p-value presheaves

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HAL Id: hal-02190029

https://hal.archives-ouvertes.fr/hal-02190029v2

Submitted on 28 Aug 2019

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p-value presheaves

Erwan Beurier, Dominique Pastor

To cite this version:

Erwan Beurier, Dominique Pastor. p-value presheaves. [Research Report] RR-2019-02-SC, IMT At- lantique. 2019. �hal-02190029v2�

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Collection des rapports de recherche d’IMT Atlantique IMTA-RR-2019-02-SC

p-value presheaves

Date d’édition : August 28, 2019 Version : 1.0

Erwan Beurier IMT Atlantique Dominique Pastor IMT Atlantique IMT Atlantique

Dépt. Signal & Communications Technopôle de Brest-Iroise - CS 83818 29238 Brest Cedex 3

Téléphone: +33 (0)2 29 00 13 04 Télécopie: +33 (0)2 29 00 10 12 URL:www.imt-atlantique.fr

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Contents

1. Introduction

In this preliminary work, we study the sheaf-ness properties of the p-value. In particular, p-values are presheaves and become sheaves if we add a bottom element to the initial topology. This trick is valid because p-values are fuzzy sets!

We also study the basic properties of the presheaf topos of p-values. Our purpose here is to exhibit the tools provided by topos theory to "approximate" p-values of Neyman-Pearson tests by p-values of other types of tests. In this report, we provide the basic tools but do not achieve the approximation yet!

2. Notation

Lbg R^d

is the set of all Lebesgue-measurable subsets ofR^d.

3. P-values

Definition 3.1 (Test). Afamily of tests, or simply test, is a functionT : [0,1] → Lbg R^d

such that a 6a⁰⇒T(a)6T(a⁰).

Remark3.2. IfT :[0,1] →Lbg R^d

is a test, then fora∈ [0,1],T(a)is the rejection region of the null hypothesis with sizea.

Definition 3.3(p-value). LetT be a test.

For x ∈Rⁿ, the p-value ofxis defined as:

pvalT(x)=inf({a∈ [0,1] | x ∈T(a)})

The p-value ofx is the minimum size of the test that putsxinto the rejection region.

Definition 3.4(Presheaf). LetC be a small category.

A presheaf is a functorC^op →Sets.

Now consider the total order([0,1],6)(in fact, it is a complete lattice). It defines a small category that we will denote byU.

Definition 3.5(P-value presheaf). LetT :[0,1] →Lbg R^d

be a test.

Thep-value presheaf forT is the following functor:

P_T :













U^op −→ Sets

a 7−→

x ∈Rⁿ|pvalT(x)> a a ⊆b 7−→

P_T(b) −→ P_T(a)

x 7−→ x

3.1. This is not a sheaf

Let us introduce the definition of a sheaf. We generally use topological spaces for that purpose.

Definition 3.6(Sheaf). Let(X,Op(X))be a topological space, we consider the category Op(X)^op. A sheafS : Op(X)^op →Setsis a presheaf with the following condition, calledsheaf condition: (sheaf condition) For all open set U ∈ Op(X), for all covering U = Ð

a∈A

U_a, and for all family (x_a)_a_∈_A ∈ Î

a∈A

S(U_a) such that for alla,b ∈ A, we have x_{a U}_a_∩U_b = x_{b U}_a_∩U_b, there exists a unique x ∈S(U)such that∀a ∈ A, x _a = x_a.

A family (U_a)_a_∈A such thatU = Ð

a∈A

U_a is called acovering ofU. A family (x_a)_a∈_A ∈ Î

a∈A

S(U_a) such that for alla,b∈ A, we havex_{a U}_a_∩U_b = x_{b U}_a_∩U_b is called amatching family ofS-sections. When

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3. P-values

x_{a U}_a_∩U_b = x_{b U}_a_∩U_b, we also say thatx_a andx_bagreeon their common domain. The uniquex ∈S(U) such that∀a∈ A, x _a= x_ais called thegluingof(x_a)_a∈_A.

However, our p-value presheaf is not defined on a topological space, but on a partial order. The notion of sheaves can be extended from a topological space to a complete Heyting algebra. Besides, our partial order is a complete Heyting algebra.

Remark3.7 (Introducing morphisms for sheaves on Heyting algebras). The following is adapted from [1, page 376, section 15.5.3].

Let(H,6,∧,∨)be a complete Heyting algebra, and letP:H^op →Setsbe a presheaf overH. Let(h_a)_a∈_Aa subset of elements ofHand leth= Ô

a∈A

h_abe their supremum (it always exists because H is complete). We denote the restriction functions byr_a^h : P(h) → P(h_a). There is a unique function r^h :P(h) →Î

a∈AP(h_a)such thatπ_P(h_a)◦r^h =r_a^h(definition of product). Similarly, for everya,b∈ A, there are functionsp^h_a,b:P(h_a) →P(h_a∧h_b)andq_a,b^h :P(h_b) →P(h_a∧h_b). These also combine into:

p^h,q^h : Î

a∈A

P(h_a) → Î

a,b∈A

P(h_a∧h_b).

We can now define sheaves on Heyting algebras.

Definition 3.8 (Sheaf on a complete Heyting algebra). Let(H,6,∧,∨,0H,1H) be a complete Heyting algebra. LetS:H^op →Setsbe a presheaf overH.

