User:IssaRice/Logical induction notation: Difference between revisions

Latest revision as of 00:51, 25 June 2019

Term	Notation	Type	Notes
$F$ -combination	$A$	$S \cup {1} \to F_{n}$	Function application of an $F$ -combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in $F$ 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 $T$ against $\bar{P}$ (a $Q$ -combination)	$T (\bar{P})$	$S \cup {1} \to Q$
Trading strategy	$T$	$S \cup {1} \to E F$
Feature	$α$	$[0, 1]^{S \times N^{+}} \to R$ or equivalently $(S \times N^{+} \to [0, 1]) \to R$ or equivalently $F$

Example of a 5-strategy given on p. 18 of the paper:

\underset{ξ_{1}}{\underset{⏟}{[(\neg \neg ϕ)^{* 5} - ϕ^{* 5}]}} \cdot (ϕ - ϕ^{* 5}) + \underset{ξ_{2}}{\underset{⏟}{[ϕ^{* 5} - (\neg \neg ϕ)^{* 5}]}} \cdot (\neg \neg ϕ - (\neg \neg ϕ)^{* 5})

Since the coefficients ( $ξ_{1}$ and $ξ_{2}$ ) are in ${E F}_{5}$ , this is an ${E F}_{5}$ -combination. Let's call this 5-strategy $T_{5}$ . We can pick out the coefficient for the $ϕ$ term like $T_{5} [ϕ] = (\neg \neg ϕ)^{* 5} - ϕ^{* 5}$ . But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence $\bar{V}$ , like this:

T_{5} (\bar{V}) = [(\neg \neg ϕ)^{* 5} (\bar{V}) - ϕ^{* 5} (\bar{V})] \cdot (ϕ - ϕ^{* 5} (\bar{V})) + [ϕ^{* 5} (\bar{V}) - (\neg \neg ϕ)^{* 5} (\bar{V})] \cdot (\neg \neg ϕ - (\neg \neg ϕ)^{* 5} (\bar{V}))

Now each coefficient is a real number, so $T_{5} (\bar{V})$ is an $R$ -combination. Note that since $T_{5} : S \cup {1} \to {E F}_{5}$ is a function that takes a sentence or the number $1$ and $\bar{V}$ is a valuation sequence (not a sentence or number), there appears to be a type error in writing $T_{5} (\bar{V})$ . What is going on is that we aren't evaluating $T_{5}$ at $\bar{V}$ ; rather, we are evaluating each coefficient of $T_{5}$ , to convert the range of $T_{5}$ from ${E F}_{5}$ to $R$ .

To summarize the types:

$T_{5} : S \cup {1} \to {E F}_{5}$
$T_{5} [ϕ] \in {E F}_{5}$ in other words $T_{5} [ϕ] : [0, 1]^{S \times N^{+}} \to R$
$T_{5} (\bar{V}) : S \cup {1} \to R$

If $T = c + ξ_{1} ϕ_{1} + \dots + ξ_{k} ϕ_{k} : S \cup {1} \to {E F}_{n}$ , then

V (T) = c + ξ_{1} V (ϕ_{1}) + \dots + ξ_{k} V (ϕ_{k}) \in {E F}_{n}

and

T (\bar{V}) = c (\bar{V}) + ξ_{1} (\bar{V}) ϕ_{1} + \dots + ξ_{k} (\bar{V}) ϕ_{k} : S \cup {1} \to R

and

W (T (\bar{V})) = c (\bar{V}) + ξ_{1} (\bar{V}) W (ϕ_{1}) + \dots + ξ_{k} (\bar{V}) W (ϕ_{k}) \in R

I think $W (T (\bar{V})) = (W (T)) (\bar{V})$ but the former notation seems to be preferred in the paper.

External links

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-This is in user space because it's not really about machine learning.
 {| class="sortable wikitable"
 |-
 ! Term !! Notation !! Type !! Definition !! Notes
 |-
-| <math>\mathcal F</math>-combination || <math>A</math> || <math>\mathcal S \cup \{0,1\} \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).
+| <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 \{1\} \to \mathbb Q</math> || ||
 |-
-| 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 \{0,1\} \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{EF}</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> || ||
 |}
 Example of a 5-strategy given on p. 18 of the paper:
-:<math>\left[(\neg\neg\phi)^{*5} -\phi^{*5}\right] \cdot (\phi - \phi^{*5}) + \left[\phi^{*5} - (\neg \neg \phi)^{*5}\right] \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 are in <math>\mathcal F_5</math>, this is an <math>\mathcal 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:
+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>
-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 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 F_5</math> to <math>\mathbb R</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{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==