For clarity I have divided my question into parts:

  1. Let [a,b] and [c,d] be two non-overlapping intervals inside the interval [0,N]. Now if I randomly select a number inside [0,N], the probability of the number selected being in either interval is proportional to the length of the interval (b-a) and (c-d). However, this seems to conflict with notions of size in set theory. If we define one set to contain the reals in the interval [a,b] and another set to contain the reals in the interval [c,d], don't these sets have the same cardinality?--in which case it seems they should yield the same probabilities of containing a random number, since they are the same "size".

  2. Now suppose that instead of simply considering the interval [0,N] we consider the entire real number line. It is impossible to randomly select any real number in the intuitive sense like we did in (1) (assigning "equal chance" to equally sized intervals) because then the probability density would not be normalizable. Thus we must define a probability distribution which determines how the "random" number is selected. So it is no longer the case that larger intervals have a higher probability of containing the "randomly" selected number. Rather, the probabilities that a number lies in any particular interval depends on the probability distribution chosen. For example, a continous probability distribution can be chosen such that the probability for both intervals is 0, or one is greater than the other, or vice versa. In any case--intuitively, the probability distribution "biases" the random number towards certain parts of the real line. (I guess this wasn't really a question but just wanting to verify I am correct).

  3. Given (2), I can't help but feel like mathematics has failed me, for it seems at least conceivable to choose a random real number, each with equal probability. Moreover it seems that if this did occur, then given two intervals of different size, we would be justified in saying that it is more likely the randomly selected number falls in the larger interval. Alternatively, it seems we could also say the probability is zero for any finite interval since there are so many reals. Yet there appears to be no mathematical basis for any such claim. Is developing such a mathematical theory just not of interest to mathematicians, or is it impossible?

  4. Is there any probability distribution over the entire real numbers that is close to being uniform?

  5. If I wanted to explore these mathematical questions rigorously, myself, should I be looking to pick up an intro probability text or what? Would an intro-level text answer all these questions?



1 Answer 1


Well in some sense the size of a set is an intrinsic and unalienable property of it. A measure on the other hand is a structure you overlay on top of a collection of sets which must follow a set of rules and which assigns to each set in the collection a non-negative real number. You can measure a collection of sets in many ways (overlay many different measures on the collection), but for each set the size is the size is the size.

Now when working with collections of finite sets size can function as a measure, but as you saw with the real numbers size can't function as a consistent measure in this case.

Yes it's simply not possible to get a consistent probability measure on $\mathbb{R}$ the way you're suggesting in (3). Although in a weird way this is consistent with our experience of the real world. Real numbers take an infinite number of bits to precisely specify, and so in practice we always end up with a little interval due to numerical error.

It depends on what you mean by close to uniform. The normal distribution has the most entropy, there are distributions on $\mathbb{R}$ which have infinite variance, you could take the limit as $n\rightarrow\infty$ of the distribution $\frac{\bf{1}_{(-n,n)}}{2n}$, etc.

If you're interested in probability just start studying it, doing lots of problems, learning the theory, etc. Philosophical thoughts will naturally arise as you learn more and over time you'll be able to form your own opinions on them. I guess that's all I've got.

  • $\begingroup$ Could you explain what it means to take the limit as $n\rightarrow\infty$ of $\frac{\bf{1}_{(-n,n)}}{2n}$ $\endgroup$
    – Anonymous
    Jan 22, 2022 at 11:07

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