The presheafSis called a sheaf when the following condition is verified:

for all(h_a)_a∈_A∈H^A, ifh= Ô

a∈A

h_aandr^h :S(h) → Î

a∈A

S(h_a),p^h : Î

a∈A

S(h_a) → Î

a,b∈A

S(h_a∧h_b) andq^h: Î

a∈A

S(h_a) → Î

a,b∈A

S(h_a∧h_b)are functions as defined in Remark 3.7, then the following diagram is an equaliser:

S(h) Î

a∈A

S(h_a) Î

a,b∈A

S(h_a∧h_b)

r^h p^h

q^h

For the sake of understandability, we will keep the same terminology as introduced with topological spaces, namely:

A family(h_a)_a∈_Asuch thath= Ô

a∈A

h_ais called acovering ofh.

A family(x_a)_a_∈A ∈ Î

a∈A

S(h_a)such that for alla,b∈ A, we havex_{a h}_a∧hb = x_{b h}_a∧hb, is called a matching family ofS-sections. Whenx_{a h}_a_∧h_b = x_{b h}_a_∧h_b, we also say that x_aandx_bagreeon their common domain.

The uniquex ∈S(h)such that∀a∈ A, x _h_a =x_ais called thegluingof(x_a)_a_∈_A. Lemma 3.9. IfS:H→Setsis a sheaf, thenS(0H) 1.

Proof. Consider the empty covering 0H = Ô

a∈ ∅h_a. The products Î

a∈A

S(h_a)and Î

a,b∈A

S(h_a∧h_b)are empty: Î

a∈A

S(h_a) Î

a,b∈A

S(h_a∧h_b) 1. Thus, the arrowsp⁰andq⁰are the same and are the identity id₁. The equaliser of:

S(0H) ^r⁰ 1 ^p⁰ 1

q⁰

is obviously a terminal object 1; in other words, ifSis a sheaf, thenS(0H) 1.

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3. P-values

Proposition 3.10. LetT :[0,1] →Lbg R^d

be a test and letP_T be its p-value presheaf.

Then,P_T is not a sheaf.

Proof. SupposeP_T is a sheaf. By Lemma 3.9,P_T(0)1. However:

P_T(∅)=

x ∈Rⁿ|pvalT (x) >0 =R 1

Remark3.11. The fact thatP_T is not a sheaf also means that there are coveringsh= Ô

a∈A

h_athat do not lead to a unique gluing (either there is none, or there is more than one). In our case, the empty covering is the unique covering that does not find its gluing. In fact, the p-value presheaf only fails to be a sheaf on one covering.

Let h=Ô

a∈Ah_abe a cover such that A, ∅. Let(x_a)_a_∈_Abe a matching family ofP_T-sections over (h_a)_a_∈A: x_a ∈P_T(h_a)andx_{a h}_a_∧h_b = x_{b h}_a_∧h_b.

By definition ofP_T, we have:

x_{a h}_a_∧h_b =P_T (h_a∧h_b 6 h_a) (x_a)= x_a

which yields, for alla,b∈ A,x_a= x_b. Letx= x_a; then∀a ∈A, we havex ∈P_T(h_a), so pvalT(x)> h_a and pvalT(x) > sup

a∈A

(h_a) = hand x ∈ P_T(h) = P_T

Ô

a∈A

h_a

. Consequently, x is the unique gluing of (x_a)_a∈_A, andP_T satisfies the sheaf condition whenever the index family Ais not empty.

When Ais empty, then the gluingx ∈ Adoesn’t exist.

This problem was already known in [1, p401, part. 15.6.6], not for p-values, but in the context of fuzzy logic. We consider a Heyting algebraH. A fuzzy set (overH) is a pair(S,s)such thatS∈Setsand s :S →H. The category of fuzzy sets overHis simply the sliceSets/H(whereHis a lattice).

Let(S,s)be a fuzzy set. We define the presheaf:

P:

H −→ Sets h 7−→ {x∈S | s(x) > h}

This presheaf is very similar to that of our p-value presheaf. However, both suffer from the same problem: they are almost sheaves, and the part that fails is the image of the empty set or least element, which contains the whole set instead of being the terminal object. The solution proposed by [1] is to change the Heyting algebra for another Heyting algebra, adding a new element⊥that becomes the new least element, and by forcingP(⊥)=1.

LetU⁺=U ∪ {⊥}. Then, let6U⁺ be the smallest partial order that contains:

(⊥,U) ∈U⁺×U⁺ ∪ ⊂_U where⊂_U is the inclusion of open subsets inU.

It is easy to see that:

Proposition 3.12. U⁺is a complete Heyting algebra.

In fact, the pair(U⁺,6_U⁺)is simplyU with a new initial element. As it is still a Heyting algebra, we can define a sheaf on it.

Definition 3.13(P-value sheaf). LetT :[0,1] →Lbg R^d

be a test.

Thep-value sheaf forTis the following functor:

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4. First random results

P_T :













U⁺^op −→ Sets a 7−→

x ∈Rⁿ|pvalT(x) >a ifa ∈ [0,1]

{∅} ifa=⊥

a⊆ b 7−→

P_T(b) −→ P_T(a)

x 7−→ x

Proposition 3.14. The p-value sheaf is an actual sheaf.

