User:IssaRice/Logical induction notation: Difference between revisions
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! Term !! Notation !! Type !! Definition !! Notes | ! Term !! Notation !! Type !! Definition !! Notes | ||
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| <math>\mathcal F</math>-combination || <math>A</math> || <math>\mathcal S \cup \{ | | <math>\mathcal F</math>-combination || <math>A</math> || <math>\mathcal S \cup \{1\} \to \mathcal F_n</math> || || Function application of an <math>\mathcal F</math>-combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in <math>\mathcal F</math> so is itself a function. This means we have two senses of "application": we can pick out the specific coefficient we want (square brackets), or we can apply each coefficient to return something (parentheses). | ||
|- | |- | ||
| Holdings from <math>T</math> against <math>\overline{\mathbb P}</math> (a <math>\mathbb Q</math>-combination)|| <math>T(\overline{\mathbb P})</math> || <math>\mathcal S \cup \{ | | Holdings from <math>T</math> against <math>\overline{\mathbb P}</math> (a <math>\mathbb Q</math>-combination)|| <math>T(\overline{\mathbb P})</math> || <math>\mathcal S \cup \{1\} \to \mathbb Q</math> || || | ||
|- | |- | ||
| Trading strategy || <math>T</math> || <math>\mathcal S \cup \{1\} \to \mathcal{E\!F}</math> || || | | Trading strategy || <math>T</math> || <math>\mathcal S \cup \{1\} \to \mathcal{E\!F}</math> || || | ||
|- | |||
| Feature || <math>\alpha</math> || <math>[0,1]^{\mathcal S \times \mathbb N^{+}} \to \mathbb R</math> or equivalently <math>(\mathcal S \times \mathbb N^{+} \to [0,1]) \to \mathbb R</math> or equivalently <math>\mathcal F</math> || || | |||
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:<math>\underbrace{\left[(\neg\neg\phi)^{*5} -\phi^{*5}\right]}_{\xi_1} \cdot (\phi - \phi^{*5}) + \underbrace{\left[\phi^{*5} - (\neg \neg \phi)^{*5}\right]}_{\xi_2} \cdot \left(\neg\neg\phi - (\neg\neg\phi)^{*5}\right)</math> | :<math>\underbrace{\left[(\neg\neg\phi)^{*5} -\phi^{*5}\right]}_{\xi_1} \cdot (\phi - \phi^{*5}) + \underbrace{\left[\phi^{*5} - (\neg \neg \phi)^{*5}\right]}_{\xi_2} \cdot \left(\neg\neg\phi - (\neg\neg\phi)^{*5}\right)</math> | ||
Since the coefficients (<math>\xi_1</math> and <math>\xi_2</math>) are in <math>\mathcal | Since the coefficients (<math>\xi_1</math> and <math>\xi_2</math>) are in <math>\mathcal{E\!F}_5</math>, this is an <math>\mathcal{E\!F}_5</math>-combination. Let's call this 5-strategy <math>T_5</math>. We can pick out the coefficient for the <math>\phi</math> term like <math>T_5[\phi] = (\neg\neg\phi)^{*5} -\phi^{*5}</math>. But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence <math>\overline{\mathbb V}</math>, like this: | ||
:<math>T_5(\overline{\mathbb V}) = \left[(\neg\neg\phi)^{*5}(\overline{\mathbb V}) -\phi^{*5}(\overline{\mathbb V})\right] \cdot (\phi - \phi^{*5}(\overline{\mathbb V})) + \left[\phi^{*5}(\overline{\mathbb V}) - (\neg \neg \phi)^{*5}(\overline{\mathbb V})\right] \cdot \left(\neg\neg\phi - (\neg\neg\phi)^{*5}(\overline{\mathbb V})\right)</math> | :<math>T_5(\overline{\mathbb V}) = \left[(\neg\neg\phi)^{*5}(\overline{\mathbb V}) -\phi^{*5}(\overline{\mathbb V})\right] \cdot (\phi - \phi^{*5}(\overline{\mathbb V})) + \left[\phi^{*5}(\overline{\mathbb V}) - (\neg \neg \phi)^{*5}(\overline{\mathbb V})\right] \cdot \left(\neg\neg\phi - (\neg\neg\phi)^{*5}(\overline{\mathbb V})\right)</math> | ||
