Special case: scalar variable
Z∼CN(0,1)
fZ(z)=1πe−|z|2f|Z|,Θ(|z|,θ)=1π|z|e−|z|2
Note that the distribution is a function of the magnitude of
Z, and therefore constant over all angles
θ. To get the distribution of the magnitude of Z, we integrate over all angles
−π≤θ≤π.
f|Z|(|z|)=∫f|Z|,Θ(|z|,θ)dθ=∫π−πdθ1π|z|e−|z|2=2|z|e−|z|2,|z|>0
Define
Y=|Z| then,
fY(y)=2ye−y2,y>0
For magnitude squared distribution, we can perform a change of variable, with
X=Y2=g(Y), since
P({−√x,x>0})=0, we only have one region (the set
{√x,x>0} ) to worry about.
fX(x)=fY(g−1(x))|∂g−1(x)∂x|=fY(√x)|∂√x∂x|=(2x12e−x)12x−12=e−x,x>0
Note:
Here we assume the probability spaces
(X,X,PX),
(Y,Y,PY)and that a mapping T exists that maps
A∈X to
B∈Y.
Start from the joint cdf in the X-space, define
A≡{√X2+Y2≤|z|,tan−1(Y/X)≤θ0}
F|Z|,Θ(|z|,θ0)=P(A)=∬(x,y)∈Af(x,y)dxdy
With a change of variable, define
B≡{|Z|≤|z|,Θ≤θ0}
F|Z|,Θ(|z|,θ0)=P(B)=∬(r,θ)∈Bf(x(r,θ),y(r,θ))rdrdθ=∫|z|0∫θ00f(x(r,θ),y(r,θ))rdrdθ
Differentiate wrt
|z| and
θ to get the pdf.
Theorem. Given any function
g:R2→R and T that maps the
A∈X to
B∈Y then
∬Ag(x1,x2)dx1dx2=∬Bg(x1(y1,y2),x2(y1,y2))|J(y1,y2)|dy1dy2 where
J(y1,y2) is the Jacobian.