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2 Dimensions

Suppose both x and y ~ N0,σ.
N0,σpdf(x) = 1/(√(2π)σ) e-x2/(2σ2)
N0,σpr([x-ε/2,x+ε/2]) = N0,σpdf(x) . ε, if ε<<σ.
(pdf = probability density function,   pr = probability.)
For x and y,
pdf2(x & y) = 1/(2πσ2) e-(x2+y2)/(2σ2)
= 1/(2πσ2) e-r2/(2σ2),  where r2=x2+y2
pr([x-ε/2,x+ε/2] × [y-ε/2,y+ε/2]) = 1/(2πσ2) e-(x2+y2)/(2σ2) . ε2
Let r2 = x2 + y2, the radius r>0, x = r cosθ, y = r sinθ, and c = rθ.  (c is the (arc) distance, swept by θ, along the circumference of a circle of radius r.)
A uniform distribution on θ has
pdf3(θ) = 1/(2π),   0≤θ<2π.
A uniform distribution on c has
pdf4(c|r) = 1/(2πr),   0≤c<2πr.
To know c±ε/2 is to know θ±ε/(2r).
Now (x, y) ↔ (r, c) does not involve any stretching so it must be that
pdf5(r & c) = pdf6(r) pdf4(c|r) = pdf2(x & y)   (see #1)
= 1/(2πσ2) e-r2/(2σ2)
pdf6(r) = r/σ2  e-r2/(2σ2)   (a χ-distribution, k=2)
pdf7(r & θ) = r/(2πσ2) e-r2/(2σ2) = 1/(2π) pdf6(r) = r . pdf5(r & c).
(And the distribution of r2 is a 'χ2-distribution', chi-squared.)

3 Dimensions +

(... exercise.-)
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