Proof. An application of [1, Proposition 15.6.8].

3.2. Study of the presheaf

The use of sheaves is beyond our purposes. We actually only need a topos, and that topos doesn’t need to be a sheaf topos. A presheaf topos is enough. Consider the presheaf topos based on the complete Heyting algebra[0,1]:PSh([0,1]). Its subobject classifier is:

Ω:













U⁺ −→ Sets

a 7−→ {a⁰|a⁰ 6 a}=[0,a]

a6 b 7−→

Ω(b) −→ Ω(a) c 7−→ a∧c=min(a,c)

LetT be a test andP_T its associated p-value presheaf. A presheaf morphismp:P_T →Ωmakes the following diagram commute for alla6 b:

a P_T(b) Ω(b)

{ X

b P_T(a) Ω(a)

u PT(u)

pb

Ω(u)

pa

For x ∈P_T(b), we have:

p_a◦P_T (u) (x)=Ω(u) ◦p_b(x) p_a(x)= p_b(x) ∧a The canonical example is the following natural transformation:

p_a :

P_T(a) −→ Ω(a)

x 7−→ a

4. First random results

4.1. About p-value sheaves

We have already concluded that[0,1], with the usual∧and∨operators, is a complete Heyting algebra.

Considering it as a proset categoryU, being a complete Heyting algebra says that:

Proposition 4.1. U has all small limits and colimits.

Proof. LetD:I →U^op be a diagram inU^op. Then:

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Colim(D)= sup

i∈ObI

D(i)= Û

i∈ObI

D(i) Lim(D)= inf

i∈ObI

D(i)= Ü

i∈ObI

D(i)

Note that the sup becomes aÓ, and the inf becomes aÔ, because we are consideringU^op and notU. Also note that the infima and suprema always exist because we are in a complete Heyting algebra.

Proposition 4.2. For any testT, its p-value presheafP_T is continuous and cocontinuous.

Proof. LetD:I →U^op be any (small) diagram inU^op.

We only consider the case of a limit; the proof is very similar for colimits.

P_T(Lim(D))= P_T Ü

i∈I

D(i)

!

= (

x ∈R^d

pvalT(x)> Ü

i∈I

D(i) )

=

x ∈R^d

∀i∈I, pval_T (x) > D(i)

= Ù

i∈I

x ∈R^d

pvalT(x) > D(i)

= Ù

i∈I

P_T (D(i))

We now have to check that Ñ

i∈IP_T(D(i))Lim(P_T ◦D).

For all i → j ∈ I, we have P_T(D(i)) ⊂ P_T(D(j)). We denote by ιi,j = P_T(D(j) →D(i)) = P_T(D(i) ⊂ D(j)): P_T (D(i)) → P_T (D(j))the inclusion mapping betweenP_T (D(i))’s. We also denote byι_i : Ñ

i∈IP_T (D(i)) →P_T (D(i))the inclusion mapping of the intersection. For alli→ j ∈I, we have ιi,j◦ι_i =ι_j, so thatι=

ι_i : Ñ

i∈IP_T (D(i)) →P_T (D(i))

i∈I is a cone toP_T ◦D.

Let(A, α)be any cone toP_T ◦D. We denote theP_T(D(i))-components ofαbyα_i : A→P_T(D(i)). For alli→ j∈I, and for allx ∈X, we haveιi,j◦α_i(x)=α_j(x)= α_i(x).

P_T(D(i)) A

P_T (D(j))

ιi,j

αi

αj

So in fact,α_i(X)=α_j(X) ⊂ P_T(D(i)), which yields that, for alli∈I,α_i(X) ⊂ Ñ

i∈IP_T(D(i)). Letube such that:

u:

( X −→ Ñ

i∈IP_T (D(i)) x 7−→ α_i(x)

for any of thei∈I, becauseα_j(x)=α_i(x). Then for alli∈I, we haveι_i◦u=α_i. It is also easy to check the unicity of thatu: supposeu,u⁰:X → Ñ

i∈IP_T (D(i)), then for allx ∈ X, we have:

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ι_i◦u(x)=α_i(x)=ι_i◦u⁰(x) u(x)=u⁰(x)

which leads tou=u⁰. Consequently, Ñ

i∈IP_T(D(i))is the limit ofP_T ◦D.

We know that all p-value presheaves are continuous. The question is: is every continuous presheaf a p-value presheaf? In other words, is any continuous presheafP:U^op →SetsaP_T for some testT? We have to restrict our search for now.

Definition 4.3(Lebesgue presheaf). LetS:U^op →Setsbe a presheaf.

The presheafSis called aLebesgue presheaf when for alla∈U^op,S(a) ∈Lbg R^d . Definition 4.4(Inclusive presheaf). LetS :U^op →Setsbe a presheaf.

The presheaf S is called inclusivewhen it sends every mapping a 6 b to the inclusion mapping S(a) ⊂S(b).

Definition 4.5(Presheaf from a test). LetS:U^op →Setsbe a presheaf and letT :[0,1] →Lbg R^d test. be a

We say thatScomes fromT if there exists a natural isomorphismα:S→P_T. We say thatScomes from a testif there exists a testT such thatScomes fromT.

Lemma 4.6. Every continuous inclusive Lebesgue presheaf comes from a continuous test.

