[PPT] - Generalized Ruuds Theorem Mariusz Kubkowski a , Jan Mielniczuk a , b PowerPoint Presentation

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Generalized Ruud’s Theorem

Mariusz Kubkowskia, Jan Mielniczuka,b

aWarsaw University of Technology

Faculty of Mathematics and Information Science,

bInstitute of Computer Science

Polish Academy of Sciences

Będlewo, December 1st 2016

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Introduction

p, k ∈ N, k ≤ p,

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Introduction

p, k ∈ N, k ≤ p, (X, Y ) ∈ Rp × {0, 1} - random vector, X ∼ PX,

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Introduction

p, k ∈ N, k ≤ p, (X, Y ) ∈ Rp × {0, 1} - random vector, X ∼ PX, q : Rk → [0, 1] - unknown response function,

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Introduction

p, k ∈ N, k ≤ p, (X, Y ) ∈ Rp × {0, 1} - random vector, X ∼ PX, q : Rk → [0, 1] - unknown response function, β1, . . . , βk ∈ Rp, - true coefficients, B =

β1

. . . βk

∈ Rp×k.
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Introduction

p, k ∈ N, k ≤ p, (X, Y ) ∈ Rp × {0, 1} - random vector, X ∼ PX, q : Rk → [0, 1] - unknown response function, β1, . . . , βk ∈ Rp, - true coefficients, B =

β1

. . . βk

∈ Rp×k.

Conditional distribution of Y |X : P(Y = 1|X) = q(βT

1 X, . . . , βT k X) =: q(BTX).

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Introduction

β01, . . . , β0k ∈ R, q(BTX) = ˜ q(β01 + βT

1 X, . . . , β0k + βT k X).

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Introduction

β01, . . . , β0k ∈ R, q(BTX) = ˜ q(β01 + βT

1 X, . . . , β0k + βT k X).

Special case (k = 1, B = β1): P(Y = 1|X) = q(βT

1 X).

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Misspecification problem

We fit the model: P(Y = 1|X) = qL(β∗

0 + β∗TX) - model M0 (with intercept)

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Misspecification problem

We fit the model: P(Y = 1|X) = qL(β∗

0 + β∗TX) - model M0 (with intercept)

r

P(Y = 1|X) = qL(β∗TX) - model M (without intercept),

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Misspecification problem

We fit the model: P(Y = 1|X) = qL(β∗

0 + β∗TX) - model M0 (with intercept)

r

P(Y = 1|X) = qL(β∗TX) - model M (without intercept), where: qL(x) = 1 1 + e−x , x ∈ R.

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Misspecification problem

Log-likelihood function (b0 ∈ R, b ∈ Rp): l(b0, b, X, Y ) = Y (b0 + bTX) − log

1 + exp
b0 + bTX
.
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Misspecification problem

Log-likelihood function (b0 ∈ R, b ∈ Rp): l(b0, b, X, Y ) = Y (b0 + bTX) − log

1 + exp
b0 + bTX
.

We want to find coefficients (in the model M0): (β∗

0, β∗) =

arg max

(b0,b)∈R×Rp E(X,Y )l(b0, b, X, Y )

r (in the model M):

β∗ = arg max

b∈Rp

E(X,Y )l(0, b, X, Y ).

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Misspecification problem

Theorem (Li, Duan, 1989)

If q(BTX) ∈ (0, 1) a.s. and E||X|| < ∞, then there exist solutions for models M0 and M.

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Misspecification problem

Theorem (Li, Duan, 1989)

If q(BTX) ∈ (0, 1) a.s. and E||X|| < ∞, then there exist solutions for models M0 and M.

Theorem (Li, Duan, 1989)

If q(BTX) ∈ (0, 1) a.s., E||X||2 < ∞, and EXXT > 0, then the solutions for models M0 and M are unique.

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Misspecification problem

Theorem (Li, Duan, 1989)

If q(BTX) ∈ (0, 1) a.s. and E||X|| < ∞, then there exist solutions for models M0 and M.

Theorem (Li, Duan, 1989)

If q(BTX) ∈ (0, 1) a.s., E||X||2 < ∞, and EXXT > 0, then the solutions for models M0 and M are unique. Normal equations (model M0): E(X,Y )D(b0,b)l(β∗

0, β∗, X, Y ) = E

1

XTT (q(BTX)−qL(β∗

0+β∗TX)) = 0,

equivalently: EqL(β∗

0 + β∗TX)

1

XTT = Eq(BTX)

1

XTT .

