“Conservation of Information Made Simple” at ENV

_{William Dembski

August 28, 2012

Informatics, Intelligent Design}

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Evolution News & Views just posted a long article I wrote on conservation of information.

EXCERPT: “In this article, I’m going to follow the example of these books, laying out as simply and clearly as I can what conservation of information is and why it poses a challenge to conventional evolutionary thinking. I’ll break this concept down so that it seems natural and straightforward. Right now, it’s too easy for critics of intelligent design to say, ‘Oh, that conservation of information stuff is just mumbo-jumbo. It’s part of the ID agenda to make a gullible public think there’s some science backing ID when it’s really all smoke and mirrors.’ Conservation of information is not a difficult concept and once it is understood, it becomes clear that evolutionary processes cannot create the information required to power biological evolution.” MORE

TEASER: The article quotes some interesting email correspondence that I had with Richard Dawkins and with Simon Conway Morris, now going back about a decade, but still highly relevant.

Comments

Joe:
I'll ask you a few of the questions that Joe hasn't answered: - If I define ? for a die as {even numbers, odd numbers}, is that improperly defined?
I answered that one- yes it is improperly defined as 8 is an even number but 8 isn't in the search space.
Sorry, I had forgotten that you had already answered this. And my intent was {even numbers on a die, odd numbers on a die}, but I wasn't explicit about that. Regardless, when you say that "8 isn't in the search space", I assume you mean that there is no way to get an 8 from rolling a die. But you have already acknowledged that zero-probability outcomes can be included in Ω, so why can't all numbers be included?

- If I define Ω for poker hands as {royal flush, everything else}, is that improperly defined?
Yes
And yet Dembski has used that exact sample space in at least three of his works.R0bb_{September 10, 2012
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Joe:

I'll ask you a few of the questions that Joe hasn't answered: - If I define ? for a die as {even numbers, odd numbers}, is that improperly defined?
I answered that one- yes it is improperly defined as 8 is an even number but 8 isn't in the search space.
Sorry, I had forgotten that you had already answered this. And my intent was {even numbers on a die, odd numbers on a die}, but I wasn't explicit about that. Regardless, when you say that "8 isn't in the search space", I assume you mean that there is no way to get an 8 from rolling a die. But you have already acknowledged that zero-probability outcomes can be included in Ω, so why can't all numbers be included?

