Question

设x,y相互独立,都服从N(0,1)分布,试求E(根号(x2＋y2))

Answer 1

φ(x) =[1/(根号2π)]e^[-(x^2)/2]

故:f(x,y) =φ(x) *φ(y) =[1/(2π)]e^[-(x^2+y^2)/2].

故:E((X^2+Y^2)^(1/2))=∫∫[(x^2+y^2)^(1/2)]*f(x,y)dxdy (积分区域D:xoy平面)

=∫∫(x^2+y^2)^(1/2){[1/(2π)]e^[-(x^2+y^2)/2].}dxdy

=[1/(2π)]*∫∫(r){e^[-(r^2/2].}rdrdθ ( 化为极坐标系下的二重积分,D表示为:0