Adaptive Bayes Test for monotonicity - ENSAI-ENSAEsalomond/doc/presENSAI.pdf · 2013. 4. 16. ·...
Transcript of Adaptive Bayes Test for monotonicity - ENSAI-ENSAEsalomond/doc/presENSAI.pdf · 2013. 4. 16. ·...
Programme Introduction A Bayesian approach Main results Simulation study
Adaptive Bayes Test for monotonicityENSAI-ENSAE
J-B. Salomond
CREST - Dauphine
22-23 Mars 2012
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Table des matières
1 Introduction
Testing for Monotonicity
Motivations
2 A Bayesian approach
Our Model
The testing procedure
3 Main results
4 Simulation study
Useful remark
Experimental design
Results
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Testing for Monotonicity
Introduction
We consider the regression model with Gaussian residuals
Yi = f (i/n) + ǫi
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Testing for Monotonicity
Introduction
We consider the regression model with Gaussian residuals
Yi = f (i/n) + ǫi
Aim
We want to test f is monotone non increasing versus f is not
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Testing for Monotonicity
Introduction
We consider the regression model with Gaussian residuals
Yi = f (i/n) + ǫi
Aim
We thus construct a Bayesian testing procedure which
has good asymptotic properties
is easy to implement and does not require heavy
computations, even for large datasets
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Motivations
Why ?
Monotonicity appears in many applications (drug response models
for instance)
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Motivations
Why ?
Monotonicity appears in many applications (drug response models
for instance)
Many results in the frequentist literature
but no Bayesian results are known
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Our Model
Our Model
Given a partition (Ii )i of (0, 1) in k steps, we consider a piecewise
constant approximation of the regression function
fω,k(.) =
k∑
i=1
ωi1Ii (.)
and put a prior on f by choosing a prior on ω and k
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Our Model
Our Model cnt’d
When f is monotone, fω,k will be monotone
Idea for a test
We will thus test the monotonicity of the sequence (ωi )i and thus
need
A criteria for monotonicity
Conditions on the prior such that fω,k concentrates around f
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The testing procedure
The testing procedure
We consider H(ω, k) = maxj>i (ωj − ωi). We thus have a test
δπn = 1
{
π(
H(ω, k) > Mkn |Yn
)
> 1/2}
where Mkn is a threshold such that our test is consistent.
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The testing procedure
Required asymp. properties
Let F be the set of monotone non increasing functions. We
would like our test to be consistent
supf ∈F
En0(δπn ) = o(1) (1a)
supf ,d(f ,F)>ρ,f ∈Hα(L)
En0(1− δ
πn ) = o(1) (1b)
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The testing procedure
Required asymp. properties cnt’d
We would like our test to achieve an optimal separation rate
supf ∈F
En0(δπn ) = o(1) (2a)
supf ,d(f ,F)>ρn,f ∈Hα(L)
En0(1 − δ
πn ) = o(1) (2b)
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Main Theorem
Theorem
Let Mkn = M0
√
k log(n)/n and let π a prior on fω,k such that
ωi |kiid∼ g and k ∼ πk . Assume that g puts mass on R and that πk
is such that their exist positive constants Cd and Cu such that
eCdkL(k) ≤ πk (k) ≤ eCukL(k)
Where L(k) is either log(k) or 1. Consider the test
H0 : f ∈ F versus H1 : f 6∈ F , f ∈ Hα(L)
let ρn = M(
n/ log(n))−α/(2α+1)
, then δπn is consistent and
achieve the separation rate ρn
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Useful remark
Some useful results
Some specific choices for the prior can be handy.
Prior
We take k ∼ P(λ) and ω|k ∼ N (m, v2). Then, if ǫi ∼ N (0, σ2)
we get
πk(k |Yn) = C (Y n)e
1/2∑
i
(
Yi/σ2+m/v2
ni/σ2+1/v2
)2k∏
i=1
(
ni/σ2 + 1/v2
)1/2πk(k)
ωi |Yn, k ∼ N
(
m/v2 + Yi/σ2
ni/σ2 + 1/v2;
1
ni/σ2 + 1/v2
)
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Experimental design
Experimental design
We choose nine regression functions and generate N independent
samples yi , i = 1 . . . n. For each sample we perform our test and
approximate the proportion of rejection.
x
Fun
ctio
ns
0.0
0.5
1.0
1.5
2.0
−0.4
−0.2
0.0
0.2
0.4
−1.8
−1.6
−1.4
−1.2
−1.0
F1
F4
F7
0.0 0.2 0.4 0.6 0.8 1.0
0.000.020.040.060.080.100.120.14
−0.15
−0.10
−0.05
0.00
0.05
0.10
−0.5
−0.4
−0.3
−0.2
−0.1
0.0
F2
F5
F8
0.0 0.2 0.4 0.6 0.8 1.0
0.05
0.10
0.15
−0.6
−0.4
−0.2
0.0
0.2
0.4
0
F3
F6
F9
0.0 0.2 0.4 0.6 0.8 1.0
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Experimental design
Experimental design cnt’d
For some empirically realistic value for the parameters, simulate
N = 25 samples for n = 100, 250, 500, 1000.� �Still performing simulation for larger values of N and n
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Results
Results
Good results even for small sample size
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Results
Results cnt’d
We now compare our result with the existing procedures
Function σ2 Bayes Sregn TB
f1 0.01 1.00 0.99 0.99
f2 0.01 1.00 1.00 0.99
f3 0.01 1.00 0.98 0.99
f4 0.01 1.00 0.99 1.00
f5 0.004 1.00 0.99 0.99
f6 0.006 1.00 0.99 0.98
f7 0.01 0.68 0.68 0.76
Table: Comparison with the existing procedures for n = 100
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Results
Results cnt’d
Comparison with the
Bayes Factor approach
We compute the following
Bayes Factor
BF01 =
π(H(ω, k) ≤ 0|Y n)
π(H(ω, k) > 0|Y n)×
π(H(ω, k) > 0)
π(H(ω, k) ≤ 0)
n
100 250 500 1000
f1 0.00 0.00 0.00 0.00
f2 0.00 0.00 0.00 0.00
f3 0.02 0.00 0.00 0.00
f4 0.00 0.00 0.00 0.00
f5 0.00 0.00 0.00 0.00
f6 0.00 0.00 0.00 0.00
f7 0.00 0.00 0.00 0.00
f8 0.94 0.37 0.23 0.10
f9 0.11 0.02 0.00 0.00
Table: B01
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Results
Results cnt’d
Comparison with the
Bayes Factor approach
Satisfying results under
H1, but our procedure
perform better under H0
n
100 250 500 1000
f1 0.00 0.00 0.00 0.00
f2 0.00 0.00 0.00 0.00
f3 0.02 0.00 0.00 0.00
f4 0.00 0.00 0.00 0.00
f5 0.00 0.00 0.00 0.00
f6 0.00 0.00 0.00 0.00
f7 0.00 0.00 0.00 0.00
f8 0.94 0.37 0.23 0.10
f9 0.11 0.02 0.00 0.00
Table: B01
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