|
|
2 Dimensions
- Suppose both x and y ~ N0,σ.
- N0,σpdf(x)
= 1/(√(2π)σ)
e-x2/(2σ2)
- and
- 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
- 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.-)
|
|