Home » Intelligent Design » Information-Theoretic Conjecture — $1000 Cash Prize

Information-Theoretic Conjecture — $1000 Cash Prize

This problem has been resolved by a professor of mathematics. The solution is elegant. The contest is closed. I love the Internet! –WmAD

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I’m offering the first person who completely resolves the following conjecture $1000 cash. I need a complete error-free proof and I need to be able to use it in my writings (of course, I’ll give full credit to the mathematician who proves it).

Information-Theoretic Conjecture

This conjecture is relevant to the conservation of information.

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33 Responses to Information-Theoretic Conjecture — $1000 Cash Prize

  1. I am tempted to offer $1000 cash to anyone who can just resolve that into plain English =P.

  2. Ah! Now I remember why I’ve chosen to take a major in history…

  3. Was never good @ math so i got no idea / no klue. To be honest i was completely konfuesd from the first line :(.

    “I am tempted to offer $1000 cash to anyone who can just resolve that into plain English”
    -good one Mung :)

    Charlie

  4. 42!

  5. oh, wait, wrong question!

  6. It has been a rainy day in Flatland, so I have been inside squaring my thoughts (fie, fie…)

    My best efforts got me here:
    http://www-ee.stanford.edu/~gray/it.pdf

    using keywords like:

    ergodic, shannon, relative entropy, KL entropy, divergence, radon-nikodym, gelfand
    markov, central limit, information measures, random variation

    I am having difficultly expressing this challange as a supremum, rather than the infimum that is given. Most of the similar problems are expressed in in the supremum form.

    Perhaps the old wisdom of throwing the net on the other side of the boat (Jhn 21:6) might yeild some better results.

    Also, if you could toss out some keywords, it might help in the “search”. It has been about fifteen years since I tried anything like this (but the internet was just a dreamcatcher back then)

  7. Marcos: you are majoring is history? Get out while you still can…just kidding, so did I. Love it. Have never earned a dime from it, but I wouldn’t change a thing!

  8. What’s that really tall, skinny “S” anyway?

  9. I don’t know about your screens, but on my screen the ‘blogads’ are covering over some of the equations.

    Is there a way to get rid of (or around) the ‘blogads’ along the right margin?

  10. An integral sign.

  11. PaV: The text of interest is a “picture”. You have two options, If you have activity on the left of your screen, such as “favorites” or “history”, then slide ‘em to the left. That does it on my screen. If that doesn’t work, try upping the resolution of your display.

    Another option is to right-click on the picture and select “save picture as”.

  12. tinabrewer: It´s too late now… =( I finished it 2 years ago. Ain´t made a dime out of it too. But at least we are Historians! HURRAY!

  13. “What’s that really tall, skinny “S” anyway?”
    This really tall skinny “S” is an integrand sign meaning you want to integrate the function that is denoted afterwards. Example:
    ∫f(x)dx, means integrate f(x) or find the anti-derivative of f(x), with respect to variable ‘x’ (that’s what dx tells you). As for the numbers after ∫, it is the interval you want to find the integral or area under the curve.

  14. What’s that really tall, skinny “S” anyway?

    lol! I think that means the summation of what follows the skinny S over the range of the values at the value at the bottom and the value at the top of the skinny S =P. What I can’t tell you though is what that means.

  15. The tall skinny S is an integral sign

  16. The summation symbol is the Capital Sigma in greek, which looks a bit like a pointy E.

  17. in·fi·mum (plural in·fi·ma)

    noun
    Definition:

    largest of lowest elements of set: a number less than or equal to all elements of a set, thus a lower bound, but greater than or equal to all other lower bounds of the set

    [Mid-20th century.

    And:

    The infimum is the greatest lower bound of a set S, defined as a quantity m such that no member of the set is less than m, but if epsilon is any positive quantity, however small, there is always one member that is less than m+epsilon (Jeffreys and Jeffreys 1988). When it exists (which is not required by this definition, e.g., infR does not exist), the infimum is denoted infS or inf_(x in S)x. The infimum can be computed using the Mathematica Version 4.0 undocumented Experimental context command Infimum[f, constr, vars].

    Consider the real numbers with their usual order. Then for any set M subset= R, the infimum infM exists (in R) if and only if M is bounded from below and nonempty.

    More formally, the infimum infS for S a (nonempty) subset of the affinely extended real numbers R^_==R union {+/-infty} is the largest value y in R^_ such that for all x in S we have x>=y. Using this definition, infS always exists and, in particular, infR==-infty.

    Whenever an infimum exists, its value is unique.

    Given a sequence of real numbers a_n, the infimum limit, also called the lower limit but more often simply pronounced ‘lim-inf’ and written lim inf is the limit of
    A_n==inf_(k>n)a_k

    as n->infty. Note that by definition, A_n is nondecreasing, and so either has a limit or tends to infty. For example, suppose a_n==(-1)^n/n, then for n odd, A_n==-1/n, and for n even, A_n==-1/(n+1). Another example is a_n==sinn, in which case A_n is a constant sequence A_n==-1.