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Ruud’s theorem (k = 1)

Ruud’s condition for B and k = 1 (Ruud, 1983)

∀z ∈ R∃u0, u ∈ Rp : E(X|βT

1 X = z) = u0 + uz

Remark: Ruud’s condition is satisfied, when X has elliptically contoured distribution, in particular normal distribution.

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Ruud’s theorem (k = 1)

Ruud’s condition for B and k = 1 (Ruud, 1983)

∀z ∈ R∃u0, u ∈ Rp : E(X|βT

1 X = z) = u0 + uz

Remark: Ruud’s condition is satisfied, when X has elliptically contoured distribution, in particular normal distribution.

Theorem (Ruud, 1983)

If k = 1, q(βT

1 X) ∈ (0, 1) a.s., X satisfies Ruud’s condition for B and

k = 1, E||X||2 < ∞, and EXXT > 0, then in the models M and M0 there exists η ∈ R such that: β∗ = ηβ1.

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Generalized Ruud’s theorem

Ruud’s condition for B (Li, 1991)

∀z ∈ Rk∃u0 ∈ Rp, U ∈ Rp×k : E(X|BTX = z) = u0+Uz = u0+

k

i=1

U(i)zi

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Generalized Ruud’s theorem

Ruud’s condition for B (Li, 1991)

∀z ∈ Rk∃u0 ∈ Rp, U ∈ Rp×k : E(X|BTX = z) = u0+Uz = u0+

k

i=1

U(i)zi

Theorem

If q(BTX) ∈ (0, 1) a.s., X satisfies Ruud’s condition for B, E||X||2 < ∞, and EXXT > 0, then the coefficients of the logistic models M and M0 satisfy the following condition: ∃η ∈ Rk : β∗ = Bη =

k

i=1

ηiβi.

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Form of η

Theorem (representation of η)

If the assumptions of the generalized Ruud’s theorem are satisfied, rank B = k, and additionally X satisfies Ruud’s condition for β∗, then η from the generalized Ruud’s theorem satisfies the following equation: aβ∗ · η = aB, where aG = (Var(G TX))−1 Cov(G TX, Y ) for G ∈ Rp×l (l ∈ N) - full column rank matrix. Moreover, if Cov(BTX, Y ) = 0, then aβ∗ = 0.

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Form of η

Theorem (representation of η)

If the assumptions of the generalized Ruud’s theorem are satisfied, rank B = k, and additionally X satisfies Ruud’s condition for β∗, then η from the generalized Ruud’s theorem satisfies the following equation: aβ∗ · η = aB, where aG = (Var(G TX))−1 Cov(G TX, Y ) for G ∈ Rp×l (l ∈ N) - full column rank matrix. Moreover, if Cov(BTX, Y ) = 0, then aβ∗ = 0. Remark: If X ∼ Np(µ, Σ), Σ > 0, q - differentiable and E|Dq(BTX)| < ∞, then from the Stein’s lemma: aB = EDq(BTX), aβ∗ = Eq′

L(β∗TX).

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Algorithm for finding η

X ∼ N(µ, Σ), Σ > 0, rank B = k. Let (model M0): U = BTX, ˜ U = [1 UT]T, ˜ η = [β∗

0 ηT]T.

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Algorithm for finding η

X ∼ N(µ, Σ), Σ > 0, rank B = k. Let (model M0): U = BTX, ˜ U = [1 UT]T, ˜ η = [β∗

0 ηT]T.

Consider the function: F(˜ η) = EqL(˜ ηT ˜ U)˜ U − Eq(U)˜ U, Newton-Raphson iterations (˜ η0 - fixed): ˜ ηn+1 = ˜ ηn − (DF(˜ ηn))−1F(˜ ηn). Problem: how to compute expected values?

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Algorithm for finding η

X ∼ N(µ, Σ), Σ > 0, rank B = k. Let (model M0): U = BTX, ˜ U = [1 UT]T, ˜ η = [β∗

0 ηT]T.

