Randomness in h N versus [0, 1]

I. 3.1 (Partial) Recursive Functions and Sets

IV.2 Randomness in h N versus {0, 1} N

IV.2.2 Randomness in h N versus [0, 1]

The map X ↦0.X from {0,1}^N to [0,1] can be described in a different way. With Iσ as in the proof of Proposition IV.2.6, 0.Xis the unique element of⋂n∈NIX↾n. Said another way,[0,1]is split into two intervals of length 1/2 corresponding to⟨0⟩ and⟨1⟩, each of those intervals are split into two intervals of length 1/4 corresponding to ⟨0,0⟩, ⟨0,1⟩, ⟨1,0⟩, and ⟨1,1⟩, and so on, then we take the intersection of the intervals corresponding to the initial segments ofX to get 0.X.

Repeating this methodology forh^∗produces closed subintervals of[0,1]with rational endpoints, but not necessarily elements ofJ. For that reason, we letI be the set ofall closed intervalsI⊆ [0,1]with rational endpoints.

Definition IV.2.7. Given σ∈h^∗, defineπ^h(σ) ∶= [k/∣h^∣σ∣∣,(k+1)/∣^∣σ∣∣], wherek= ∑^∣_i^σ₌^∣−₀¹σ(i) ⋅ ∣hⁱ∣; in other words, π^h(⟨⟩) = [0,1] and, for σ ∈ h^∗ and 0≤ i < h(∣σ∣), π^h(σ^⌢⟨i⟩) is the i-th subinterval of π^h(σ) after splittingπ^h(σ)intoh(∣σ∣)-many consecutive closed subintervals of equal length 1/∣h^∣σ∣+1∣.

The mapπ^h∶h^N→ [0,1]is then defined by setting, for X∈h^N, π^h(X) ∶=unique element of ⋂

n∈N

π^h(X↾n).

Lemma IV.2.8. π^h is a measure-preserving surjection of h^N onto [0,1]. However, π^h is not injec- tive, and for distinct X, Y ∈ h^N, π^h(X) = π^h(Y) if and only if there is σ ∈ h^∗ such that {X, Y} = {σ^⌢⟨0,0, . . .⟩, σ^⌢⟨h(∣σ∣) −1, h(∣σ∣ +1) −1, . . .⟩}.

Proof. Straight-forward.

For an interval I∈ I, we wish to consider f(∣I∣), although∣I∣ ∈Nonly forI∈ J, requiring the following convention:

Convention IV.2.9. Givenf∶N→ [0,∞), we implicitly extend f to a function[0,∞) → [0,∞)by letting f(x) = (f(⌊x⌋ +1) −f(⌊x⌋))(x− ⌊x⌋) +f(⌊x⌋).

We extend the definition of dwtf toIandP (I )in the obvious manner, and the definitions off-ML tests and by extensionf-randomness can likewise be extended. We will term these extended definitions by adding the prefix ‘extended’, as in, “x∈ [0,1]is extended f-random in [0,1]if no extended f-ML test coversx.”

An additional assumption we must make regards f and the sequence ⟨f(n)/n⟩_n∈N≥1. Later we will strengthen this assumption further.

Convention IV.2.10. Given f, we assume ⟨f(n)/n⟩n∈N≥1 is nondecreasing. As such, the function x ∈ (0,∞) ↦f(x)/x∈ [0,∞)is nondecreasing as well.

Notation IV.2.11. Lets∶N→ [0,∞)be the unique nondecreasing computable function such thats(0) =1 and for which∣hⁿ∣ =2^n⋅s(n) for alln∈N. Consequently, ∣π^h(σ)∣ =n⋅s(n)for allσ∈hⁿ.

Proposition IV.2.12. For anyS ⊆h^∗, the set π^h[S] = {π^h(σ) ∣σ∈S} satisfies dwt_f(π^h[S]) ≤dwt_f(S).

Moreover, if S is r.e. thenπ^h[S]is r.e.