- If I define Ω for poker hands as {royal flush, everything else}, is that improperly defined?
Yes
And yet Dembski has used that exact sample space in at least three of his works.R0bb_{September 10, 2012
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R0bb:
- If I define ? for a die as {even numbers, odd numbers}, is that improperly defined?
No. that's a 50/50 probability distribution.
- If I define ? for poker hands as {royal flush, everything else}, is that improperly defined?
No. Don't ask me for the distribution, though I could probably figure it out given time, lol. I think you highlight an area where Sal needs additional instruction/contemplation. He sometimes confuses {1/6, 1/6, 1/6, 1/6, 1/6, 1/6} with {1/2, 1/2}Mung_{September 7, 2012
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R0bb from #30:
The problem is that Dembski and Marks don’t define Ω this way. In most of their examples, active information is the result of restricting the alternate search to a subset of Ω.
Exactly! The restricted subset does not include the zero-probability choices.
If they were to follow your reasoning, they would define ? to include only the outcomes that are accessible to the alternate search
They redefine the search space within Ω.
, and the resulting active information would be zero. So by your reasoning, their examples of active information don’t really have any active information.
LoL! The restricted subset is the active information, R0bb. THAT is the whole point behind the restriction- the search space is now limited therefor making the search easier.Joe_{September 6, 2012
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R0bb:
scordova, Chance Ratliff, Dieb, onlooker, and anyone else who might read this comment: Do any of you agree that the following is how the LCI applies to the real world?
R0bb if you doubt it why don't you just make your case?Joe_{September 6, 2012
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R0bb:
I’ll ask you a few of the questions that Joe hasn’t answered: - If I define ? for a die as {even numbers, odd numbers}, is that improperly defined?
I answered that one- yes it is improperly defined as 8 is an even number but 8 isn't in the search space.
- If I define ? for poker hands as {royal flush, everything else}, is that improperly defined?
YesJoe_{September 6, 2012
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Robb, You asked.
The second model says that the outcome “1? occurs with a probability of 1/6, and that the outcome “higher than 1? occurs with a probability of 5/6. How does that not conform to a fair die? How does it deviate from the actual statistics of what is being modeled?
I should have been more explicit. I was referring to this claim:
But we could, instead, define ? as {1, higher than 1}. In that case, I+ = log2((5/6) / (1/2)) = .7 bits.
Redefining ? in this way is not faithful to the statistics of a fair die. You gave no justification for leaving p(T) at 5/6 while letting |T|/|?| = 1/2. If you say that |T|/|?| = 1/2 because there are only two states, that's not quite correct because if they are not equiprobable states, you have to modify the way you do the accounting, the fact that P(T) is 5/6 is an indication that they are not equiproable states, and hence the ratio |T|/|?| needs to account for this. As I pointed out that ratio has to be deduced with some qualification. If the states are not equiprobable, that needs to be accounted for. A charitable reading would have perceived this. But anyway, these are the sort of discussions that need to be passed on to Bill, Mark, and the evolution informatics lab. I hope they will address your points. Salscordova_{September 6, 2012
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Since q is not a function of |?|, obviously we don’t expect to see the number 16 in its calculation. But I_+, the active information, is a function of ?, and we do see the number 16 (log-scaled) in its calculation:
R0bb, that is because that calculation is for the difference between the two figures. That 16 refers to figure 1. And that just tells you the amount of information each square contains, ie 2^4.Joe_{September 6, 2012
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R0bb, You keep equating Ω with the searchable space. The two are not the same. Ω can contain zero probability sections and those zero probability sections sections are never considered in the equation. So no, we were not talking about their inclusion in the definition of Ω, we were talking about their inclusion in the search space. At least I was and I made that very clear.Joe_{September 6, 2012
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scordova:
Thank you for your discussion, but I think this isn't the most charitable interpretation regarding the die:
I'm very glad to see you joining the discussion, scordova. I'm not sure what your point is regarding charitability. Are you saying that I'm interpreting Dembski uncharitably, or that I'm modeling the die uncharitably? WRT the way that you say it should be modeled, I think what you mean is the following: |{1}| = 1 |T|=|{2,3,4,5,6}| = 5 |Ω|=|{1}|+|{2,3,4,5,6}|=6 |T|/|Ω|=5/6 That, of course, is identical to the first model in the article. I submit that the second model, where the outcomes are individuated differently, is no less accurate than the first. You seem to dispute this:
Actually the different numbers arise because the model no longer conforms to a fair die. Then notion of the measure of information not being arbitrary, but conforming to the statistics of the probability, still holds. Deviation from the actual statistics of what is being modeled in the way that you have done is imposing an arbitrary (and incorrect) measure of information for the system being modeled.