    When lim supa_n==lim infa_n, the sequence converges to the real number
    lima_n==lim supa_n==lim infa_n.

    Otherwise, the sequence does not converge.

    Hope that clears things up!

  18. Jwrennie,

    the long horizontal line means division, and the 2 floating in the air means we have to square the result of the prevous letters. I think we are getting closer to the answer…

  19. 19

    to quote “fascinating”, or more contemporary, “intriguing”

  20. As somebody else noted, the blogads are interfering with the equations on my screen – otherwise I would solve it for you.
    But first I need to rent Good Will Hunting…

  21. If I hadn’t known beforehand, I would think that I accessed the wrong blog. Gosh, Bill , you wanna give me a heart attack? *gasps*

  22. I need to find sources on how to integrate general functions like that. My advanced cal book and intro to analysis book don’t contain sufficient material for me to know where to begin. It looks like we’re dealing with generalized probability distributions over the interval [0,1].

  23. Joke!!
    I actually studied calculus in college. However, the equation is way the heck over my head.

  24. I typically don’t attempt equations that a PhD in Mathematics can’t do. Just a personal thing, I guess. I did go through Calc in college, but never made it to differential equations.

  25. He’s just asking how small can the definite integral on the interval [0,1] of g(x)^2 divided by f(x) with respect to x be given that f and g are strictly nonnegative, and severally have definite integrals over [0,1] equal to 1, with the expected value of f on [0,1] less than or equal to some positive number p which is strictly less than 1 and the expected value of g on [0,1] greater than or equal to some positive number q which is strictly bigger than p and also less than 1?

  26. I’ll take a stab at describing Bill’s conjecture in plain English.

    The concept of an infimum is actually quite simple. Take a set of numbers and find the largest number that is less than or equal to the smallest number in the set. This is the greatest lower bound. From example, if the set of numbers is all positive integers (1, 2, 3, etc.) the infimum is 1, which is the smallest number in the set. If the set of numbers is all real numbers x such that x^2 > 2 (x squared is greater than 2, and there is no smallest number in this set because we can get arbitrarily close to 2 on the high side by using an arbitrary number of decimal places in our x) then the infimum is the square root of 2, which when squared gives 2, which is the number that x squared must always be larger than.

    Take two numbers p and q that are both greater than 0 and less than 1, and p is smaller than q. Take some function g, square it, and divide that by some function f. Plot the resulting curve and take the area under that curve between 0 and 1 on the x axis. This is the integral for which we are seeking an infimum. The functions g and f have limitations. They must have values greater than or equal to zero. When plotted, they must both have an area under the curve of 1 square unit between 0 and 1 on the x axis. When multiplied by the variable x, f must have an area under the curve less than or equal to p between 0 and 1 on the x axis. When multiplied by the variable x, g must have an area under the curve greater than or equal to q between 0 and 1 on the x axis.

    Given these constraints, Bill proposes that ((1-q)^2 / (1-p)) + q^2/p will be the largest number that is smaller than the smallest number produced by the original integral. Since probabilities lie between 0 (impossible) and 1 (certain), this looks like a search for a lower probability bound.

    One thing looks fishy. If the function f is allowed to take on a value of zero, won’t you get division by zero in the original integral? Division by zero is a no-no.

  27. This proves how important it cn be to be the ASKER of questions.

  28. I think we need Matt Daemon here. Just give him a copy of the “Good Will Hunting” script, and I’m sure he’ll solve the equation in no time.

    GilDodgen, you are clearly the closest. Kinda cool that you figured out where Dembski is heading and already found a major issue with the formula before even being able to solve it. I’m impressed.

  29. To progress, we can try a test case as an illustration.
    Suppose we choose f(x) = (n+1)x^n, then the first integral [0,1] = 1 is satisfied.
    Suppose we choose g(x) = (n+2)x^(n+1), then the second integral [0,1] = 1 is also satisfied.
    The integral xf(x) [0,1] is (n+1)/(n+2) =q.
    This implies that 0

  30. Corretion. The conjecture sums to 10/9 in the previous example (not 2), so the infimum conjecture holds for this example.

  31. “One thing looks fishy. If the function f is allowed to take on a value of zero, won’t you get division by zero in the original integral? Division by zero is a no-no.”

    Several thoughts: Almost no probability is ever exactly zero. Furthermore, there could be other conditions, such as g0 also tends to zero. Integration through singularities is quite common in mathematics; it all depends on the behaviour of the function around the singularity.

  32. Hmm. It seems I must have used some special escape characters in my last comment, because it’s missing an entire sentence. I’ll spell it out instead: “Furthermore, there could be other conditions, such as g is less than f as g tends to zero, which would ensure that g squared divided by f has an integrable limit as f tends to zero.”

  33. 33

    William:

    I’ve sent a proof of your conjecture to your ISCID email address. Is there a better place to send it?