Proof. Becauseπ^h is measure preserving,µh(σ) =λ(π^h(σ)). Given σ∈hⁿ, n≤n⋅s(n)impliesf(∣σ∣)/∣σ∣ ≤ f(∣π^h(σ)∣)/∣π^h(σ)∣, and consequentlys(n) ⋅f(n) ≤f(n⋅s(n)). In particular,

2^{−f(∣πh(σ)∣)}=2^{−f(n⋅s(n))}≤2^{−s(n)⋅f(n)}= (2^−n⋅s(n))^f(n)/n=γ(σ)^f(∣σ∣). Thus, dwtf(π^h[S]) ≤dwtf(S). Thatπ^h[S]is r.e. ifS is r.e. is immediate.

Corollary IV.2.13. Ifπ^h(X) ∈ [0,1]is extended f-random in[0,1], thenX∈h^N isf-random inh^N. Proof. The uniformity of the assignment S ↦π^h[S] implies that if⟨Si⟩i∈N is a uniformly r.e. sequence of subsets ofh^∗, then ⟨π^h[Si]⟩i∈Nis a uniformly r.e. sequence of subsets of[0,1]. With Proposition IV.2.12, it follows that if⟨Si⟩i∈Nis anf-ML test inh^N then⟨π^h[Si]⟩i∈Nis an extended f-ML test in[0,1].

The proof of Proposition IV.2.12 suggests that if wish to convert a extended f-ML test in[0,1]into an f-ML test in h^N then we want to pull intervals inI back into strings in h^∗. However, the map π^h∶h^∗ → I is not surjective, so givenI ∈ I we must instead cover I with intervals of the form π^h(σ)for σ∈h^∗. This procedure must be sufficiently regular for an extendedf-ML test in[0,1]to be pulled back to ag-ML test inh^Nfor some appropriateg.

Definition IV.2.14. Given f≤domg, then we say that the regularity condition(∗)(g, f)holds forhif sup

n∈N

exp_h(n−1)(1−f(n⋅s(n)) ⋅ (n⋅s(n))⁻¹)

exp₂(s(n) ⋅g(n) −f(n⋅s(n))) < ∞. (∗)(g, f) Remark IV.2.15. In [9], the regularity condition(∗)(g, f)is simplified by the fact thatf is linear, and hence

f(n⋅s(n))

n⋅s(n) simplifies into an expression independent ofnors(n).

Proposition IV.2.16. Suppose(∗)(g, f)holds forhand let α=3⋅sup

n∈N

exp_h(n−1)(1−f(n⋅s(n)) ⋅ (n⋅s(n))⁻¹) exp₂(s(n) ⋅g(n) −f(n⋅s(n))) .

Then for eachI∈ I there existsI˜⊆h^∗ such that I⊆ ⋃π^h[I]˜ and dwtg(I) ≤˜ α⋅2^{−f(∣I∣)}. Moreover, I˜can be uniformly computed from a code forI.

Proof. We start by setting notation. For each I∈ I, let nI be the unique n≥1 such that ∣hⁿ∣⁻¹<λ(I) ≤

∣hⁿ⁻¹∣⁻¹ andkI be the greatest integer k such thatk/∣h^nI∣ ≤λ(I). Then kI <h(nI −1) and there is a set Iˆ⊆π^h[h^nI]computable from a code ofI of size≤kI+2 such thatI⊆ ⋃I, namely the intervals inˆ π^h[h^nI] intersectingI nontrivially (i.e., intervals which intersectI at more than just an endpoint).

SupposeI∈ I is given, and letn=n_I and k=k_I. Becausek/∣hⁿ∣ ≤λ(I), we have∣hⁿ∣⁻¹/λ(I) ≤k. Given

J∈I,ˆ λ(J) ≤λ(I)and so ∣I∣ ≤ ∣J∣ =n⋅s(n). Then dwt_g(I) = ∑˜

σ∈I˜

γ_n^−g(n)≤ (k+2)2^{−s(n)⋅g(n)}

≤3k⋅exp₂(−(s(n) ⋅g(n) −f(n⋅s(n)))) ⋅ (λ(I)/k)^{f(n⋅s(n))⋅(n⋅s(n))−1}

≤3⋅

exp_k(1−f(n⋅s(n)) ⋅ (n⋅s(n))⁻¹)

exp₂(s(n) ⋅g(n) −f(n⋅s(n))) ⋅λ(I)^{f(n⋅s(n))⋅(n⋅s(n))−1}

≤3⋅

exp_h(n−1)(1−f(n⋅s(n)) ⋅ (n⋅s(n))⁻¹)

exp₂(s(n) ⋅g(n) −f(n⋅s(n))) ⋅λ(I)^{f(∣I∣)/∣I∣}

≤α⋅2^{−f(∣I∣)}.

