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You don't have any knowledge of X beyond the fact that it's between 0 and 100. It could be uniformly distributed, it could be the number of heads in 100 coin fl

You don't have any knowledge of X beyond the fact that it's between 0 and 100. It could be uniformly distributed, it could be the number of heads in 100 coin flips, or it could be the average number of yards that a kickoff was returned for in all NFL games last season. It could even just be 73. Assuming a uniform distribution is an arbitrary prior assumption that you're making with no support from any data, and it's no more or less reasonable than any other prior assumption you could make about the distribution. There's nothing special about the uniform distribution that makes it some kind of natural or default prior for a random variable on a bounded interval.

skissane 9y ago
> There's nothing special about the uniform distribution that makes it some kind of natural or default prior for a random variable on a bounded interval. I don't agree. Suppose I carry out this process: Step 1) Generate a random histogram over the bounded interval Step 2) Pick a random number according to the distribution that histogram represents If I repeat that process a large number of times (say a few million), what will be the distribution of the result? It will be approximately uniform, and the more times I repeat the process the closer to uniform I get. So the uniform distribution actually is special, as the average case of randomly chosen distributions, and hence why it should act as the default when the only thing we know about the distribution is an upper and lower bound.
- rcthompson 9y ago
  If you're trying to say that the uniform distribution represents an average of the uncountably infinite set of possible distributions on an interval, I don't see how you can justify that. "Generate a random distribution on an interval" is not an obviously well-defined procedure, so I don't think you can say anything about what the distribution produced by "averaging" every possible distribution would look like. Which matches the fact that you have no information on what the distribution of X looks like, other than that it is zero outside of the interval [0,100]. Edit: In any case, the process you've described is just begging the question (i.e. it is circular reasoning). In order to prove that the uniform distribution is a natural choice for X, you assume that choosing uniformly from the set of all possible distributions is natural. Even if there was a finite number of possible distributions to choose from such that choosing randomly from that set according to a uniform distribution was a well-defined operation, you're still assuming that it's natural to assign equal probability to all options in order to prove that it's natural to assign equal probability to all options.
  - skissane 9y ago
    > If you're trying to say that the uniform distribution represents an average of the uncountably infinite set of possible distributions on an interval, I don't see how you can justify that. So, is this X actually an arbitrary real number? Because real numbers are not known to exist in the physical world. In the physical world, we will take some measurement to a finite number of digits, and also produce an estimate of the error in that measurement. The measurement and the error bounds are all described by rational numbers, of which there are countably many. We don't know whether the set of possible "real values" (as opposed to measurements – if that concept is even meaningful) is actually uncountable – for all we know, all the "real values" could be rational, or could be drawn from some countable subset of the reals. So, if we are dealing with some kind of measured physical quantity, we actually know that the interval [0,100] is countable. We also know that the value could only possibly be measured to some finite degree of accuracy, so we actually know that the interval [0,100] is a finite set. > In any case, the process you've described is just begging the question (i.e. it is circular reasoning). Maybe it is just an axiom? At some point we just have to accept some things as axiomatic, so why not accept that? It can't be used to derive a contradiction, and it feels self-evident to me (even if not to you), so I feel justified in accepting it as such.
    - rcthompson 9y ago
      I think it's probably reasonable to say it's an axiom. That's compatible with my claim that it's an arbitrary choice that is not inherently more correct nor better supported by any information than any other choice, except perhaps in the sense that it feels better or more intuitive.
      - skissane 9y ago
        Well, what happens if we adopt alternative axioms (e.g. some sort of assumption of a normal distribution instead of a uniform one)? Are those alternative axioms consistent? Maybe we might be able to show that agents adopting the uniform axiom make better (more rewarding) decisions in practice than those who adopt a competing axiom do? I don't have a proof, but my gut says yes.
newen 9y ago
> There's nothing special about the uniform distribution that makes it some kind of natural or default prior for a random variable on a bounded interval. Okay... what's your basis for this? Uniform prior is how Bayesians do it, even in unbounded intervals. It's especially natural when you have a bounded interval. Again, the whole point is that you are not making use of any other prior information except for the interval bounds.
- rcthompson 9y ago
  Well, you have to make some prior assumption, and the uniform seems intuitively reasonable. But I don't think there's a theoretical justification that says the uniform is an inherently better choice than anything else. It's still an arbitrary choice.
  - newen 9y ago
    I am sure there has been lots of theoretical justification in favor of the uniform prior in Bayesian statistics, especially for bounded intervals. I am not a stats guy though, so I can't show you the justifications in detail. The whole argument though was that you are not using prior information. And so you cannot assign prior probabilities in a more intelligent manner than uniform. If you do use prior information like say, that lots of people have tried to solve for P=NP and have not succeeded, you can assign a higher probability to P/=NP. But that's only if you assume that people trying to solve for P=NP and not succeeding makes it more likely that P/=NP is true. It is an assumption that you have to make explicit and this is the assumption that is being argued for in the OP article. The thing is that without having priors, you can't do better than uniform.

