的积分).
(换元:x=sint, dx=cost dt)
=(1/pi)*2*{在[0,pi/2]积分 (sint)^2dt=
=(1/pi)*2*{在[0,pi/2]积分 [(1-cos2t)/2]dt=
=(1/pi)*2*{函数[t- (1/2)sin2t]在t=pi/2的值- 在t=0处的值}=(1/pi)*2*{pi/2}=1.
由此:D(X) = E(X^2)- {E(X)}^2= 1
即E(X) = 0, D(X)= 1.
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