Consider the function: F(˜ η) = EqL(˜ ηT ˜ U)˜ U − Eq(U)˜ U, Newton-Raphson iterations (˜ η0 - fixed): ˜ ηn+1 = ˜ ηn − (DF(˜ ηn))−1F(˜ ηn). Problem: how to compute expected values? Gauss - Hermite quadratures (from R package fastGHQuad).

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Additive misspecification - special case

Let: q(BTX) =

k

i=1

λiqi(βT

i X),

where for each i = 1, . . . , k : qi : R → (0, 1), λi ≥ 0,

k

i=1

λi = 1.

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Additive misspecification - special case

Let: q(BTX) =

k

i=1

λiqi(βT

i X),

where for each i = 1, . . . , k : qi : R → (0, 1), λi ≥ 0,

k

i=1

λi = 1. Assume that X ∼ Np(0, Σ), Σ > 0, qi are differentiable and E|q′

i(βT i X)| < ∞.

We assume q(0, . . . , 0) = 0.5 and fit the model M (without intercept).

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Additive misspecification - special case

It follows (Σ > 0): β∗ =

k

i=1

ηiβi, ηi = λi Eq′

i(βT i X)

Eq′

L(β∗TX).

q′

i > 0 ⇒ ηi ≥ 0,

q′

i > 0 ⇒ ηi ≤???

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Additive misspecification

Theorem (upper bounds for weighted sum of η)

Let X ∼ Np(0, Σ), Σ > 0. Consider the additive misspecification problem with qi : R → (0, 1) - strictly increasing and differentiable functions and E|q′

i(βT i X)| < ∞.

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Additive misspecification

Theorem (upper bounds for weighted sum of η)

Let X ∼ Np(0, Σ), Σ > 0. Consider the additive misspecification problem with qi : R → (0, 1) - strictly increasing and differentiable functions and E|q′

i(βT i X)| < ∞.

If matrix B is of full column rank then in the model M :

k

i=1

ηi Eq′

i(βT i X) ≤ D =

max

i=1,...,k

Di

Eq′

i(βT i X)

,

where Di are (unique) solutions of equations:

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Additive misspecification

Theorem (upper bounds for weighted sum of η)

Let X ∼ Np(0, Σ), Σ > 0. Consider the additive misspecification problem with qi : R → (0, 1) - strictly increasing and differentiable functions and E|q′

i(βT i X)| < ∞.

If matrix B is of full column rank then in the model M :

k

i=1

ηi Eq′

i(βT i X) ≤ D =

max

i=1,...,k

Di

Eq′

i(βT i X)

,

where Di are (unique) solutions of equations: EqL(DiβT

i X)βT i X = Eqi(βT i X)βT i X.

Idea of proof: implicit function theorem.

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Additive misspecification

k

i=1

ηi Eq′

i(βT i X) ≤ D =

max

i=1,...,k

Di

Eq′

i(βT i X)

,

is equivalent to: Eq′

L(ηTBTX) ≥ min Eq′ L(DiβT i X).

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Additive misspecification

k

i=1

ηi Eq′

i(βT i X) ≤ D =

max

i=1,...,k

Di

Eq′

i(βT i X)

,

is equivalent to: Eq′

L(ηTBTX) ≥ min Eq′ L(DiβT i X).

Z ∼ N(0, 1), σ ≥ 0, h(σ) = Eq′

L(σZ) - decreasing.

Hence: Var(ηTBTX) ≤ max Var(DiβT

i X).

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Additive misspecification

Theorem (shrinking property)

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL and Var(βT

1 X) = . . . = Var(βT k X), then (in the

model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi

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Additive misspecification

Theorem (shrinking property)

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL and Var(βT

1 X) = . . . = Var(βT k X), then (in the

model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi and (equivalently): η1 + . . . + ηk ≤ 1, where the equality η1 + . . . + ηk = 1 holds if and only if ∃i ∈ {1, . . . , k} : λi = 1.

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Additive misspecification

Theorem

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL, then (in the model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi Eq′

L(βT i X)

min

j=1,...,k Eq′ L(βT j X).

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Additive misspecification

Theorem

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL, then (in the model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi Eq′

L(βT i X)

min

j=1,...,k Eq′ L(βT j X).

Remark: for k = 2 (λ2 = 1 − λ1): ∂η1(λ1) ∂λ1

|λ1=0

= Eq′

L(βT 1 X)

Eq′

L(βT 2 X).

