0
$\begingroup$

I'm a bit conflicted by two answers I read.

This first one, regarding the implications of different infinite cardinalities, explains that

We often talk about a "uniform" probability over [0,1] where every singleton has probability zero. If [0,1] is countable, we can't do that.

This to me implies we can define a uniform probability distribution over the interval.

However, this question hints at the fact that we can't define a uniform probability distribution over the positive reals. Since one can create a bijection from $[0,1]$ to $\mathbb{R}^+$, how are they different? And if so, why can we define a uniform distribution on the interval but not the positive reals? I have yet to take a course with measure theory.

$\endgroup$
  • 2
    $\begingroup$ People generally take 'uniform' to mean that e.g. all intervals of the same length have the same probability. How do you create a bijection from [0,1] to $\mathbb{R}^+$ that preserves equality of lengths? $\endgroup$ – Steven Stadnicki Jun 30 '18 at 5:22
2
$\begingroup$

Bijections just care about the cardinality of the two sets. Once you have a bijection it doesn't tell you much about how measure transforms unless it is differentiable.

As an example, let us make a bijection between $[0,1]$ and $[0,2]$ as $$f(x)= \begin {cases} x&0\le x \le 0.9\\11(x-0.9)&x \gt 0.9 \end {cases}$$ If we pull a random number $x$ uniformly in $[0,1]$ the chance that $f(x)$ is in a small interval less than $0.9$ is the length of the interval. The chance that $f(x)$ is in a small interval greater than $0.9$ is $\frac 1{11}$ times the length of the interval. When both intervals are finite there is a linear relation between that keeps the measure of intervals proportional to their length. If one interval is infinite that is not possible. You can define a measure on $\Bbb R^+$ but it will not be uniform.

$\endgroup$
0
$\begingroup$

If $B\subseteq\mathbb R$ is measurable with $0<\lambda(B)<\infty$ (where $\lambda$ denotes the Lebesgue measure) then we can induce a well defined uniform probability by setting: $$\mathsf P(A)=\frac{\lambda(A)}{\lambda(B)}$$for measurable subsets of $A$.

Unfortunately this does not work if $\lambda(B)=+\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.