f(x)=(1/(根號(2π)σ)e^(-x^2/(2σ^2))
f(y)=(1/(根號(2π)σ)e^(-y^2/(2σ^2))
因為X,Y獨立,所以f(x,y)=f(x)*f(y)=(1/(2πσ^2)e^(-(x^2+y^2)/(2σ^2))
Z=根號下X^2+Y^2
當(dāng)z<0時,FZ(z)=0
當(dāng)z≥0時,FZ(z)=P{根號下X^2+Y^2≤z}=∫∫)=(1/(2πσ^2)e^(-(x^2+y^2)/(2σ^2))dxdy（積分區(qū)域為x^2+y^2≤z^2)
轉(zhuǎn)換為極坐標求積分,令x=rcosθ,y=rsinθ
FZ(z)=∫(0,2π)dθ∫(0,z)(1/(2πσ^2)e^(-r^2/(2σ^2))rdr=1-e^(-z^2/(2σ^2))
fZ(z)=FZ'(z)=(z/σ^2)e^(-z^2/(2σ^2)) z≥0
fZ(z)=0 z<0
E(z)=∫∫（根號下X^2+Y^2）*f(x,y)dxdy（積分區(qū)域是x^2+y^2≤z^2)=)=∫(0,2π)dθ∫(0,正無窮)(1/(2πσ^2)e^(-r^2/(2σ^2))r^2dr=根號π*σ/根號2
方法應(yīng)該就是這樣,至于有沒有算錯,你檢查一下吧