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Additive misspecification

Theorem

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL, then (in the model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi Eq′

L(βT i X)

min

j=1,...,k Eq′ L(βT j X).

Remark: for k = 2 (λ2 = 1 − λ1): ∂η1(λ1) ∂λ1

|λ1=0

= Eq′

L(βT 1 X)

Eq′

L(βT 2 X).

Z ∼ N(0, 1), σ ≥ 0, h(σ) = Eq′

L(σZ) - decreasing

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Additive misspecification

Theorem

If the assumptions of the upper bounds for weighted sum of η theorem hold, q1 = . . . = qk = qL, then (in the model M): ∀i ∈ {1, . . . , k} : ηi ≤ λi Eq′

L(βT i X)

min

j=1,...,k Eq′ L(βT j X).

Remark: for k = 2 (λ2 = 1 − λ1): ∂η1(λ1) ∂λ1

|λ1=0

= Eq′

L(βT 1 X)

Eq′

L(βT 2 X).

Z ∼ N(0, 1), σ ≥ 0, h(σ) = Eq′

L(σZ) - decreasing

Var(βT

1 X) < Var(βT 2 X) ⇒ ∂η1(λ1) ∂λ1 |λ1=0 > 1 ⇒ η1(λ1) > λ1 for small λ1

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Additive misspecification - example 1

Example

Let β ∈ Rp \ {0}, α ∈ [0, 1], k = 2, B =

β

−β

, q1 = q2 = qL, X ∼

Np(0, Σ), Σ > 0. Then: P(Y = 1|X) = αqL(βTX) + (1 − α)qL(−βTX). In this case rank B = 1 < 2 and |η| ≤ 1 (η from Ruud’s theorem for k = 1). From implicit function theorem (and Stein lemma): η′(α) = 2Eq′

L(U)

Eq′

L(η(α)U)U2 > 0.

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Additive misspecification - example 1

Example

Let β ∈ Rp \ {0}, α ∈ [0, 1], k = 2, B =

β

−β

, q1 = q2 = qL, X ∼

Np(0, Σ), Σ > 0. Then: P(Y = 1|X) = αqL(βTX) + (1 − α)qL(−βTX). In this case rank B = 1 < 2 and |η| ≤ 1 (η from Ruud’s theorem for k = 1). From implicit function theorem (and Stein lemma): η′(α) = 2Eq′

L(U)

Eq′

L(η(α)U)U2 > 0.

Hence −1 = η(0) ≤ η(α) ≤ η(1) = 1. Remark: |β∗

i | ≤ |βi| for i = 1, . . . , p.

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Additive misspecification - example 1

−1.0 −0.5 0.0 0.5 1.0 0.00 0.25 0.50 0.75 1.00

alpha eta sigma

0.5 1 1.5 2 2.5 3 3.5 4

Additive misspecification k=2, rank B=1, q1=q2=qL

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Additive misspecification - example 2

r > 0, ρ ∈ [−1, 1], α ∈ [0, 1], X ∼ N2

,

r2 ρr ρr 1

,

P(Y = 1|X) = αqL(X1) + (1 − α)qL(X2). We fit the model: P(Y = 1|X) = qL(η1X1 + η2X2).

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0, r = 0.5

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0, r = 1

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0, r = 2

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0, r = 4

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0.9, r = 0.5

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0.9, r = 1

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0.9, r = 2

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Additive misspecification - example 2

0.00 0.25 0.50 0.75 1.00 0.00 0.25 0.50 0.75 1.00

alpha eta Legend

eta1 eta2 s=eta1+eta2

Additive misspecification rank B=k=2, q1=q2=qL rho = 0.9, r = 4

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Conclusions and open problems

for ρ = 0.9 values of ηi are ”closer” to upper bounds,

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Conclusions and open problems

for ρ = 0.9 values of ηi are ”closer” to upper bounds, it can happen that ηi > λi,

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Conclusions and open problems

for ρ = 0.9 values of ηi are ”closer” to upper bounds, it can happen that ηi > λi,

∂ηi ∂λi > 0?

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Conclusions and open problems

for ρ = 0.9 values of ηi are ”closer” to upper bounds, it can happen that ηi > λi,

∂ηi ∂λi > 0?

η1 + . . . + ηk ≤ 1 when q1 = . . . = qk = qL and Var(βT

i X) are not all

equal?

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