User:IssaRice/Logical inductor construction

From Machinelearning

Notes from the Logical Induction paper as I walk through the construction of LIA in section 5.

Lemma 5.1.1 (Fixed Point Lemma)

"Observe that V' is equal to the natural inclusion of the finite-dimensional cube [0,1]S in the space of all valuations V=[0,1]S." -- I think what this is saying is that since S'S, we can think of [0,1]S as being sort of a subset of [0,1]S. Except it's not strictly speaking a subset, since the functions in [0,1]S and [0,1]S have different domains. How can we make it a subset? The "natural" way to do this is to set everything outside of S' to zero. But that's exactly what V'={V[0,1]S:V(ϕ)=0 whenever ϕS} is. One thing I'm still not sure about is the "finite-dimensional" part; doesn't having [0,1] make the cube infinite-dimensional?

"the compact, convex space V'" -- this intuitively makes sense, since V' basically "looks like" a cube. But I'm not sure how to verify this.

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

Writing the n-strategy as

Tn=j=1kξjϕjj=1kξjϕj*n

we have

V(Tn(Pn1,V))=j=1kξj(Pn1,V)V(ϕj)j=1kξj(Pn1,V)ϕj*n(Pn1,V)

But ϕj*n(Pn1,V)=V(ϕj) so the two sums cancel to obtain 0.

Definition/Proposition 5.1.2 (MarketMaker)

Lemma 5.1.3 (MarketMaker Inexploitability)

Definition/Proposition 5.2.1 (Budgeter)

Lemma 5.2.2 (Properties of Budgeter)

Proposition 5.3.1 (Redundant Enumeration of e.c. Traders)

Definition/Proposition 5.3.2 (TradingFirm)

Lemma 5.3.3 (Trading Firm Dominance)

See also