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#53
10-20-2005, 09:46 PM
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Re: Classic Type Game Theory Problem
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Would you rather that the players were dealt integers uniformly distributed over the range [0, N], where N is a large positive integer? [/ QUOTE ] This doesn't come close to approximating the intent of the original problem. But if that's what we want to make the problem, that's fine. At least it makes sense. 
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#55
10-20-2005, 11:12 PM
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Re: Classic Type Game Theory Problem
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This doesn't come close to approximating the intent of the original problem. [/ QUOTE ] How do you know what the intent was? You said the problem made no sense to you.
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#56
10-20-2005, 11:16 PM
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Re: Classic Type Game Theory Problem
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[ QUOTE ] This doesn't come close to approximating the intent of the original problem. [/ QUOTE ] How do you know what the intent was? You said the problem made no sense to you. [/ QUOTE ] Valid point. I would like to reiterate that the original statement does not make sense to me.
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#57
10-20-2005, 11:17 PM
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Re: Classic Type Game Theory Problem
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Suppose m is a measure on [0,1] such that m([0,1]) = 1 and m({x}) = m({y}) for any x, y in [0,1]. That's what we mean by a uniform distribution. [/ QUOTE ] No it's not. That would allow too many measures. For example, any probability measure on [0,1] which is absolutely continuous with respect to Lebesgue measure has this property. A uniform distribution is one which is invariant under translations, rotations, and reflections. [ QUOTE ] If m({x}) > 0 for some x in [0,1] choose n so large that n * m({x}) > 1. Then choose n distinct x_1, ... x_n in [0,1]. We'd have 1 = m([0,1]) >= m({x_1, ..., x_n}) = m({x_1}) + ... + m({x_n}) = n * m({x_1}) > 1 so that 1 > 1. This is a contradiction. Hence m({x}) = 0 for any x in [0,1]. Oops. [/ QUOTE ] What does "oops" mean here? Do you think you have arrived at a contradiction to the existence of m? "Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones. Are you, by chance, an algebraist?
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#58
10-20-2005, 11:30 PM
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Re: Classic Type Game Theory Problem
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[ QUOTE ] Suppose m is a measure on [0,1] such that m([0,1]) = 1 and m({x}) = m({y}) for any x, y in [0,1]. That's what we mean by a uniform distribution. [/ QUOTE ] No it's not. That would allow too many measures. For example, any probability measure on [0,1] which is absolutely continuous with respect to Lebesgue measure has this property. [/ QUOTE ] No. What I mean by "uniform distribution" is a measure that assigns nonzero equal measure to each of the singletons in the space. This is what we need so that we can pick real numbers uniformly at random. If you want a continuous uniform distribution that's a different issue. [ QUOTE ] A uniform distribution is one which is invariant under translations, rotations, and reflections. [/ QUOTE ] The notions of translations, rotations and reflections don't even make sense on a general measure space. [ QUOTE ] [ QUOTE ] If m({x}) > 0 for some x in [0,1] choose n so large that n * m({x}) > 1. Then choose n distinct x_1, ... x_n in [0,1]. We'd have 1 = m([0,1]) >= m({x_1, ..., x_n}) = m({x_1}) + ... + m({x_n}) = n * m({x_1}) > 1 so that 1 > 1. This is a contradiction. Hence m({x}) = 0 for any x in [0,1]. Oops. [/ QUOTE ] What does "oops" mean here? Do you think you have arrived at a contradiction to the existence of m? [/ QUOTE ] We want the event {x} to have nonzero probability. That's the oops. We concluded that there is no measure assigning nonzero equal weights to all the points. [ QUOTE ] "Uniform distribution on [0,1]" is standard terminology for Lebesgue measure. And if m is Lebesgue measure, then m([0,1])=1 and m({x})=0 for all x. The statement of the problem makes perfect sense and the existence of the uniform distribution on [a,b] is proven in any first-year graduate real analysis course, and some undergraduate ones. [/ QUOTE ] You're joking, right? I just proved for you that you can't assign nonzero equal mass to each of the points. You can't pick real numbers uniformly at random. [ QUOTE ] Are you, by chance, an algebraist? [/ QUOTE ] No. Don't insult me. I'm a complex analyst. I study very serious problems which to a certain extent involve about trying to find very special measures on certain compact subsets of the complex place. The notions of Hausdorff measure, doubling measures, Calderon-Zygmund theory of singular integrals, capacity, etc. all play very major roles in what I study. Capacity as you might know arises by assigning a very special probability measure to the boundary of a compact set. cf. the Painleve problem, the Denjoy conjecture, the Vitushkin conjecture and Tolsa's theorem.
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#59
10-20-2005, 11:39 PM
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Re: Classic Type Game Theory Problem
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Don't insult me... I study very serious problems [/ QUOTE ] Nit! [img]/images/graemlins/laugh.gif[/img] Also, congratulations to Jerrod Ankenman for solving a problem which apparently makes no sense.
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#60
10-21-2005, 01:33 AM
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typed solution
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if you are right [/ QUOTE ] [img]/images/graemlins/shocked.gif[/img] I typed a proof of my solution, but it's a 3-page pdf file. If you or anyone wants it, they can PM me an email address and I will email it. I may not be able to respond until Fri night / Sat morning. I also estimate the overall edge to player A as 8%, though I didn't proofread that part. That is lower than I expected. The document makes it clear where the alThor constant comes from. (Though I didn't call it that [img]/images/graemlins/smile.gif[/img] ) alThor
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Linear Mode