The second model says that the outcome "1" occurs with a probability of 1/6, and that the outcome "higher than 1" occurs with a probability of 5/6. How does that not conform to a fair die? How does it deviate from the actual statistics of what is being modeled? I'll ask you a few of the questions that Joe hasn't answered: - If I define Ω for a die as {even numbers, odd numbers}, is that improperly defined? - If I define Ω for poker hands as {royal flush, everything else}, is that improperly defined? You say that the measure of information in the example is incorrect -- can you point to the error? You also say that it's arbitrary, which sounds like you're agreeing with the point of the article, since the information measure is Dembski's active information. Dembski used to make this same point often. As quoted in the article, he said that an information measure is ill-defined if it depends on how the possibilities are individuated. He later invented an information measure, namely active information, that depends on how the possibilities are individuated.R0bb_{September 6, 2012
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Joe:
Then the numbers higher than 6 are NOT included.
We're talking about inclusion in the definition of Ω. Are you under the impression that if an outcome has zero probability, then it's automatically excluded from the definition of the sample space? I thought we were now in agreement that Ω can contain zero-probability outcomes, since it does so in the example from the S4S paper. If not, there are some questions that I've already asked that will settle this if you'll attempt to answer them. Will you?R0bb_{September 6, 2012
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Joe:
Thus equation for figure 2 does not contain the number 16
The section regarding figure 2 contains calculations of q, I_S, and I_+. When you say "the equation for figure 2", I assume you're referring to the calculation of q. Is that right? Since q is not a function of |Ω|, obviously we don't expect to see the number 16 in its calculation. But I_+, the active information, is a function of Ω, and we do see the number 16 (log-scaled) in its calculation:
I_+ = 4.00-2.59 = 1.41 bits.
Without the negative log-scaling, this says that the active probability = endogenous probability / exogenous probability = (1/16) / (1/6). Note that it's 1/16, not 1/13, even though there are only 13 non-zero-probability choices. By exactly the same token, the active probability in my random walk example is (1/5) / (1/2), not (1/2) / (1/2), even though there are only 2 non-zero-probability choices. Are we in agreement now?R0bb_{September 6, 2012
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scordova, Chance Ratliff, Dieb, onlooker, and anyone else who might read this comment: Do any of you agree that the following is how the LCI applies to the real world?
As for how the LCI applies in to the real world- It tells us that having directions or a recipe is an easier way to have a successful search than to just do stuff until you get what you want. If you have directions they narrow your search grid, ie they provide active information. The same with a recipe.
R0bb_{September 6, 2012
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Sal, Please see Chance's post #42 above for how to post ΩsJoe_{September 6, 2012
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Sorry to the reader's my omegas are appearing as "?" after I made the previous post. Some unwanted transformation occurred after I hit the "Post Comment" button.scordova_{September 6, 2012
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R0b, Thank you for your discussion, but I think this isn't the most charitable interpretation regarding the die:
" What we’re modeling hasn’t changed, but we’ve gained active information by making a different modeling choice. "
The ratio |T|/|?| should be based on the space of outcomes adjusted for probability of those outcomes, it should not be based on merely counting the discrete states (unless the states are equiprobable). Hence if you define ? in only two states: State 1: 1 State 2: 2,3,4,5, or 6 The corresponding space of outcomes need to be adjusted such that |T|/|?| is 5/6, otherwise the model would not be faithful to the statistics of the object (a fair die) that is being modeled. That would be the charitable reading. You are of course free to complain that Bill could have said things otherwise to avoid such uncharitable readings. |{1}| = 1/6 |T|=|{2,3,4,5,6}| = 5/6 |?|=|{1}|+|{2,3,4,5,6}|=1 |T|/|?|=5/6
"What we’re modeling hasn’t changed, but we’ve gained active information by making a different modeling choice "
Actually the different numbers arise because the model no longer conforms to a fair die. Then notion of the measure of information not being arbitrary, but conforming to the statistics of the probability, still holds. Deviation from the actual statistics of what is being modeled in the way that you have done is imposing an arbitrary (and incorrect) measure of information for the system being modeled.scordova_{September 6, 2012
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R0bb:
Do you see now that the new search grid, as you call it, has 13 squares, not 16 as you said in #94?
The search grid consists of 13 squares however it is made of of two sets, one containing 12 and the otehr containing 4. 12 + 4 = 16. That 3 squares overlap is of no concern to what I said in #94. Ya see R0bb, well before 94 I exposed your nonsense. The nonsense that you still refuse to address.Joe_{September 6, 2012
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R0bb:
The probability of rolling a 6 would still be 1/6, and every number higher than 6 would have a probability of zero.
Then the numbers higher than 6 are NOT included.Joe_{September 6, 2012
