Borel's paradox

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Borel–Kolmogorov paradox

Suppose that a random variable has a uniform distribution on a unit sphere. What is its conditional distribution on a great circle? Because of the symmetry of the sphere, one might expect that the distribution is uniform and independent of the choice of coordinates. However, two analyses give contradictory results. First, note that choosing a point uniformly on the sphere is equivalent to choosing the longitude $ \lambda $ uniformly from $ [-\pi ,\pi ] $ and choosing the latitude $ \varphi $ from $ {\textstyle [-{\frac {\pi }{2}},{\frac {\pi }{2}}]} $ with density $ {\textstyle {\frac {1}{2}}\cos \varphi } $.^[1] Then we can look at two different great circles:

1. If the coordinates are chosen so that the great circle is an equator (latitude $ \varphi =0 $), the conditional density for a longitude $ \lambda $ defined on the interval $ [-\pi ,\pi ] $ is $ {\displaystyle f(\lambda \mid \varphi =0)={\frac {1}{2\pi }}.} $
2. If the great circle is a line of longitude with $ \lambda =0 $, the conditional density for $ \varphi $ on the interval $ {\textstyle [-{\frac {\pi }{2}},{\frac {\pi }{2}}]} $ is $ {\displaystyle f(\varphi \mid \lambda =0)={\frac {1}{2}}\cos \varphi .} $

One distribution is uniform on the circle, the other is not. Yet both seem to be referring to the same great circle in different coordinate systems.

Many quite futile arguments have raged — between otherwise competent probabilists — over which of these results is 'correct'.

— E.T. Jaynes^[1]

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Conditional probability distribution

The concept of the conditional distribution of a continuous random variable is not as intuitive as it might seem: Borel's paradox shows that conditional probability density functions need not be invariant under coordinate transformations.

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A8LectureNotes_MT22_24Sep2022.pdf
Example 4.8 (Borel's paradox). Consider the uniform distribution on the half-disc $C=$ $\left\{(x, y): y \geq 0, x^2+y^2 \leq 1\right\}$. The joint density of $X$ and $Y$ is given by
\[
f(x, y)= \begin{cases}\frac{2}{\pi} & (x, y) \in C, \\ 0 & \text { otherwise. }\end{cases}
\]
What is the conditional distribution of $Y$ given $X=0$ ? Its density is given by
\[
f_{Y \mid X=0}(y)=\frac{2 / \pi}{f_X(0)}
\]
for $y \in[0,1]$, and 0 elsewhere. So the distribution is uniform on $[0,1]$ (we do not need to calculate $f_X(0)$ to see this, since it is only a normalising constant).

We could change variables and represent the same distribution in polar coordinates. Then $R$ and $\Theta$ are independent; $R$ has density $2 r$ on $[0,1]$ and $\Theta$ is uniform on $[0, \pi)$. (See first question on problem sheet 2 for the transformation to polar coordinates. But in this case where the density of $X, Y$ is uniform on a set, one can also easily derive the joint distribution of $R$ and $\Theta$ directly by considering areas of subsets of the set $C$ ).
Note that the events $\{X=0\}$ and $\{\Theta=\pi / 2\}$ are the same.
What is the conditional distribution of $R$ given $\Theta=\pi / 2$ ? Since $R$ and $\Theta$ are independent, it still has density $2 r$ on $[0,1]$. This is not uniform on $[0,1]$.

But when $X=0$, i.e. when $\Theta=\pi / 2, R$ and $Y$ are the same thing. So the distribution of $R$ given $\Theta=\pi / 2$ ought to be the same as the distribution of $Y$ given $X=0$, should it not?

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When $X$ is a discrete random variable, everything works fine; since $\mathbb{P}(X=x)$ will be positive, we can use the approach above.

However, if $X$ is continuous, then $\mathbb{P}(X=x)$ will be 0 for every $x$. Now we have a problem, since if the event $A$ in (4.2) has probability 0 , then the definition makes no sense.

4.4.2 Definition of conditional probability density functions, and an example

To resolve this problem, rather than conditioning directly on $\{X=x\}$, we look at the distribution of $Y$ conditioned on $\{x \leq X \leq x+\epsilon\}$. If the joint distribution is well-behaved (as it will be in all the cases that we wish to consider), we can obtain a limit as $\epsilon \downarrow 0$, which we define as the distribution of $Y$ given $X=x$.
As $\epsilon \rightarrow 0$, we have (if $f_{X, Y}$ is sufficiently smooth)
\begin{align*}
\mathbb{P}(Y \leq y \mid x \leq X \leq x+\epsilon) &=\frac{\int_{v=-\infty}^y \int_{u=x}^{x+\epsilon} f_{X, Y}(u, v) d u d v}{\int_{u=x}^{x+\epsilon} f_X(u) d u} \\
& \sim \frac{\epsilon \int_{v=-\infty}^y f_{X, Y}(x, v) d v}{\epsilon f_X(x)} \\
&=\int_{v=-\infty}^y \frac{f_{X, Y}(x, v)}{f_X(x)} d v\tag{4.3}
\end{align*}
Irrespective of the approximation argument, we define $F_{Y \mid X=x}(y)$, the conditional distribution function of $Y$ given $X=x$, as the right-hand side of (4.3), whenever this is well-defined.

In this case, we also define the conditional density function of $Y$ given $X=x$, written as $f_{Y \mid X=x}(y)$ :
\[
f_{Y \mid X=x}(y)=\frac{f_{X, Y}(x, y)}{f_X(x)} \text {. }
\]
Note that this relates to the (conditional) cumulative distribution function in the usual way.
These definitions make sense whenever $f_X(x)>0$. In that case, note that $f_{Y \mid X=x}$ is indeed a density function, because we have defined $f_X(x)=\int_{-\infty}^{\infty} f_{X, Y}(x, y) d y$. (Notice that the denominator $f_X(x)$ does not involve $y$ at all; it is just a normalising constant).

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当我们将 $X$ 限制在 0 的 $ε$ 附近内时，我们将限制为一个近似于矩形的集合（下图左侧），但是，当我们将 $Θ$ 限制在 $π/2$ 附近时，我们限制为一个薄扇形，这大约是一个三角形（下图右侧）。在第二种情况下，我们将选择的点偏向上方，当 $ε → 0$，这种偏向持续存在；两个极限不相等！

What this “paradox” illustrates is that conditioning for continuous random variables involves a limit, and that it can be important exactly how the limit is taken. The procedure whereby we generate $X$ from $f_X$ and then $Y$ from $f_{Y|X}$ makes sense in terms of a particular set of variables; but the conditional densities involved are not robust to a change of variables.

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hbghlyj 发表于 2022-10-29 16:15
$R$ has density $2 r$ on $[0,1]$

这句话我才明白：
$\Pr(R\le r)=\frac{\frac\pi2r^2}{\frac\pi21^2}=r^2$
$f_R(r)=\frac\rmd{\rmd r}r^2=2r$