Proof. LetSbe a continuous inclusive Lebesgue presheaf. DefineT :[0,1] →Lbg R^d as:

T(a)=R^d\S(a) The continuity ofSgives the continuity ofT. LetA⊂ [0,1]:

Ø

a∈A

T(a)= Ø

a∈A

R^d\S(a)

=R^d\Ù

a∈A

S(a)

=R^d\S(sup(A))

=T(sup(A)) (The proof is the same for intersections.)

Then, let us compareSwith the p-value presheaf associated withT:

P_T(a)=

x ∈R^d

pval_T (x)> a

=R^d\Ø

b<a

T(b)

=R^d\T(a)

= S(a)

AsSis inclusive, we know thatS(a 6 b)=P_T(a 6b). Consequently,Scomes fromT.

Corollary 4.7. For every p-value presheafP_T associated toT, there exists a continuous testT⁰such that P_T =S_T⁰.

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The following result is a small addition to the previous lemma.

Definition 4.8(Mono-preserving functor). The functorF:C →Setsis calledmono-preservingwhen for all monomorphismm: A→B,F(m)is also monic. (The image of a mono is a mono.)

In our case, the presheafS:U^op →Setswill be mono-preserving when for alla,b∈ [0,1]such that a 6 b, we haveS(a 6b)monic. (The image of an inequality is a monomorphism.)

Lemma 4.9. Continuous mono-inducing Lebesgue sheaves come from a test.

Proof. LetSbe a continuous mono-inducing Lebesgue presheaf. Fora < b(a,b∈ [0,1]), we denote by σ_a,bthe arrowσ_a,b= S(a< b).

AsSis mono-inducing,σ_a,bis monic. InSets, every function f : A→Bcan be written as a f =m◦e wheree: A→ f(A)is epic, andm: f(A) → Bis the canonical inclusion. But, if f is monic, theneis monic too, and inSets,ebecomes an isomorphism.

Then, for all a ∈ [0,1], the arrow σ₀_,a : S(a) → S(0) decomposes to σ₀_,a = m◦ e_a, where e_a:S(a) →σ₀,a(S(a))is an isomorphism andm:σ₀,a(S(a)) → S(0)is the canonical inclusion. Note that fora=0, we haveσ₀_,₀=idS(0)= e₀.

LetFbe the following functor:

F :











U^op −→ Sets a 7−→ σ₀,a(S(a)) a 6 b 7−→ F(b) ⊂F(a) Let us studye=(e_a)_a∈[₀_,₁_]:

S(b) F(b)

S(a) F(a)

S(0) F(0)

eb

σ_a,b

σ_0,b

6 ea 6

σ_0,a 6

e₀

We have to prove that the square(S(b),F(b),F(a),S(a))commutes, while we know that the other squares and triangles commute. Forx ∈S(b), we have:

e_a◦σ_a,b(x)=F(06a) ◦e_a◦σ_a,b(x)

=e₀◦σ₀_,a◦σ_a,b(x)

=e₀◦σ₀_,b(x)

=F(06 b) ◦e_b(x)

=e_b(x)

=F(a6 b) ◦e_b(x)

Soe :S →F is a natural transformation, and each component is an isomorphism, soeis a natural isomorphism. Then, some properties ofS transfer toF: F is a continuous presheaf. It is by definition inclusive and Lebesgue, soF comes from a test and so doesS.

However, other presheaves unlikely come from a test. For example, any presheafU^op →Setssuch that card(S(a))> 2^ℵ⁰for somea∈U will have trouble being isomorphic toR.

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4.2. About sheaf morphisms

Let us first study the sheaf morphisms between p-value sheavesp:P_R →P_T. Proposition 4.10(Nesting property). LetR,T :[0,1] →Lbg R^d

be two tests (of the same size or not).

Letp:R^d →R^d be a function.

The functionpdefines a natural transformationp¯:P_R →P_T = p _P_R₍_a₎

a∈[0,1]

⇔for alla∈ [0,1], p(P_R(a)) ⊂ P_T(a).

In other words,pdefines a function between nested open sets, as in Figure 1.

𝑃𝑅𝑎

𝑃_𝑅𝑏 𝑅^𝑑

𝑃𝑇 𝑎

𝑃_𝑇 𝑏 𝑅^𝑑

𝑝 𝑃_𝑅(𝑏)

𝑝 𝑃𝑅(𝑎)

𝑝

Figure 1: Illustration of the nesting property ofp. Ifp :R^d →R^dverifies this nesting property, then it defines a natural transformation between two p-value sheaves.

Proof. Easy deduction from the following natural transformation diagram:

a P_R(b) P_T(b)

{ X

b P_R(a) P_T(a)

⊂ ⊂

p_PR(b)

⊂

p_PR_(a)

Corollary 4.11. Ifp¯=(p_a)_a∈[₀_,₁_]is a natural transformationP_R →P_T, thenp= Ð

a∈A

p_ais a function that verifies the nesting property.

Proposition 4.12. For alla∈ [0,1],P_T(a)=R^d\

Ð

b<aT(b)

. Proof. By computation:

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P_T(a)= x ∈R^d

pval_T(x)> a

=

x ∈R^d inf

x∈T(b)

(b∈ [0,1])> a

=

x ∈R^d

∀b<a,x <T(b)

= Ù

b<a

R^d\T(b)

=R^d\ Ø

b<a

T(b)

!

