User:IssaRice/Logical induction notation: Difference between revisions

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I think <math>\mathbb W(T(\overline{\mathbb V})) = (\mathbb W(T))(\overline{\mathbb V})</math>.
I think <math>\mathbb W(T(\overline{\mathbb V})) = (\mathbb W(T))(\overline{\mathbb V})</math>.
The following is used in the Fixed Point Lemma (5.1.1):
Writing the <math>n</math>-strategy as
:<math>T_n = \sum_{j=1}^k \xi_j \phi_j - \sum_{j=1}^k \xi_j\phi_j^{*n}</math>
we have
:<math>\mathbb V(T_n(\mathbb P_{\leq n-1}, \mathbb V)) = \sum_{j=1}^k \xi_j(\mathbb P_{\leq n-1}, \mathbb V)\cdot \mathbb V(\phi_j) - \sum_{j=1}^k \xi_j(\mathbb P_{\leq n-1}, \mathbb V) \cdot \phi_j^{*n}(\mathbb P_{\leq n-1}, \mathbb V)</math>
But <math>\phi^{*n}(\mathbb P_{\leq n-1}, \mathbb V) = \mathbb V(\phi_j)</math> so the two sums cancel to obtain <math>0</math>.


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

Revision as of 17:43, 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
Feature α [0,1]S×N+→R or equivalently (S×N+→[0,1])→R or equivalently F

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

[(¬¬ϕ)*5−ϕ*5]⏟ξ1⋅(ϕ−ϕ*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)*5−ϕ*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¯)]⋅(ϕ−ϕ*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 in other words T5[ϕ]:[0,1]S×N+→R
  • T5(V¯):S∪{1}→R

If T=c+ξ1ϕ1+⋯+ξkϕk:S∪{1}→EFn, then

V(T)=c+ξ1V(ϕ1)+⋯+ξkV(ϕk)∈EFn

and

T(V¯)=c(V¯)+ξ1(V¯)ϕ1+⋯+ξk(V¯)ϕk:S∪{1}→R

and

W(T(V¯))=c(V¯)+ξ1(V¯)W(ϕ1)+⋯+ξk(V¯)W(ϕk)∈R

I think W(T(V¯))=(W(T))(V¯).

The following is used in the Fixed Point Lemma (5.1.1):

Writing the n-strategy as

Tn=∑j=1kξjϕj−∑j=1kξjϕj*n

we have

V(Tn(P≤n−1,V))=∑j=1kξj(P≤n−1,V)⋅V(ϕj)−∑j=1kξj(P≤n−1,V)⋅ϕj*n(P≤n−1,V)

But ϕ*n(P≤n−1,V)=V(ϕj) so the two sums cancel to obtain 0.

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