User:IssaRice/Logical induction notation

From Machinelearning
Jump to: navigation, search
Term Notation Type Definition Notes
\mathcal F-combination A \mathcal S \cup \{1\} \to \mathcal F_n Function application of an \mathcal F-combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in \mathcal 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 \overline{\mathbb P} (a \mathbb Q-combination) T(\overline{\mathbb P}) \mathcal S \cup \{1\} \to \mathbb Q
Trading strategy T \mathcal S \cup \{1\} \to \mathcal{E\!F}
Feature \alpha [0,1]^{\mathcal S \times \mathbb N^{+}} \to \mathbb R or equivalently (\mathcal S \times \mathbb N^{+} \to [0,1]) \to \mathbb R or equivalently \mathcal F

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

\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)

Since the coefficients (\xi_1 and \xi_2) are in \mathcal{E\!F}_5, this is an \mathcal{E\!F}_5-combination. Let's call this 5-strategy T_5. We can pick out the coefficient for the \phi term like T_5[\phi] = (\neg\neg\phi)^{*5} -\phi^{*5}. But since each coefficient is a feature (which is a function), we can also apply each coefficient to some valuation sequence \overline{\mathbb V}, like this:

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)

Now each coefficient is a real number, so T_5(\overline{\mathbb V}) is an \mathbb R-combination. Note that since T_5\colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_5 is a function that takes a sentence or the number 1 and \overline{\mathbb V} is a valuation sequence (not a sentence or number), there appears to be a type error in writing T_5(\overline{\mathbb V}). What is going on is that we aren't evaluating T_5 at \overline{\mathbb V}; rather, we are evaluating each coefficient of T_5, to convert the range of T_5 from \mathcal{E\!F}_5 to \mathbb R.

To summarize the types:

  • T_5 \colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_5
  • T_5[\phi] \in \mathcal{E\!F}_5 in other words T_5[\phi] \colon [0,1]^{\mathcal S\times \mathbb N^{+}} \to \mathbb R
  • T_5(\overline{\mathbb V}) \colon \mathcal S \cup \{1\} \to \mathbb R

If T = c + \xi_1\phi_1 + \cdots + \xi_k\phi_k \colon \mathcal S \cup \{1\} \to \mathcal{E\!F}_n, then

\mathbb V(T) = c + \xi_1\mathbb V(\phi_1) + \cdots + \xi_k\mathbb V(\phi_k) \in \mathcal{E\!F}_n

and

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

and

\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

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

See also

External links