Corollary 4.13. For alla ∈ [0,1], if Ð

b<aT(b)=T(a)thenP_T(a)=R^d\T(a).

Example4.14. Note that we are considering very general testsT :[0,1] →Lbg R^d , so

Ð

b<aT(b)

has no reason to be equal toT(a). Let us give a counterexample.

Consider the following functions:

f :











[0,1] −→ [0,1] x 7−→

x ifx < ¹₂

12x+ ¹₂ ifx > ¹₂ g:

]0,1[ −→ R x 7−→ tan πx− ^π₂ T :











[0,1] −→ Lbg R^d

a 7−→

]−g◦ f(a),g◦ f(a)[ ifa <1 R^d ifa=1 The function f is strictly increasing and establishes a bijection[0,1] →

0,¹₂

∪₃

4,1

. Then g is bijective and strictly increasing, sog◦ f is striclty increasing and injective, and finallyTis stricly increasing in the Lebesgue set, so it is injective, and has a left inverse Lbg R^d

→ [0,1](which is a size). In other words,T is a test. However:

Ø

a<¹₂

T(a)= Ø

a<¹₂

]−g◦ f(a),g◦ f(a)[

=

#

−lim

a<¹₂g◦ f(a),lim

a<¹₂g◦ f(a)

"

=

#

−g lim

a<¹₂ f(a)

!

,g lim

a<¹₂ f(a)

! "

=

−g 1

2

,g 1

2 (T

1 2

=

−g 3

4

,g 3

4

Definition 4.15(Continuous test). LetTbe a test. We callTacontinuous testwhen, for allA⊂ [0,1], we have Ð

a∈A

T(a)=T(sup(A))and Ñ

a∈A

T(a)=T(inf(A)).

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Note that this notion of continuity is closer to the categorical notion than the analytic notion.

In the following, we give an example of such a continuous test.

Definition 4.16(Likelihood ratio). Let f₀,f₁ : R^d → R⁺be two probability density functions, and let x ∈R^d.

Thelikelihood ratio of xbeing fromX ∼ f₁instead ofX ∼ f₀, writtenΛ^f⁰^,^f¹(x)or simplyΛ(x)when there is no ambiguity, is the following function:

Λ:













R^d −→ R⁺∪ {∞}

x 7−→











 f₁(x)

f₀(x) if f₀(x),0

∞ if f₀(x)=0 and f₁(x)>0 0 if f₀(x)=0 and f₁(x)=0

In the following, we consider X ∼ f₀, and we restrict ourselves to the cases whereΛ(X)is an absolutely continuous random variable. This happens, using Jacobi’s transformation formula [2], when f₀ >0 and when both f₀and f₁ are differentiable, for example when f₀ and f₁are both Gaussian distributions. In this case, the functionk →P[Λ(X)> k | X ∼ f₀](which is the false alarm probability) is continuous and stricly decreasing, and has a continuous and strictly decreasing inverse which is described in the following definition.

Definition 4.17(NP-threshold function). Let f₀,f₁:R^d →R⁺be two probability density functions such that, ifX ∼ f₀thenΛ(X)is absolutely continuous.

TheNP-threshold function, writtenk_NPor simplyk in the following, is the function that assigns to each a ∈ [0,1], the thresholdk(a)such that:

a=

∫

Λ(x)>k(a)

f₀(x)dx=P[Λ(X)> k(a) | X ∼ f₀]

Remark4.18. The thresholdk(a)is the unique threshold whose false alarm probability isa.

As the functionk →P[Λ(X)> k | X∼ f₀]is continuous and stricly decreasing, it is easy to see that:

Proposition 4.19. The NP-threshold function is continuous and strictly decreasing.

Definition 4.20(Neyman-Pearson test). Let f₀,f₁ :R^d →R⁺be two probability density functions such that, ifX ∼ f₀thenΛ(X)is absolutely continuous.

TheNeyman-Pearson testis the following test:

NP :

[0,1] −→ Lbg R^d

a 7−→

x ∈R^d

Λ(x) > k(a) Proposition 4.21. The testNPis continuous.

Proof. This is due to the monotonicity of the threshold functionk. LetA⊂ [0,1]: Ø

a∈A

NP(a)= Ø

a∈A

x∈R^d

Λ(x) > k(a)

= x ∈R^d

∃a ∈ A, Λ(x)> k(a)

=

x ∈R^d

Λ(x)> inf

a∈A(k(a))

= x ∈R^d

Λ(x)> k(sup(A))

=NP(sup(A)) The proof for the intersection is roughly the same.

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Corollary 4.22. S_NP(a)=R\NP(a).

Definition 4.23(RDT context). An RDT-context is a tuple(d, θ₀,C, τ)where:

1. d ∈Nis thedimension; 2. θ₀ ∈R^d is called themodel;

3. C∈Matd,d(R)is a positively definite squared×d real matrix;

4. τ∈R^∗₊is thetolerance.

Definition 4.24(RDT problem). Let(d, θ₀,C, τ)be an RDT-context.

TheRDT problem with context(θ₀,C, τ)consists in the following testing problem:

Observation Y =Θ+XwhereX ∼ N 0,C²

,Θis any random vector inR^d andΘandXare independent.

