2 Dimensions

Suppose both x and y ~ N_0,σ.

N_0,σpdf(x) = ¹/_(√(2π)σ) e^-x2/(2σ2)

and

N_0,σpr([x-ε/2,x+ε/2]) = N_0,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

= ¹/_(2πσ²) e^-r2/(2σ2),  where r²=x²+y²

and

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)

so

pdf₆(r) = ^r/_σ²  e^-r2/(2σ2)   (a χ-distribution, k=2)

and

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.-)