Now each coefficient is a real number, so <math>T_5(\overline{\mathbb V})</math> is an <math>\mathbb R</math>-combination. Note that since <math>T_5\colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_5</math> is a function that takes a sentence or the number <math>1</math> and <math>\overline{\mathbb V}</math> is a valuation sequence (''not'' a sentence or number), there appears to be a type error in writing <math>T_5(\overline{\mathbb V})</math>. What is going on is that we aren't evaluating <math>T_5</math> at <math>\overline{\mathbb V}</math>; rather, we are evaluating ''each coefficient'' of <math>T_5</math>, to convert the range of <math>T_5</math> from <math>\mathcal | Now each coefficient is a real number, so <math>T_5(\overline{\mathbb V})</math> is an <math>\mathbb R</math>-combination. Note that since <math>T_5\colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_5</math> is a function that takes a sentence or the number <math>1</math> and <math>\overline{\mathbb V}</math> is a valuation sequence (''not'' a sentence or number), there appears to be a type error in writing <math>T_5(\overline{\mathbb V})</math>. What is going on is that we aren't evaluating <math>T_5</math> at <math>\overline{\mathbb V}</math>; rather, we are evaluating ''each coefficient'' of <math>T_5</math>, to convert the range of <math>T_5</math> from <math>\mathcal{E\!F}_5</math> to <math>\mathbb R</math>. | ||
To summarize the types: | |||
* <math>T_5 \colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_5</math> | |||
* <math>T_5[\phi] \in \mathcal{E\!F}_5</math> in other words <math>T_5[\phi] \colon [0,1]^{\mathcal S\times \mathbb N^{+}} \to \mathbb R</math> | |||
* <math>T_5(\overline{\mathbb V}) \colon \mathcal S \cup \{1\} \to \mathbb R</math> | |||
If <math>T = c + \xi_1\phi_1 + \cdots + \xi_k\phi_k \colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_n</math>, then | |||
:<math>\mathbb V(T) = c + \xi_1\mathbb V(\phi_1) + \cdots + \xi_k\mathbb V(\phi_k) \in \mathcal{E\!F}_n</math> | |||
and | |||
:<math>T(\overline{\mathbb V}) = c(\overline{\mathbb V})+ \xi_1(\overline{\mathbb V})\phi_1 + \cdots + \xi_k(\overline{\mathbb V})\phi_k \colon \mathcal S \cup \{1\} \to \mathbb R</math> | |||
and | |||
:<math>\mathbb W(T(\overline{\mathbb V})) = c(\overline{\mathbb V})+ \xi_1(\overline{\mathbb V})\mathbb W(\phi_1) + \cdots + \xi_k(\overline{\mathbb V})\mathbb W(\phi_k) \in \mathbb R</math> | |||
I think <math>\mathbb W(T(\overline{\mathbb V})) = (\mathbb W(T))(\overline{\mathbb V})</math> but the former notation seems to be preferred in the paper. | |||
==See also== | |||
* https://machinelearning.subwiki.org/wiki/User:IssaRice/Logical_inductor_construction | |||
==External links== | ==External links== |
Latest revision as of 00:51, 25 June 2019
Term | Notation | Type | Definition | Notes |
---|---|---|---|---|
-combination | Function application of an -combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in so is itself a function. This means we have two senses of "application": we can pick out the specific coefficient we want (square brackets), or we can apply each coefficient to return something (parentheses). | |||
Holdings from against (a -combination) | ||||
Trading strategy | ||||
Feature | or equivalently or equivalently |
Example of a 5-strategy given on p. 18 of the paper:
Since the coefficients ( and ) are in , this is an -combination. Let's call this 5-strategy . We can pick out the coefficient for the term like . But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence , like this:
Now each coefficient is a real number, so is an -combination. Note that since is a function that takes a sentence or the number and is a valuation sequence (not a sentence or number), there appears to be a type error in writing . What is going on is that we aren't evaluating at ; rather, we are evaluating each coefficient of , to convert the range of from to .
To summarize the types:
- in other words
If , then
and
and
I think but the former notation seems to be preferred in the paper.