H₀ kΘ−Yk_C 6 τ H₁ kΘ−Yk_C > τ

Definition 4.25(RDT threshold function). Let(d, θ₀,C, τ)be an RDT-context and leta ∈ ]0,1[. TheRDT threshold function with false alarma, writtenλ_a, is the following function:

λ_a :











R⁺ −→ R⁺

τ 7−→ the unique solution inηto the equation: a=1−F_χ²

d(τ²) η² Definition 4.26(RDT test). Let(d, θ₀,C, τ)be an RDT-context.

TheRDT test with context(d, θ₀,C, τ)is the following test:

RDT :













[0,1] −→ Lbg R^d

a 7−→











∅ ifa=0

R^d\{θ₀} ifa=1 x ∈R^d

λ_a(τ)< kx−θ₀k_C otherwise Proposition 4.27. The testRDTis continuous.

Proof. Let A⊂ [0,1]. The proof is similar to the one for NP, however we have a few details to take care before.

Ø

a∈A

RDT(a)= Ø

a∈A

x ∈R^d

λ_a(τ) < kx−θ₀k_C

= x∈R^d

∃a∈ A, λa(τ)< kx−θ₀k_C

=

x ∈R^d inf

a∈A(λ_a(τ))< kx−θ₀k_C

Then,a7→λ_a(τ)is decreasing and continuous ina, so:

a∈infA(λ_a(τ))= lim

x∈A x→sup(A)

λ_a(τ)=

λ_sup_(A)(τ) if 1,sup(A)

0 otherwise

If 1,sup(A)then:

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Ø

a∈A

RDT(a)=

x ∈R^d inf

a∈A(λ_a(τ))< kx−θ₀k_C

= x ∈R^d

λ_sup(A)(τ)< kx−θ₀k_C

=RDT(sup(A)) Otherwise, if 1=sup(A)then:

Ø

a∈A

RDT(a)=

x ∈R^d inf

a∈A(λa(τ))< kx−θ₀k_C

= x ∈R^d

0< kx−θ₀k_C

=R^d\{θ₀}

=RDT(sup(A))

Proposition 4.28.

4.3. About predicates

Definition 4.29(Predicate). LetSbe a sheaf inPSh(U). Apredicateis a sheaf morphismp:S →Ω.

Computing predicates

LetS:U^op →Setsbe a sheaf, and letp₁,p₂:S →Ωbe predicates. The predicatesp₁∧p₂,p₁∨p₂,p₁→ p₂,¬p₁are all predicatesS→Ω.

Leta∈ [0,1]. For x ∈S(a), definea₁= p₁_,a(x)anda₂= p₂_,_a(x). We have the trivial results:

(p₁∧p₂)_a(x)=min(a₁,a₂) (p₁∨p₂)_a(x)=max(a₁,a₂) Also:

(p₁→p₂)_a(x)=p₁_,a(x) →p₂_,a(x)

= Ü

R∧a₁6a₂

R

= Ü

b∈[0,a]

b∧a16a2

b

= Ü

b∈[0,a] min(a1,b)6a2

b

= Ü

b∈[0,a] min(a1,b)6a2

b

=U_c

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for somec∈ [0,a]. Let us take a look at thatc. We have:

c= sup

b∈[0,a]

min(a1,b)6a2

(b)=

a ifa₁ 6a₂ a₂ otherwise The negation¬p₁corresponds to the special case wherea₂=U₀ =∅:

(¬p₁)_a(x)= p₁_,_a(x) → ∅

= Ü

b∈[0,a] min(a1,b)60

b

=U_c Withcbeing:

c= sup

b∈[0,a]

min(a1,b)60

(b)=

a ifa₁ =0 0 ifa₁ >0 In summary:

Formula Condition Result

p₁_,a(x) a₁

p₂_,a(x) a₂

(p₁∧p₂)_a(x) min(a₁,a₂) (p₁∨p₂)_a(x) max(a₁,a₂) (p₁ →p₂)_a(x) ifa₁ 6 a₂ a

ifa₁ >a₂ a₂ (¬p₁)_a(x) ifa₁ =0 a

ifa₁ >0 U₀=∅ Construction of a predicate

How to write "x ∈B(θ₀, ρ)" in a predicate? (We are considering an open ball)

We first need to find the right test. Then it will give us a sheaf, and we will design the right predicate.

The predicate should be a sheaf morphism: p:P_T →Ω, such that, for alla∈ [0,1], it takesx∈ P_T(a) and returns the subset ofafor whichx ∈B(θ₀, ρ).

As a first answer, we choose ρto be a parameter for the test. We define the following test:

T(a)=B θ₀, aρ

1−a

ThenT is a continuous test, andP_T(a)= R\B θ₀, aρ

1−a

. Then the predicate should bep:S →Ω such that p_a(x) = ∅ if x < B(θ₀, ρ), and p_a(x) = a∧U1

2 where ¹₂ is the solution to the equation in a:

aρ 1−a = ρ.

Fora <b, we havea 6 band we need the following diagram to commute:

S(b) Ω(b)

S(a) Ω(a)

pb

6 ”∧a”

pa

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5. Study of the category of p-values Ifx ∈S(b), then:

”∧a”◦p_b(x)=

∅ ∧a ifx<B(θ₀, ρ) b∧a∧U1

2 ifx∈B(θ₀, ρ)

=p_a(x)

5. Study of the category of p-values

Definition 5.1(CategoryPVal(C)). LetC be a category representing a complete Heyting algebra.

