Editing Borel–Kolmogorov paradox (section)

=== Proof of contradiction ===
{{Original research|paragraph|discuss=Talk:Borel–Kolmogorov_paradox#Error_in_"Mathematical_Explication"|date=March 2021}} 

Consider a random vector <math>(X,Y,Z)</math> that is uniformly distributed on the unit sphere <math>S^2</math>.

We begin by parametrizing the sphere with the usual [[spherical polar coordinates]]:
:<math>\begin{aligned}
  x &= \cos(\varphi) \cos (\theta) \\
  y &= \cos(\varphi) \sin (\theta) \\
  z &= \sin(\varphi)
\end{aligned}</math>
where <math display="inline">-\frac{\pi}{2} \le \varphi \le \frac{\pi}{2}</math> and <math>-\pi \le \theta \le \pi</math>.

We can define random variables <math>\Phi</math>, <math>\Theta</math> as the values of <math>(X, Y, Z)</math>
under the inverse of this parametrization, or more formally using the [[arctan2|arctan2 function]]:
:<math>\begin{align}
    \Phi &= \arcsin(Z) \\
  \Theta &= \arctan_2\left(\frac{Y}{\sqrt{1 - Z^2}}, \frac{X}{\sqrt{1 - Z^2}}\right)
\end{align}</math>

Using the formulas for the surface area [[spherical cap]] and the [[spherical wedge]], the surface of a spherical cap wedge is given by
:<math>
\operatorname{Area}(\Theta \le \theta, \Phi \le \varphi) = (1 + \sin(\varphi)) (\theta + \pi)
</math>

Since <math>(X,Y,Z)</math> is uniformly distributed, the probability is proportional to the surface area, giving the [[Joint probability distribution#Joint cumulative distribution function|joint cumulative distribution function]]
:<math>
F_{\Phi, \Theta}(\varphi, \theta) = P(\Theta \le \theta, \Phi \le \varphi) = \frac{1}{4\pi}(1 + \sin(\varphi)) (\theta + \pi)
</math>

The [[Joint probability distribution#Joint density function or mass function|joint probability density function]] is then given by
:<math>
  f_{\Phi, \Theta}(\varphi, \theta) = 
  \frac{\partial^2}{\partial \varphi \partial \theta} F_{\Phi, \Theta}(\varphi, \theta) = 
  \frac{1}{4\pi} \cos(\varphi) 
</math>
Note that <math>\Phi</math> and <math>\Theta</math> 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 <math>\mathbb{P}(A|B)</math> with
:<math>\begin{aligned}
  A &= \left\{ 0 < \Theta < \frac{\pi}{4} \right\} &&= \{ 0 < X < 1, 0 < Y < X \}\\
  B &= \{ \Phi = 0 \} &&= \{ Z = 0 \}
\end{aligned}</math>

We attempt to evaluate the conditional probability as a limit of conditioning on the events
:<math>B_\varepsilon = \{ | \Phi | < \varepsilon \}</math>

As <math>\Phi</math> and <math>\Theta</math> are independent, so are the events <math>A</math> and <math>B_\varepsilon</math>, therefore
:<math>
  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}.
</math>

Now we repeat the process with a different parametrization of the sphere:
:<math>\begin{align}
  x &=  \sin(\varphi) \\
  y &=  \cos(\varphi) \sin(\theta) \\
  z &= -\cos(\varphi) \cos(\theta)
\end{align}</math>
This is equivalent to the previous parametrization [[Rotation matrix#Basic rotations|rotated by 90 degrees around the y axis]].

Define new random variables
:<math>\begin{align}
    \Phi' &= \arcsin(X) \\
  \Theta' &= \arctan_2\left(\frac{Y}{\sqrt{1 - X^2}}, \frac{-Z}{\sqrt{1 - X^2}}\right).
\end{align}</math>

Rotation is [[measure-preserving transformation|measure preserving]] so the density of <math>\Phi'</math> and  <math>\Theta'</math> is the same:
:<math> f_{\Phi', \Theta'}(\varphi, \theta) = \frac{1}{4\pi} \cos(\varphi) </math>.

The expressions for {{mvar|A}} and {{mvar|B}} are:
:<math>\begin{align}
  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{align}</math>

Attempting again to evaluate the conditional probability as a limit of conditioning on the events
:<math>B^\prime_\varepsilon = \left\{ \left|\Theta' + \frac{\pi}{2}\right| < \varepsilon \right\} \cup \left\{ \left|\Theta'-\frac{\pi}{2}\right| < \varepsilon \right\}.</math>

Using [[L'Hôpital's rule]] and [[Leibniz integral rule|differentiation under the integral sign]]:
:<math>\begin{align}
  P(A \mid B) &\mathrel{\stackrel{?}{=}} \lim_{\varepsilon \to 0} \frac{P(A \cap B^\prime_\varepsilon )}{P(B^\prime_\varepsilon )}\\
    &=  \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{align}</math>

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]].