Advice for mathematicians:
Randomly choosing a direction
Cole Zmurchok
September 22, 2015
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How do you randomly choose a direction in 2-dimensions?
y x
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How do you uniformly choose a direction in 2-dimensions?
y
θ x
▶ Uniformly choose θ
∈
[0 , 2 π ]
▶ Polar coordinates x = cos θ y = sin θ give a unit vector in
R 2
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How do you uniformly choose a direction in 2-dimensions?
This works!
1 1
0 0
-1
-1 0 x
1
-1
-1 0 x
1
Why?
Want any small “arc length” on the circle to contain the same number of points: ds =
√ dx 2 + dy 2 =
√ d θ 2 = d θ
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How do you uniformly choose a direction in 3-dimensions?
▶
▶
Uniformly choose two angles: θ
∈
[0 , 2 π ] and ϕ
∈
[0 , π ]
Spherical coordinates: x = cos( θ ) sin( ϕ ), y = sin( θ ) sin( ϕ ), z = cos( ϕ )
Clustering at poles!
1
1
0 0
-1
1 y
0
-1 -1
0 x
1 -1
-1 0 x
1
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How do you uniformly choose a direction in 3-dimensions?
▶ Independently choosing two angles from uniform distribution does not take into account that not all latitudes have the same circumference !
▶ Want any small area on the unit sphere to contain the same number of points, but the area element d Ω = sin ϕ d θ d ϕ is a function of ϕ .
▶ You could design a transformation so that this works.
▶ See “Sphere Point Picking” on Wolfram Mathworld.
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Uniformly choosing directions Normally
Density function for a normally distributed random variable X p ( x ; µ, σ
2
) =
√
1
2 πσ 2 exp
(
−
( x
−
µ )
2
)
2 σ 2
Density function for a normally distributed vector-valued random variable X p ( x ; µ, Σ) =
√
(2 π )
1 n det(Σ) exp
(
−
1
2
( x
−
µ )
T
Σ
−
1
( x
−
µ )
) with µ
∈ R n
, Σ a positive definite n
× n covariance matrix
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Uniformly choosing directions Normally
▶ If Σ = I and µ = p ( x ; 0 , I ) =
=
√
√
1
2
1
(2
π
π ) n exp exp
(
−
( x
2
1
−
1
2
)
2 i =1 x
·
. . .
· i
2
)
√
1
2 π exp
(
− x
2 n
)
2
In general a n -dimensional Gaussian with mean µ
∈ R n and diagonal covariance matrix Σ = diag( σ
2
1
, . . . , σ
2 n
)
= a collection of n independent Gaussian random variables with mean
µ i and variance σ i
2
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So what?
Uniform direction distribution in n -dimensions
1.
Generate n standard normal random variables x i
2.
The distribution of vectors
√ x
2
1
1
+
· · ·
+ x 2 n
( x
1
, . . . , x n
)
T is uniform over the unit sphere in n -dimensions
Why?
On the unit sphere in p ( x ; 0 , I ) = n
√
-dimensions,
1
(2 π ) n exp
∑ n i =1
(
−
1
2 i =1 x i
2
= 1, so
) x i
2
=
√ e
− 1
2
(2 π ) n
⇒ is spherically symmetric
⇒ distribution is constant
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Example: 3-dimensions
No clustering!
1
1
0
-1
1 y
0
-1 -1
0 x
1
0
-1
-1 0 x
1
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Example: 2-dimensions
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