The category of p-values onC, denoted byPVal(C), is the full subcategory ofPSh(C)such that every P∈PVal(C)comes from a test.

In the following we will study the properties of that category.

Proposition 5.2. (Conjecture)

1. PVal(U)doesn’t have all exponentials.

2. PVal(U)has all finite products.

3. PVal(U)has no terminal object.

The problem with these conjectures is that the properties of PSh([0,1]) don’t transfer to their subcategories. A first guess was that, ifPVal(U)had all exponentials, then the exponentials inPVal(U) should be isomorphic to the exponentials inPSh(U)given thatPVal(U)is a full subcategory ofPSh(U). In summary, it is thinkable that, if[B→C]is the exponential inPVal(U)andC^B is the exponential in PSh(U), then:

∀A,B,C ∈PVal(U),Hom^PSh(U)(A,[B→C])Hom^PSh(U)

A,C^B

⇒ [B→C]C^B

However, this is not the case. Consider a topological space(X,T)as a Heyting algebra. Then take any non-trivial topological subspace(Y,TY)ofX, also viewed as a Heyting algebra. The exponential in a topological space is defined from the exponential in a Heyting algebra and is defined as:

U∩V 6W ⇔V 6 U→W

| {z }

exponential

A more explicit version of this exponential is a follows [3, Section 1.1.4., p21]:

U →W

| {z }

inT

=max({Z ∈T | Z∩U 6W}) Which is the interior ofW∪ (X\U)inX.

IfUandWare open subsets ofY then, then the exponentialU→WinY is the interior ofW∪ (Y\U) which may be different to the interior ofW∪ (X\U)(cf. Figure 2).

Finally, we can suspect there is no terminal object, because we don’t see which object, other than∆(1), where∆:U →Sets^U, could be the terminal object, and∆(1)is not inPVal(U).

Otherwise, the product seems like an easy result.

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6. Natural transformation between NP and RDT

𝑋

𝑌

𝑊 𝑈

Figure 2: Illustration of what may happen when comparing the exponentials of a topological space with one of its subspaces. Here,Yis a topological subspace of X,UandWare open subsets ofY (and thus of X), and the exponentialU →W inY could be the exponential(U →W)_X ∩Y where(U→W)_X is the exponential inX.

Proof. (PVal(U) has all products) Let d₁,d₂ ∈ N, and letT₁ : [0,1] → Lbg R^d¹

andT₂ : [0,1] → Lbg R^d²

be two tests.

Then, for everya∈ [0,1],T₁(a) ×T₂(a)is also in Lbg R^d¹⁺^d². Then,T₁×T₂:[0,1] →Lbg R^d¹⁺^d² is also a test.

Thus:

S_T₁(a) ×S_T₂(a)= R^d¹\Ø

b<a

T₁(b)

!

× R^d²\Ø

b<a

T₂(b)

!

=R^d¹⁺^d²\Ø

b<a

(T₁(b) ×T₂(b))

= S_T₁×S_T₂ (a) Thus, the product inPVal(U)is the same as inPSh(U).

6. Natural transformation between NP and RDT

We will have to reformulate the problems they both answer to, in order to have two comparable tests.

LetCbe a positive-definited×dmatrix. Letµ∈R^d. The NP problem amounts to testing:

Observation : Y ∼ N (ε,C)=ε+N (0,C) H₀: ε=0

H₁: ε= µ While the RDT problem amounts to:

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Observation : Y =ε+X whereX ∼ N (0,C) H₀:

ε− ¹₂µ C 6

¹₂µ C =τ H₁:

ε− ¹₂µ C >

¹₂µ C =τ

We assume here thatεcan only take 0 orµas values. The RDT problem is more general (as seen above in Definition 4.24) in thatεcould be any real random variable. The NP problem is also more general, because it could compare any two distributions. We also note that, in this formulation, the two versions of the null hypothesis are equivalent, and the two versions of the alternative hypothesis too. With this formulation, both problems are exactly the same. However, this doesn’t mean that RDT and NP will give the same answer to a caught signal.

The goal of this section is to present the computations that lead to a natural transformation betweenS_NP andS_RDT, so that NP and RDT do give the same answer to the same caught signal.

Leta∈ [0,1].

x∈S_NP(a) ⇔Λ(x) 6 k(a)

⇔ exp

−¹₂kx− µk²_C exp

−¹₂kx−0k_C² 6 k(a)

⇔ 1

2kxk_C² − 1

2kx−µk²_C 6ln(k(a))

⇔ x^tC⁻¹x−x^tC⁻¹x+2x^tC⁻¹µ−µ^tC⁻¹µ62 ln(k(a))

⇔ x^tC⁻¹µ6 1

2µ^tC⁻¹µ+ln(k(a))

| {z }

K(a)

As for RDT:

x ∈S_RDT(a) ⇔

x− 1 2µ

C

6 λ_a(τ)

The NP problem amounts to comparing a scalar product to a threshold: x^tC⁻¹µ6 K(a), while the RDT problem amounts to comparing a distance to a threshold:

x− ¹₂µ

C 6λ_a(τ). The expression ofK(a)and λ_a(τ)are not important here.

