| Nolecaj Dnomsed Poster Post Number: 16 |
If X is uniformly distributed on [0,1], what is the probability that X is rational? Does it even make sense to ask this question? Thanks, Nolecaj |
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| David Loeffler Veteran poster Post Number: 918 |
Yes, it does make sense, and the probability that X is rational is zero. Have you come across the idea of countable and uncountable sets? If you have, let me know and I'll explain why "almost all numbers are irrational". If not, say so and I'll see if I can find a suitable explanation of that concept. |
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| Nolecaj Dnomsed Poster Post Number: 17 |
Yes, I know about (un)countable sets and I see what you mean. So the only non-negative probabilities are associated with uncountable subsets of [0,1]? Is the probability that X is an irrational in(0,1/2) equal to 1/2? |
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| Nolecaj Dnomsed Poster Post Number: 19 |
What about an arbitrary uncountable set B of irrationals in [0,1]? I mean, say you take all the irrationals and form B by removing uncountably many of them in an arbitrary way. Can you do that? If so, what is the probability that X is in B? |
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| David Loeffler Veteran poster Post Number: 923 |
Yes, the probability that X is in (0,1/2) and is irrational is the same as the probability that X is in (0,1/2), which is 1/2, since the rationals in (0,1/2) are countable and hence occur with probability zero in a uniform distribution (we say they are a set of measure zero). As for the second question, it can be 0, 1, anything in between, or undefined. (There are sets that are so nasty that it's impossible to associate a consistent probability with them). For example, the probability that X lies in the Cantor set is zero, although the Cantor set is uncountable. |
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| Nolecaj Dnomsed Poster Post Number: 20 |
I see. How can you tell if your set is nasty or not? Suppose defined by if and otherwise. So it is possible that is not defined? |
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| David Loeffler Veteran poster Post Number: 924 |
Yes, it is possible that P(f(n, X) = 1) could be undefined. To spell out the correct definition of "non-nasty" (the usual term is measurable ) is difficult; it's essentially what the large area of mathematics called measure theory is all about. (But I believe that any set you can construct without using the axiom of choice is going to be measurable.) David |
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| Nolecaj Dnomsed Poster Post Number: 21 |
Thanks very much for your help; measure theory must be quite interesting. Nolecaj |