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.
Suppose we have two random variables, X and Y, with joint probability density pX,Y(x,y). We can form the conditional density for Y given X,
- <math>p_{Y|X}(y|x) = \frac{p_{X,Y}(x,y)}{p_{X}(x)}</math>
where pX(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.
- <math>p_{V|U}(v|u) = \frac{p_{U,V}(u,v)}{p_{U}(u)}</math>
Given a particular condition on X and the equivalent condition on U, intuition suggests that the conditional densities pY|X(y|x) and pV|U(v|u) should also be equivalent. This is not the case in general.
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A concrete example
A uniform distribution
We are given the joint probability density
- <math>p_{X,Y}(x,y) =\left\{\begin{matrix} 1, & 0 < y < 1, \quad -y < x < 1 - y \\ 0, & \mbox{otherwise} \end{matrix}\right. </math>
The marginal density of X is calculated to be
- <math>p_X(x) =\left\{\begin{matrix} 1+x, & -1 < x \le 0 \\ 1 - x, & 0 < x < 1 \\ 0, & \mbox{otherwise}\end{matrix}\right. </math>
So the conditional density of Y given X is
- <math>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. </math>
which is uniform with respect to y.
Reparametrization
Now, we apply the following transformation:
- <math>U = \frac{X}{Y} + 1 \qquad \qquad V = Y.</math>
Using the substitution rule, we obtain
- <math>p_{U,V}(u,v) =\left\{\begin{matrix} v, & 0 < v < 1, \quad 0 < u \cdot v < 1 \\ 0, & \mbox{otherwise} \end{matrix}\right. </math>
The marginal distribution is calculated to be
- <math>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. </math>
So the conditional density of V given U is
- <math>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. </math>
which is not uniform with respect to v.
The unintuitive result
Now we pick a particular condition to demonstrate Borel's paradox. The conditional density of Y given X = 0 is
- <math>p_{Y|X}(y|x=0) = \left\{\begin{matrix} 1, & 0 < y < 1 \\ 0, & \mbox{otherwise}\end{matrix}\right. </math>
The equivalent condition in the u-v coordinate system is U = 1, and the conditional density of V given U = 1 is
- <math>p_{V|U}(v|u=1) = \left\{\begin{matrix} 2v, & 0 < v < 1 \\ 0, & \mbox{otherwise}\end{matrix}\right. </math>
Paradoxically, V = Y and X = 0 is equivalent to U = 1, but
- <math>p_{Y|X}(y|x = 0) \ne p_{V|U}(v|u = 1).\;</math>