Borel–Kolmogorov paradox

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 = ${\textstyle \left[-{\frac {\pi }{2}},{\frac {\pi }{2}}\right]}$ 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

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

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

${\displaystyle {\begin{aligned}\omega _{r}(\lambda )&=\left\|{\partial (x,y,z) \over \partial \varphi }\right\|=r\ ,\quad {\text{respectively}}\\[3pt]\omega _{r}(\varphi )&=\left\|{\partial (x,y,z) \over \partial \lambda }\right\|=r\cos \varphi \ .\end{aligned}}}$

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

${\displaystyle {\begin{aligned}\mu _{\Phi \mid \Lambda }(d\varphi \mid \lambda )&={\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda ) \over \mu _{\Lambda }(d\lambda )}={\frac {1}{2r}}\omega _{r}(\varphi )\,d\varphi \ ,\quad {\text{and}}\\[3pt]\mu _{\Lambda \mid \Phi }(d\lambda \mid \varphi )&={\mu _{\Phi ,\Lambda }(d\varphi ,d\lambda ) \over \mu _{\Phi }(d\varphi )}={\frac {1}{2r\pi }}\omega _{r}(\lambda )\,d\lambda \ .\end{aligned}}}$

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.

Consider a random vector ${\displaystyle (X,Y,Z)}$ that is uniformly distributed on the unit sphere ${\displaystyle S^{2}}$.

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 ${\textstyle -{\frac {\pi }{2}}\leq \varphi \leq {\frac {\pi }{2}}}$ 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:

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

Using the formulas for the surface area spherical cap and the spherical wedge, the surface of a spherical cap wedge is given by

${\displaystyle \operatorname {Area} (\Theta \leq \theta ,\Phi \leq \varphi )=(1+\sin(\varphi ))(\theta +\pi )}$

Since ${\displaystyle (X,Y,Z)}$ is uniformly distributed, the probability is proportional to the surface area, giving the joint cumulative distribution function

${\displaystyle F_{\Phi ,\Theta }(\varphi ,\theta )=P(\Theta \leq \theta ,\Phi \leq \varphi )={\frac {1}{4\pi }}(1+\sin(\varphi ))(\theta +\pi )}$

The joint probability density function is then given by

${\displaystyle f_{\Phi ,\Theta }(\varphi ,\theta )={\frac {\partial ^{2}}{\partial \varphi \partial \theta }}F_{\Phi ,\Theta }(\varphi ,\theta )={\frac {1}{4\pi }}\cos(\varphi )}$

Note that ${\displaystyle \Phi }$ and ${\displaystyle \Theta }$ are independent random variables.

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

${\displaystyle {\begin{aligned}A&=\left\{0<\Theta <{\frac {\pi }{4}}\right\}&&=\{0<X<1,0<Y<X\}\\B&=\{\Phi =0\}&&=\{Z=0\}\end{aligned}}}$

We attempt to evaluate the conditional probability as a limit of conditioning on the events

${\displaystyle B_{\varepsilon }=\{|\Phi |<\varepsilon \}}$

As ${\displaystyle \Phi }$ and ${\displaystyle \Theta }$ are independent, so are the events ${\displaystyle A}$ and ${\displaystyle B_{\varepsilon }}$, therefore

${\displaystyle P(A\mid B)\mathrel {\stackrel {?}{=}} \lim _{\varepsilon \to 0}{\frac {P(A\cap B_{\varepsilon })}{P(B_{\varepsilon })}}=\lim _{\varepsilon \to 0}P(A)=P\left(0<\Theta <{\frac {\pi }{4}}\right)={\frac {1}{8}}.}$

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

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

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

Define new random variables

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

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:

${\displaystyle {\begin{aligned}A&=\left\{0<\Theta <{\frac {\pi }{4}}\right\}&&=\{0<X<1,\ 0<Y<X\}&&=\left\{0<\Theta '<\pi ,\ 0<\Phi '<{\frac {\pi }{2}},\ \sin(\Theta ')<\tan(\Phi ')\right\}\\B&=\{\Phi =0\}&&=\{Z=0\}&&=\left\{\Theta '=-{\frac {\pi }{2}}\right\}\cup \left\{\Theta '={\frac {\pi }{2}}\right\}.\end{aligned}}}$

Attempting again to evaluate the conditional probability as a limit of conditioning on the events

${\displaystyle B_{\varepsilon }^{\prime }=\left\{\left|\Theta '+{\frac {\pi }{2}}\right|<\varepsilon \right\}\cup \left\{\left|\Theta '-{\frac {\pi }{2}}\right|<\varepsilon \right\}.}$

Using L'Hôpital's rule and differentiation under the integral sign:

${\displaystyle {\begin{aligned}P(A\mid B)&\mathrel {\stackrel {?}{=}} \lim _{\varepsilon \to 0}{\frac {P(A\cap B_{\varepsilon }^{\prime })}{P(B_{\varepsilon }^{\prime })}}\\&=\lim _{\varepsilon \to 0}{\frac {1}{\frac {4\varepsilon }{2\pi }}}P\left({\frac {\pi }{2}}-\varepsilon <\Theta '<{\frac {\pi }{2}}+\varepsilon ,\ 0<\Phi '<{\frac {\pi }{2}},\ \sin(\Theta ')<\tan(\Phi ')\right)\\&={\frac {\pi }{2}}\lim _{\varepsilon \to 0}{\frac {\partial }{\partial \varepsilon }}\int _{{\pi }/{2}-\epsilon }^{{\pi }/{2}+\epsilon }\int _{0}^{{\pi }/{2}}1_{\sin(\theta )<\tan(\varphi )}f_{\Phi ',\Theta '}(\varphi ,\theta )\mathrm {d} \varphi \mathrm {d} \theta \\&=\pi \int _{0}^{{\pi }/{2}}1_{1<\tan(\varphi )}f_{\Phi ',\Theta '}\left(\varphi ,{\frac {\pi }{2}}\right)\mathrm {d} \varphi \\&=\pi \int _{\pi /4}^{\pi /2}{\frac {1}{4\pi }}\cos(\varphi )\mathrm {d} \varphi \\&={\frac {1}{4}}\left(1-{\frac {1}{\sqrt {2}}}\right)\neq {\frac {1}{8}}\end{aligned}}}$

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.