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In the textbook from which I am teaching a Discrete Math course, the authors propose randomly generating an infinite sequence of decimal digits $d_1, d_2, \dots$. We are to think of this as the decimal expansion of a real number in the unit interval.

They propose that "it is overwhelmingly likely" that the resulting sequence is non-periodic, and thus that the number in question is irrational. This is intended to motivate the idea that there are more irrational numbers in [0,1] than rationals in [0,1].

This argument is appealing until you realize that it's nonsense, since we are somehow taking a uniform probability distribution on an uncountable set.

My question is whether there is any way to interpret this so that it is not nonsense.

Let me make a couple of quick points, in the hopes that this question will not be closed out of hand. I know the irrationals are uncountable, and I know the rationals are a set of measure zero. What I am asking is whether the above argument can be put on a legitimate footing. I don't insist that it be possible to put it on a legitimate footing which I could explain to my Discrete Math students --that seems clearly out of reach-- but whether or not there is some legitimate footing will alter how I want to present this argument.

The textbook in question, if anyone is interested, is A Discrete Transition to Advanced Mathematics, by Richmond and Richmond. I like it, and will probably use it again.

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    $\begingroup$ There's nothing wrong with the argument in principle, but doing it carefully in full generality takes some measure theory. The key topic is infinite products of probability spaces (see, for example, Theorem 2.4.4 in terrytao.files.wordpress.com/2011/01/measure-book1.pdf). In this special case Lebesgue measure is all you need. $\endgroup$ – Henry Cohn Nov 6 '13 at 3:02
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    $\begingroup$ Not only is there nothing wrong with the argument, but the sense in which the random sequence is almost surely not periodic is the same sense in which the strong law of large numbers tells us that the average value of the digits (i.e., $\lim_{n\to\infty} (\sum_{i=1}^n d_i)/n$) is almost surely 4.5. This kind of probability distribution is fundamental to probability theory. $\endgroup$ – Timothy Chow Nov 7 '13 at 1:51
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You can make sense of the uniform probability distribution on lots of infinite sets, notably any compact topological group $G$, where "uniform probability distribution" should mean "normalized Haar measure." Here the group is an infinite direct product of copies of $\mathbb{Z}/10\mathbb{Z}$, which is compact by Tychonoff's theorem. You've probably also considered the uniform probability distribution on $S^1$ at some point in your life.

The failure of groups like $\mathbb{Z}$ and $\mathbb{R}$ to have a uniform probability distribution isn't just due to infinitude, it's also due to lack of compactness.

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Short answer: The countable product of probability measures is a well-defined object so the countability of periodic sequences is enough to conclude that the probability of drawing a periodic sequence is zero.

Slightly longer answer: Let $\mu$ be the uniform measure on $\{0,1,2,3,4,5,6,7,8,9\}$ since we want to talk about ordinary decimal numbers. In this situation we want to treat the sequences $09999...$ and $100000....$ as distinct since they are different outcomes in the experiment. This won't worry us because there are again only countably many such pairs so we will only be disregarding a probability zero event. The crux of the matter is the product $\prod_{i=1}^{\infty} \mu_i$ where all the $\mu_i = \mu$. But this is the conclusion of the Kolmogorov Extension Theorem which has two main conditions which are satisfied by the assumption that $\mu_i=\mu$ and that we are taking the limit of the product measures, $\prod_{i=1}^{n} \mu_i$.

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