skissane 9y ago

> There's nothing special about the uniform distribution that makes it some kind of natural or default prior for a random variable on a bounded interval. I don't agree. Suppose I carry out this process: Step 1) Generate a random histogram over the bounded interval Step 2) Pick a random number according to the distribution that histogram represents If I repeat that process a large number of times (say a few million), what will be the distribution of the result? It will be approximately uniform, and the more times I repeat the process the closer to uniform I get. So the uniform distribution actually is special, as the average case of randomly chosen distributions, and hence why it should act as the default when the only thing we know about the distribution is an upper and lower bound.

rcthompson 9y ago
If you're trying to say that the uniform distribution represents an average of the uncountably infinite set of possible distributions on an interval, I don't see how you can justify that. "Generate a random distribution on an interval" is not an obviously well-defined procedure, so I don't think you can say anything about what the distribution produced by "averaging" every possible distribution would look like. Which matches the fact that you have no information on what the distribution of X looks like, other than that it is zero outside of the interval [0,100]. Edit: In any case, the process you've described is just begging the question (i.e. it is circular reasoning). In order to prove that the uniform distribution is a natural choice for X, you assume that choosing uniformly from the set of all possible distributions is natural. Even if there was a finite number of possible distributions to choose from such that choosing randomly from that set according to a uniform distribution was a well-defined operation, you're still assuming that it's natural to assign equal probability to all options in order to prove that it's natural to assign equal probability to all options.
- skissane 9y ago
  > If you're trying to say that the uniform distribution represents an average of the uncountably infinite set of possible distributions on an interval, I don't see how you can justify that. So, is this X actually an arbitrary real number? Because real numbers are not known to exist in the physical world. In the physical world, we will take some measurement to a finite number of digits, and also produce an estimate of the error in that measurement. The measurement and the error bounds are all described by rational numbers, of which there are countably many. We don't know whether the set of possible "real values" (as opposed to measurements – if that concept is even meaningful) is actually uncountable – for all we know, all the "real values" could be rational, or could be drawn from some countable subset of the reals. So, if we are dealing with some kind of measured physical quantity, we actually know that the interval [0,100] is countable. We also know that the value could only possibly be measured to some finite degree of accuracy, so we actually know that the interval [0,100] is a finite set. > In any case, the process you've described is just begging the question (i.e. it is circular reasoning). Maybe it is just an axiom? At some point we just have to accept some things as axiomatic, so why not accept that? It can't be used to derive a contradiction, and it feels self-evident to me (even if not to you), so I feel justified in accepting it as such.
  - rcthompson 9y ago
    I think it's probably reasonable to say it's an axiom. That's compatible with my claim that it's an arbitrary choice that is not inherently more correct nor better supported by any information than any other choice, except perhaps in the sense that it feels better or more intuitive.
    - skissane 9y ago
      Well, what happens if we adopt alternative axioms (e.g. some sort of assumption of a normal distribution instead of a uniform one)? Are those alternative axioms consistent? Maybe we might be able to show that agents adopting the uniform axiom make better (more rewarding) decisions in practice than those who adopt a competing axiom do? I don't have a proof, but my gut says yes.

newen 9y ago

> There's nothing special about the uniform distribution that makes it some kind of natural or default prior for a random variable on a bounded interval. Okay... what's your basis for this? Uniform prior is how Bayesians do it, even in unbounded intervals. It's especially natural when you have a bounded interval. Again, the whole point is that you are not making use of any other prior information except for the interval bounds.

rcthompson 9y ago
Well, you have to make some prior assumption, and the uniform seems intuitively reasonable. But I don't think there's a theoretical justification that says the uniform is an inherently better choice than anything else. It's still an arbitrary choice.
- newen 9y ago
  I am sure there has been lots of theoretical justification in favor of the uniform prior in Bayesian statistics, especially for bounded intervals. I am not a stats guy though, so I can't show you the justifications in detail. The whole argument though was that you are not using prior information. And so you cannot assign prior probabilities in a more intelligent manner than uniform. If you do use prior information like say, that lots of people have tried to solve for P=NP and have not succeeded, you can assign a higher probability to P/=NP. But that's only if you assume that people trying to solve for P=NP and not succeeding makes it more likely that P/=NP is true. It is an assumption that you have to make explicit and this is the assumption that is being argued for in the OP article. The thing is that without having priors, you can't do better than uniform.