estimating 0 estimating the proportion of true null hypotheses with the method of moments by jose m...
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![Page 1: Estimating 0 Estimating the proportion of true null hypotheses with the method of moments By Jose M Muino. Email: jmui@igr.poznan.pl](https://reader036.vdocument.in/reader036/viewer/2022070401/56649f165503460f94c2ceb3/html5/thumbnails/1.jpg)
Estimating 0
Estimating the proportion of true null hypotheses with the method of moments
By Jose M Muino. Email: [email protected]
![Page 2: Estimating 0 Estimating the proportion of true null hypotheses with the method of moments By Jose M Muino. Email: jmui@igr.poznan.pl](https://reader036.vdocument.in/reader036/viewer/2022070401/56649f165503460f94c2ceb3/html5/thumbnails/2.jpg)
The objective
Objective To obtain some information (0 and moments)
to help in the construction of the critical region in a multiple hypotheses problem
The situation: Low sample size The distribution under the null hypothesis is
unknown But the expectation of the null distribution is
known
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Definitions
-2 2 4 6
0.1
0.2
0.3
0.4
t
)(1 Tf Tf0 )()1()( 1000 TfTfTg
Let Ti, i = 1, . . . ,m, be the test statistics for testing null hypotheses H0,i based on observable random variables.
Assume that H0,i is true with probability 0 and false with probability (1- 0)
Assume Ti follows a density function f0(T) under H0,i and f1(T) if H0,i is false.
Assume that the first m0 =m*0 H0,i are true, and the next m1 =m*(1-0 ) H0,i are false
)(on distributi theofmoment central thedenotes
)(on distributi theofmoment raw thedenotes
)(,
)(,
xjc
xjth
xj
thxj
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The idea
)()1()( 1000 TfTfTg
)(,10,10)(,1 10)1( TfTfTg
)(,20,202 10)1( TfTf ccd
Define:
20
20)(,22 )(1,10,1
)1(TfTfTgd
Then:
)(,2)(,1)(,12
)(,12
)(,22
)(,120
002 TgTfTgTf
TgTg
d
d
)(,1)(,1
)(,2)(,1)(,12)(,1
0
0
1
TgTf
TgTfTgTf
d
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The estimators
)(,2)(,1)(,12
)(,12
)(,22
)(,120
002 TgTfTgTf
TgTg
d
d
)(,1)(,1
)(,2)(,1)(,12)(,1
0
0
1
TgTf
TgTfTgTf
d
)(,20,202 10)1( TfTf ccd
m
iicm
d1
,22 ˆ1ˆ
)(,1 0 TfAssumed known
)(,2
)(,1
Tg
Tg
2)(,2
)(,1
1ˆ
1ˆ
iTg
iTg
Tm
Tm
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Any moment
Because:
Then:
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Estimators
Sample levelTest value level
m
ijij c
md
1,ˆ
1ˆ
m
i
jiTgj T
m 1)(,
1̂
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Example: The mean value as test statistic
The properties of the estimators can be studied by taking Taylor series.
The properties will be illustrated with the example of the mean value as test statistic
Testing m hypotheses regarding m observed samples xi,j, i=1,…m, j=1,…n, using as test statistic the mean of the observations
ii xT 0:
0:
1
0
i
i
H
H
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Properties
Assuming independence
m
ijij c
md
1,ˆ
1ˆ if
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Properties
Assuming independence
0ˆ if jd
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Properties
Assuming independence
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Numerical Simulations
m0=450,m1=50, H0->N(0,1), H1->N(1,1)(2000 simulations)
0 10 20 30 40sample size
-0.025
0
0.025
0.05
0.075
0.1
0.125
0.15
saib
a
0B
0S
0
0
0
0 10 20 30 40sample size
0
0.01
0.02
0.03
0.04
0.05
dradnatsnoitaived
b
0B
0S
0
0
0
)/5,2(
)/5,0(
40,...,2
500
1000
1
0
1
0
nNTf
nNTf
n
m
m
5000 simulations
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Numerical Simulations
0 10 20 30 40sample size
0
0.05
0.1
0.15
0.2
0.25
saib
a
B
S
3
2
1
0 10 20 30 40sample size
0
0.02
0.04
0.06
0.08
dradnatsnoitaived
b
B
S
3
2
1
)5,2(
)5,0(
40,...,2
500
1000
1
0
1
0
NTf
NTf
n
m
m
5000 simulations
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From moments to quantiles
A family of distributions (eg: Pearson family) can be used to calculate the quantiles
)/1,4( )/1,0( 5000 15000 1010 nNTfnNTfmm
error type I
n=3 n=3 n=4 n=4 n=5 n=5
MM Classical MM Classical MM Classical
0,5 0,487 0,499 0,493 0,5 0,496 0,499
0,1 0,096 0,099 0,097 0,099 0,098 0,099
0,05 0,056 0,049 0,052 0,049 0,051 0,049
0,01 0,022 0,009 0,014 0,01 0,012 0,009
0,001 0,01 0,001 0,003 0,001 0,002 0,0009
100 simulations
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From moments to quantiles
)/1,4( )/1,0( 10000 10000 1010 nNTfnNTfmm error type I
n=3 n=3 n=4 n=4 n=5 n=5
MM Classical MM Classical MM Classical
0,5 0,476 0,499 0,489 0,5 0,494 0,5
0,1 0,096 0,099 0,096 0,099 0,097 0,1
0,05 0,063 0,049 0,054 0,049 0,052 0,05
0,01 0,038 0,009 0,019 0,01 0,015 0,01
0,001 0,029 0,0009 0,007 0,001 0,003 0,001
error type I
n=3 n=3 n=4 n=4 n=5 n=5
MM Classical MM Classical MM Classical
0,5 0,489 0,5 0,495 0,5 0,496 0,5
0,1 0,097 0,1 0,098 0,099 0,098 0,1
0,05 0,054 0,05 0,052 0,049 0,051 0,05
0,01 0,018 0,01 0,013 0,01 0,012 0,01
0,001 0,006 0,0009 0,002 0,001 0,002 0,001
)/1,4( )/1,0( 2000 18000 1010 nNTfnNTfmm
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Advantages
Combine information from sample and test level.
No assumptions about the shape of the distribution (finite moments required)
Analytical solution Properties can be obtained Estimator can be improved
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Disadvantages
Different estimator for different test statistic
Estimators of the central moments of the test statistics are required
The estimation can be outside of the parameter space
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Thanks for your attention!!
Questions?, Suggestions?
Or write me at:[email protected]
Work funded by Marie Curie RTN: “Transistor” (“Trans-cis elements regulating key switches in plant development”)