User:IssaRice/Logical induction notation: Difference between revisions

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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>.
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] \colon \mathcal{E\!F}_5</math>
* <math>T_5(\overline{\mathbb V}) \colon \mathcal S \cup \{1\} \to \mathbb R</math>


==External links==
==External links==

Revision as of 04:16, 3 August 2018

This is in user space because it's not really about machine learning.

Term Notation Type Definition Notes
F-combination A S{0,1}Fn 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 P¯ (a Q-combination) T(P¯) S{0,1}Q
Trading strategy T S{1}EF

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

[(¬¬ϕ)*5ϕ*5]ξ1(ϕϕ*5)+[ϕ*5(¬¬ϕ)*5]ξ2(¬¬ϕ(¬¬ϕ)*5)

Since the coefficients (ξ1 and ξ2) are in EF5, this is an EF5-combination. Let's call this 5-strategy T5. We can pick out the coefficient for the ϕ term like T5[ϕ]=(¬¬ϕ)*5ϕ*5. But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence V¯, like this:

T5(V¯)=[(¬¬ϕ)*5(V¯)ϕ*5(V¯)](ϕϕ*5(V¯))+[ϕ*5(V¯)(¬¬ϕ)*5(V¯)](¬¬ϕ(¬¬ϕ)*5(V¯))

Now each coefficient is a real number, so T5(V¯) is an R-combination. Note that since T5:S{1}EF5 is a function that takes a sentence or the number 1 and V¯ is a valuation sequence (not a sentence or number), there appears to be a type error in writing T5(V¯). What is going on is that we aren't evaluating T5 at V¯; rather, we are evaluating each coefficient of T5, to convert the range of T5 from EF5 to R.

To summarize the types:

  • T5:S{1}EF5
  • T5[ϕ]:EF5
  • T5(V¯):S{1}R

External links