Borel–Kolmogorov paradox

Consider a random vector ${\displaystyle (X,Y,Z)}$ that is uniformly distributed on the unit sphere ${\displaystyle S^{2}}$. The uniform distribution is defined in terms of surface integrals: for a given region ${\displaystyle D\subset S^{2}}$,

${\displaystyle \mathbb {P} ((X,Y,Z)\in D)={\frac {1}{4\pi }}\int _{D}\mathrm {d} S.}$

This means that for any parametrization ${\displaystyle p(u,v)}$ that covers D, we have

${\displaystyle \mathbb {P} ((X,Y,Z)\in D)={\frac {1}{4\pi }}\int _{p^{-1}(D)}\left|\left|{\frac {\partial p}{\partial u}}\times {\frac {\partial p}{\partial v}}\right|\right|\mathrm {d} u\mathrm {d} v}$

For simplicity, we won't calculate the full conditional distribution on a great circle, only the probability that it lies in the first quadrant. That is to say, we will attempt to calculate the conditional probability ${\displaystyle \mathbb {P} (A|B)}$ with

${\displaystyle {\begin{aligned}A&=\{0<X<1,0<Y<1\}\\B&=\{Z=0\}.\end{aligned}}}$

We begin by parametrizing the sphere with the usual spherical polar coordinates:

${\displaystyle {\begin{aligned}x&=\cos(\varphi )\cos(\theta )\\y&=\cos(\varphi )\sin(\theta )\\z&=\sin(\varphi )\end{aligned}}}$

where ${\displaystyle -{\frac {\pi }{2}}\leq \varphi \leq {\frac {\pi }{2}}}$ and ${\displaystyle -\pi \leq \theta \leq \pi }$.

We can define random variables ${\displaystyle \Phi }$, ${\displaystyle \Theta }$ as the values of ${\displaystyle (X,Y,Z)}$ under the inverse of this parametrization, or more formally using the arctan2 function:

${\displaystyle {\begin{aligned}\Phi &=\arcsin(Z)\\\Theta &=\arctan _{2}\left({\frac {Y}{\sqrt {1-Z^{2}}}},{\frac {X}{\sqrt {1-Z^{2}}}}\right)\end{aligned}}}$

Via the formula above, the probability distribution on the sphere is

${\displaystyle \mathbb {P} ((X,Y,Z)\in D)={\frac {1}{4\pi }}\int _{p^{-1}(D)}\cos(\varphi )\mathrm {d} \varphi \mathrm {d} \theta }$

This equation in effect defines the joint density ${\displaystyle f_{\Phi ,\Theta }(\varphi ,\theta )}$ of ${\displaystyle \Phi }$ and ${\displaystyle \Theta }$:

${\displaystyle f_{\Phi ,\Theta }(\varphi ,\theta )={\frac {1}{4\pi }}\cos(\varphi )}$.

We can rewrite A and B in polar coordinates:

${\displaystyle {\begin{aligned}A&=\{0<X<1,0<Y<1\}&&=\left\{0<\Theta <{\frac {\pi }{2}}\right\}\\B&=\{Z=0\}&&=\{\Phi =0\}\end{aligned}}}$

We attempt to evaluate the conditional probability using the density

${\displaystyle \mathbb {P} (A\mid B){\stackrel {?}{=}}\int _{0}^{\frac {\pi }{2}}f_{\Theta |\Phi }(\theta ,0)\mathrm {d} \theta =\int _{0}^{\frac {\pi }{2}}{\frac {1}{2\pi }}\mathrm {d} \theta ={\frac {1}{4}}}$

Now we repeat the process with a different parametrization of the sphere:

${\displaystyle {\begin{aligned}x&=\cos(\varphi )\cos(\theta )\\y&=\sin(\varphi )\\z&=-\cos(\varphi )\sin(\theta )\end{aligned}}}$

This is equivalent to the previous parametrization rotated by 90 degrees around the x axis.

Define new random variables

${\displaystyle {\begin{aligned}\Phi '&=\arcsin(Y)\\\Theta '&=\arctan _{2}\left({\frac {-Z}{\sqrt {1-Y^{2}}}},{\frac {X}{\sqrt {1-Y^{2}}}}\right).\end{aligned}}}$

Rotation is measure preserving so the density of ${\displaystyle \Phi '}$ and ${\displaystyle \Theta '}$ is the same:

${\displaystyle f_{\Phi ',\Theta '}(\varphi ,\theta )={\frac {1}{4\pi }}\cos(\varphi )}$.

