Question

一道概率论问题,急!!!

X和Y相互独立且都服从N(0,σ^2),求随机变量Z=根号下X^2+Y^2的概率密度,还有它的期望值,请写出详细步骤~ 谢谢!

Answer 1

f(x)=(1/(根号(2π)σ)e^(-x^2/(2σ^2))
f(y)=(1/(根号(2π)σ)e^(-y^2/(2σ^2))
因为X,Y独立，所以f(x,y)=f(x)*f(y)=(1/(2πσ^2)e^(-(x^2+y^2)/(2σ^2))
Z=根号下X^2+Y^2
当z<0时，FZ(z)=0
当z≥0时，FZ(z)=P{根号下X^2+Y^2≤z}=∫∫)=(1/(2πσ^2)e^(-(x^2+y^2)/(2σ^2))dxdy（积分区域为x^2+y^2≤z^2)
转换为极坐标求积分，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)=∫∫（根号下X^2+Y^2）*f(x,y)dxdy（积分区域是x^2+y^2≤z^2)=)=∫(0,2π)dθ∫(0,正无穷)(1/(2πσ^2)e^(-r^2/(2σ^2))r^2dr=根号π*σ/根号2
方法应该就是这样，至于有没有算错，你检查一下吧