Borel's paradox

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Borel's paradox (sometimes known as the Borel-Kolmogorov paradox) is a paradox of probability theory relating to conditional probability density functions. The paradox lies in fact that, contrary to intuition, conditional probability density functions are not invariant under coordinate transformations.

Suppose we have two random variables, X and Y, with joint probability density p_X,Y(x,y). We can form the conditional density for Y given X,

$p_{Y|X}(y|x) = \frac{p_{X,Y}(x,y)}{p_{X}(x)}$

where p_X(x) is the appropriate marginal distribution.

Using the substitution rule, we can reparametrize the joint distribution with the functions U= f(X,Y), V = g(X,Y), and can then form the condition density for V given U.

$p_{V|U}(v|u) = \frac{p_{U,V}(u,v)}{p_{U}(u)}$

Given a particular condition on X and the equivalent condition on U, intuition suggests that the conditional densities p_Y|X(y|x) and p_V|U(v|u) should also be equivalent. This is not the case in general.

1 A concrete example
2 See also

[edit] A concrete example

[edit] A uniform distribution

We are given the joint probability density

$p_{X,Y}(x,y) =\left\{\begin{matrix} 1, & 0 < y < 1, \quad -y < x < 1 - y \\ 0, & \mbox{otherwise} \end{matrix}\right.$

The marginal density of X is calculated to be

$p_X(x) =\left\{\begin{matrix} 1+x, & -1 < x \le 0 \\ 1 - x, & 0 < x < 1 \\ 0, & \mbox{otherwise}\end{matrix}\right.$

So the conditional density of Y given X is

$p_{Y|X}(y|x) =\left\{\begin{matrix} \frac{1}{1+x}, & -1 < x \le 0, \quad -x < y < 1 \\ \\ \frac{1}{1-x}, & 0 < x < 1, \quad 0 < y < 1 - x \\ \\ 0, & \mbox{otherwise}\end{matrix}\right.$

which is uniform with respect to y.

[edit] Reparametrization

Now, we apply the following transformation:

$U = \frac{X}{Y} + 1 \qquad \qquad V = Y.$

Using the substitution rule, we obtain

$p_{U,V}(u,v) =\left\{\begin{matrix} v, & 0 < v < 1, \quad 0 < u \cdot v < 1 \\ 0, & \mbox{otherwise} \end{matrix}\right.$

The marginal distribution is calculated to be

$p_U(u) =\left\{\begin{matrix} \frac{1}{2}, & 0 < u \le 1 \\ \\ \frac {1}{2u^2}, & 1 < u < +\infty \\ \\ 0, & \mbox{otherwise}\end{matrix}\right.$

So the conditional density of V given U is

$p_{V|U}(v|u) =\left\{\begin{matrix} 2v, & 0 < u \le 1, \quad 0 < v < 1 \\ 2u^2v, & 1 < u < +\infty, \quad 0 < v < \frac{1}{u} \\ 0, & \mbox{otherwise}\end{matrix}\right.$