The expressions for A and B are:

${\displaystyle {\begin{aligned}A&=\{0<X<1,0<Y<1\}&&=\left\{0<\Theta <{\frac {\pi }{2}}\right\}&&=\left\{-{\frac {\pi }{2}}<\Theta '<{\frac {\pi }{2}},0<\Phi '<{\frac {\pi }{2}}\right\}\\B&=\{Z=0\}&&=\{\Phi =0\}&&=\{\Theta '=0\}\end{aligned}}}$

Attempting again to evaluate the conditional probability using the density

${\displaystyle \mathbb {P} (A\mid B){\stackrel {?}{=}}\int _{0}^{\frac {\pi }{2}}f_{\Phi '|\Theta '}(\varphi ,0)\mathrm {d} \varphi =\int _{0}^{\frac {\pi }{2}}{\frac {1}{2}}\cos(\varphi )\mathrm {d} \varphi ={\frac {1}{2}}.}$

This shows that the conditional density cannot be treated as conditioning on an event of probability zero, as explained in Conditional probability#Conditioning on an event of probability zero.

To understand the problem we need to recognize that a distribution on a continuous random variable is described by a density f only with respect to some measure μ. Both are important for the full description of the probability distribution. Or, equivalently, we need to fully define the space on which we want to define f.

Let Φ and Λ denote two random variables taking values in Ω₁ = [−π/2, π/2] respectively Ω₂ = [−π, π]. An event {Φ = φ, Λ = λ} gives a point on the sphere S(r) with radius r. We define the coordinate transform

${\displaystyle {\begin{aligned}x&=r\cos \varphi \cos \lambda \\y&=r\cos \varphi \sin \lambda \\z&=r\sin \varphi \end{aligned}}}$

for which we obtain the volume element

${\displaystyle \omega _{r}(\varphi ,\lambda )=\left|\left|{\partial (x,y,z) \over \partial \varphi }\times {\partial (x,y,z) \over \partial \lambda }\right|\right|=r^{2}\cos \varphi \ .}$

Furthermore, if either φ or λ is fixed, we get the volume elements

${\displaystyle {\begin{aligned}\omega _{r}(\lambda )&=\left|\left|{\partial (x,y,z) \over \partial \varphi }\right|\right|=r\ ,\quad \mathrm {respectively} \\\omega _{r}(\varphi )&=\left|\left|{\partial (x,y,z) \over \partial \lambda }\right|\right|=r\cos \varphi \ .\end{aligned}}}$

Let

${\displaystyle \mu _{\Phi ,\Lambda }(d\varphi ,d\lambda )=f_{\Phi ,\Lambda }(\varphi ,\lambda )\omega _{r}(\varphi ,\lambda )\,d\varphi \,d\lambda }$

denote the joint measure on ${\displaystyle {\mathcal {B}}(\Omega _{1}\times \Omega _{2})}$, which has a density ${\displaystyle f_{\Phi ,\Lambda }}$ with respect to ${\displaystyle \omega _{r}(\varphi ,\lambda )\,d\varphi \,d\lambda }$ and let

${\displaystyle {\begin{aligned}\mu _{\Phi }(d\varphi )&=\int _{\lambda \in \Omega _{2}}\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda )\ ,\\\mu _{\Lambda }(d\lambda )&=\int _{\varphi \in \Omega _{1}}\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda )\ .\end{aligned}}}$

If we assume that the density ${\displaystyle f_{\Phi ,\Lambda }}$ is uniform, then

${\displaystyle {\begin{aligned}\mu _{\Phi \mid \Lambda }(d\varphi \mid \lambda )&={\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda ) \over \mu _{\Lambda }(d\lambda )}={\frac {1}{2r}}\omega _{r}(\varphi )\,d\varphi \ ,\quad {\text{and}}\\\mu _{\Lambda \mid \Phi }(d\lambda \mid \varphi )&={\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda ) \over \mu _{\Phi }(d\varphi )}={\frac {1}{2r\pi }}\omega _{r}(\lambda )\,d\lambda \ .\end{aligned}}}$

Hence, ${\displaystyle \mu _{\Phi \mid \Lambda }}$ has a uniform density with respect to ${\displaystyle \omega _{r}(\varphi )\,d\varphi }$ but not with respect to the Lebesgue measure. On the other hand, ${\displaystyle \mu _{\Lambda \mid \Phi }}$ has a uniform density with respect to ${\displaystyle \omega _{r}(\lambda )\,d\lambda }$ and the Lebesgue measure.