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OK my apologies to R0bb and UD- ? does NOT change. The search grid within ? change. The original search grid included all 16 blocks. The new search grid, the one with active information, includes ONLY: • A is the uniform distribution over the rightmost four squares in the search space. • B is the uniform distribution over the bottom twelve squares in the search space.
Thanks, Joe. Do you see now that the new search grid, as you call it, has 13 squares, not 16 as you said in #94?R0bb_{September 6, 2012
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As for including numbers 7 – infinity (by which I assume you mean all integers greater than 6, none of which is actually infinity), is there any reason not to do so, other than inconvenience?
Yes, if you include them then you could never figure out the odds of rolling a “6? with a fair die
The probability of rolling a 6 would still be 1/6, and every number higher than 6 would have a probability of zero. The mean, median, variance, etc. are unaffected by inclusion of the higher numbers in Ω. Can you explain what the problem is?R0bb_{September 6, 2012
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As for how the LCI applies in to the real world- It tells us that having directions or a recipe is an easier way to have a successful search than to just do stuff until you get what you want. If you have directions they narrow your search grid, ie they provide active information. The same with a recipe.Joe_{September 5, 2012
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OK my apologies to R0bb and UD- Ω does NOT change. The search grid within Ω changes. The original search grid included all 16 blocks. The new search grid, the one with active information, includes ONLY: • A is the uniform distribution over the rightmost four squares in the search space. • B is the uniform distribution over the bottom twelve squares in the search space. Thus equation for figure 2 does not contain the number 16 And that means the new search grid excludes R0bb's 1.1, 1.2, 1.3. And that has been my point all along.Joe_{September 5, 2012
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Anyone can have a look: The Search for a Search: Measuring the Information Cost ofHigher Level Search- page 478Joe_{September 5, 2012
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R0bb:
Because the example explicitly defines ? as all the squares in the grid, and it never says that ? changes.
Yes, it says Ω changes(from figure 1 to figure 2) and provides those changes. That you can't even be honest about that exposes your agenda.Joe_{September 5, 2012
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Joe:
How can that be given: • A is the uniform distribution over the rightmost four squares in the search space. • B is the uniform distribution over the bottom twelve squares in the search space. Explain yourself, I dare you…
Because the example explicitly defines Ω as all the squares in the grid, and it never says that Ω changes. The two sentences you quote say nothing about Ω. I've tried to make that clear throughout; I apologize if I failed to do so. Since I've answered that, I'll try once again with the following question, and I'll make it even more concise: Given: |A| = 4 |B| = 12 |A∩B| = 3 Ω = A∪B What is |Ω|? Two keystrokes is all it takes. You can answer it faster than you can tell me that you're not going to answer it.R0bb_{September 5, 2012
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Robb:
If we label the squares in the grid from the example as follows: 1.1 1.2 1.3 1.4 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 4.1 4.2 4.3 4.4 I claim that in the example we’re discussing, Ω = {1.1, 1.2, 1.3, 1.4, 2.1, 2.2, 2.3, 2.4, 3.1, 3.2, 3.3, 3.4, 4.1, 4.2, 4.3, 4.4}.
How can that be given: • A is the uniform distribution over the rightmost four squares in the search space. • B is the uniform distribution over the bottom twelve squares in the search space. Explain yourself, I dare you...Joe_{September 5, 2012
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R0bb, Why are you even asking me? Everything is defined in the paper you are referencing. And if you can't understand what is in the paper, as obvioulsy you do not, then you should not try to discuss it. So no, I have indulged you enough- no, more than enough.Joe_{September 5, 2012
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Joe, I know you're very busy, but the following question can be answered in only two keystrokes: Given the following: – Set A has 4 elements – Set B has 12 elements – 3 of the elements in A are also in B – Ω = A∪B What is |Ω|? Also, when I say "in set notation", I mean in a form like "Ω = {1.1, 1.2, 1.3, 1.4, 2.1, 2.2, 2.3, 2.4, 3.1, 3.2, 3.3, 3.4, 4.1, 4.2, 4.3, 4.4}". Can you tell me, in that form, what you're claiming Ω to be? (This will require more than a few keystrokes, but maybe you'll indulge me.) Likewise, the questions about Marks' egg recipe and Dembski's treasure map can be answered in only a couple of keystrokes, so maybe you can take a few seconds and answer them. Thanks in advance.R0bb_{September 5, 2012
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onlooker- R0bb's post on TSZ are a failure- a failure to comprehend what Dembski and Marks are saying which led to a failure to properly address what they said. And one reason that you don't see progress here is you and your ilk prevent it. This thread is a great case in point...Joe_{September 5, 2012
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Chance,
onlooker @104, my #97 illustrates what I see as the difference.
At least we've clearly articulated our differences on this small point. That's more progress than is often seen here! If you have the time and inclination, I am interested in your thoughts on the larger issue of the validity of Dembski's "Law", in particular R0b's posts at The Skeptical Zone.onlooker_{September 5, 2012
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