We are looking for a set of functions f_a: S_RDT(a) →S_NP(a)org_a :S_NP(a) →S_RDT(a)that associate a ball to a hyperplane. Due to Proposition 4.10, f_aandgashould verify the nesting property. We thus look functions f :R^d →R^dorg:R^d →R^dthat respect this nesting property, and we will deduce the natural transformations from restrictions of these functions.

Namely, we are looking for agsuch that:

x^tC⁻¹µ=K(a) ⇔

g(x) − 1 2µ

C

=λ_a(τ) Let us start with dimensiond =2.

For a given µ,0, the set ofx^tC⁻¹µ=K(a)defines a straight line that is perpendicular to µ. Letµ₁be a vector orthogonal toµwith same length. We havex = x^tC⁻¹µ·µ+x^tC⁻¹µ₁·µ₁.

Consider now the circle B

12µ, λ_a(τ)

. We consider this circle as a trigonometric-like circle with radius λ_a(τ). As such, define the vector

12 +λ_a(τ)

µas the trigonometric origin on this circle. Consider the vector with trigonometric coordinateα(x)=arctan x^tC⁻¹µ₁; it is the following vectorg(x):

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g(x)= 1

2µ+λ_a(τ)

cos(α(x))

kµk_C µ+sin(α(x)) kµ₁k_C µ₁

𝐾(𝑎) 𝜇

𝜇1

𝑥^𝑡𝐶⁻¹𝜇1

Figure 3: Illustration of the action of a 2-dimensional natural transformation between NP and RDT.NP compares the projection of x overµto a thresholdK(a). As we are in finite dimension, there existsµ₁ orthogonal toµwith the same length. The coordinates ofxin the basis(µ, µ₁)are x^tC⁻¹µ,x^tC⁻¹µ₁

. The goal of the natural transformation we are building is to sendx^tC⁻¹µ₁to a trigonometric coordinate (see Figure4)

In dimensiond, it amounts to sending ad-vector to another vector in thed−1-sphere with spherical coordinates. In dimensiond, the spherical coordinates are(r, ϕ₁, . . . , ϕ_d−1), such that, ifxhas Cartesian coordinatesx₁, . . . ,x_d:

x₁=rcos(ϕ₁)

x₂=rsin(ϕ₁)cos(ϕ₂)

x₃=rsin(ϕ₁)sin(ϕ₂)cos(ϕ₃) . . .

x_d−₁=rsin(ϕ₁). . .sin(ϕ_d−₂)cos(ϕ_d−₁) x_d =rsin(ϕ₁). . .sin(ϕ_d−₂)sin(ϕ_d−₁)

Then, in dimensiond, given a positive-definite matrixCand a vectorµ, we decide of a unique orthogonal basis(µ, µ₁, . . . , µ_d−₁). For practical reasons, we do not assume that they have the same length 1 butkµk_C (it doesn’t change the result, only the easiness of the computations).

For a vector x=(x₁, . . . ,x_d)such thatx^tC⁻¹µ=K(a), we define a temporary vectorv(x)(which will play the role of the green one in Figure 4);v(x)has spherical coordinates(r, ϕ₁, . . . , ϕ_d−₁)where:

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References

1 𝜇

2𝜇 𝜇₁

Figure 4: Illustration of the action of a 2-dimensional natural transformation between NP and RDT.RDT compares the distance betweenxand¹₂µto a thresholdλa. So it defines a ball. In dimension 2, consider it as a trigonometric circle, in the sense that we want to place vectors on it. The vector with coordinates α=arctan x^tC⁻¹µ₁

on the circle is the green vector. It gives the expression ofg(x)(in red) as the sum

12µ+λ_a(τ)

cos(α(x))

kµk_C µ+ ^sin_kµ^(α(x))₁_k

C µ₁ .

r =λ_a(τ) ϕ₁=arccos

x^tC⁻¹µ₁ kxk_Ckµ₁k_C

ϕ₂=arccos

x^tC⁻¹µ₂ kxk_Ckµ₂k_C

. . . ϕ_d−₁=arcsin

x^tC⁻¹µ_d−₁ kxk_C kµ_d−₁k_C

and then:

g(x)= 1

2µ+v(x)

Finally, the natural transformation ¯g:S_NP→S_RDTis the following:

¯

g:S_NP →S_RDT =

g _S_NP_(a):

S_NP(a) −→ S_RDT(a) x 7−→ g(x)

a∈[0,1]

References

[1] M. Barr and C. Wells,Category Theory for Computing Science. Upper Saddle River, NJ, USA:

Prentice-Hall, Inc., 1998.

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References

[2] J. Jacod and P. Protter,Probability Essentials, 2nd ed., ser. Universitext. Springer-Verlag Berlin Heidelberg, 2004.

[3] “Chapter 1 - getting started,” in Residuated Lattices: An Algebraic Glimpse at Substructural Logics, ser. Studies in Logic and the Foundations of Mathematics, N. GALATOS, P. JIPSEN, T. KOWALSKI, and H. ONO, Eds. Elsevier, 2007, vol. 151, pp. 13 – 73. [Online]. Available:

http://www.sciencedirect.com/science/article/pii/S0049237X07800061

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