Corollary IV.2.17. Suppose(∗)(g, f) holds forhand let X∈h^N. If X isg-random inh^N thenπ^h(X)is generalized f-random in[0,1].

Proof. Letαbe as in the statement of Proposition IV.2.16. GivenI∈ I, let ˜I and ˆIbe as in the statement and proof of Proposition IV.2.16. GivenS⊆ I r.e., let ˆS= ⋃{Iˆ∣I∈S}and ˜S= ⋃{I˜∣I∈S}. Then ˜S is r.e.

and dwtg(S) =˜ dwtg(S) ≤ˆ α⋅dwtf(S).

Suppose for the sake of a contradiction that π^h(X) is not generalized f-random in [0,1], and so let

⟨Si⟩i∈N be a generalized f-ML test covering π^h(X). Let m∈ Nsatisfy α≤2^m. That ˜I can be computed uniformly from a code for I implies⟨S˜i+m⟩i∈N is uniformly r.e. Then ⟨S˜i+m⟩i∈N is a g-ML test coveringX, contradicting the hypothesis thatX isg-random inh^N.

Corollary IV.2.18. Supposex∈ [0,1]. Thenxisf-random in[0,1]if and only ifxis generalizedf-random in[0,1].

Proof. Being generalizedf-random in[0,1]clearly implies beingf-random in[0,1].

In the opposite direction, suppose x is f-random in [0,1], so that bin(x) is f-random in {0,1}^N by Proposition IV.2.6. Withh(n) ∶=2 for alln∈N we have s(n) =1, so ∣π^h(σ)∣ = ∣σ∣ for all σ∈h^∗ = {0,1}^∗. The condition(∗)(g, f)forhis then the statement that

sup

n∈N

exp₂(1−f(n)/n) exp₂(g(n) −f(n))< ∞.

Then we may observe that(∗)(f, f)holds forh, and so Corollary IV.2.17 impliesπ^h(bin(x)) =xis generalized f-random in[0,1].

Corollary IV.2.19. Supposelimn→∞s(n−1)

s(n) =1 andε>0. Iff(n) =n−j(n)andg(n) =n− (1−ε)^j(n_s^⋅₍^s_n⁽₎ⁿ⁾⁾ then(∗)(g, f)holds forh. Consequently, ifX isg-random inh^N thenπ^h(X)isf-random in[0,1].

Proof. If∣hⁿ∣ =n⋅s(n), thenh(n−1) = ∣hⁿ∣/∣hⁿ⁻¹∣ =2^{n⋅s(n)−(n−1)⋅s(n−1)}. Then log₂(

exp_h(n−1)(1−f(n⋅s(n)) ⋅ (n⋅s(n))⁻¹)

exp₂(s(n) ⋅g(n) −f(n⋅s(n))) ) = (n⋅s(n) − (n−1) ⋅s(n−1)) ⋅ (1−f(n⋅s(n)) n⋅s(n) )

−s(n) ⋅g(n) +f(n⋅s(n))

=n⋅s(n) − (n−1) ⋅s(n−1) −f(n⋅s(n)) +

(n−1) ⋅s(n−1)

n⋅s(n) ⋅ (n⋅s(n) −j(n⋅s(n)))

−s(n) ⋅ (n− (1−ε)j(n⋅s(n))

s(n) ) +f(n⋅s(n))

= ((1−ε) −n−1 n

s(n−1)

s(n) ) ⋅j(n⋅s(n)).

Because limn→∞s(n−1)

s(n) =1 by hypothesis, for all sufficiently large n we have 1−ε< ⁿ_n^−1s(_sⁿ_(n⁻¹₎⁾ and hence ((1−ε) −ⁿ_n^−1s(_sⁿ_(n⁻¹₎⁾) ⋅j(n⋅s(n)) <0. It follows that(∗)(g, f)holds forh.

Remark IV.2.20. The condition that limn→∞s(n−1)

s(n) = 1 is equivalent to limn→∞ log₂∣hⁿ∣

log₂∣hⁿ⁻¹∣ = 1, which is equivalent to limn→∞log₂h(n−1)

log₂∣hⁿ∣ =0.