1. Estimating  0 Estimating the proportion of true null hypotheses with the method of moments By Jose M Muino. Email: jmui@igr.poznan.pl

2. 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

3. Definitions t Let T i, i = 1,...,m, be the test statistics for testing null hypotheses H 0,i based on observable random variables. Assume that H 0,i is true with probability  0 and false with probability (1-  0 ) Assume T i follows a density function f 0 (T) under H 0,i and f 1 (T) if H 0,i is false. Assume that the first m 0 =m*  0 H 0,i are true, and the next m 1 =m*(1-  0 ) H 0,i are false

4. The idea Define: Then:

5. The estimators Assumed known

6. Any moment Because: Then:

7. Estimators Sample levelTest value level

8. 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 x i,j, i=1,…m, j=1,…n, using as test statistic the mean of the observations

9. Properties Assuming independence

10. Properties Assuming independence

11. Properties Assuming independence

12. Numerical Simulations m0=450,m1=50, H0->N(0,1), H1->N(1,1) (2000 simulations) 5000 simulations

13. Numerical Simulations 5000 simulations

14. From moments to quantiles A family of distributions (eg: Pearson family) can be used to calculate the quantiles error type I n=3 n=4 n=5 MMClassicalMMClassicalMMClassical 0,50,4870,4990,4930,50,4960,499 0,10,0960,0990,0970,0990,0980,099 0,050,0560,0490,0520,0490,0510,049 0,010,0220,0090,0140,010,0120,009 0,0010,010,0010,0030,0010,0020,0009 100 simulations

15. From moments to quantiles error type I n=3 n=4 n=5 MMClassicalMMClassicalMMClassical 0,50,4760,4990,4890,50,4940,5 0,10,0960,0990,0960,0990,0970,1 0,050,0630,0490,0540,0490,0520,05 0,010,0380,0090,0190,010,0150,01 0,0010,0290,00090,0070,0010,0030,001 error type I n=3 n=4 n=5 MMClassicalMMClassicalMMClassical 0,50,4890,50,4950,50,4960,5 0,10,0970,10,0980,0990,0980,1 0,050,0540,050,0520,0490,0510,05 0,010,0180,010,0130,010,0120,01 0,0010,0060,00090,0020,0010,0020,001

16. 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

17. 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

18. Thanks for your attention!! Questions?, Suggestions? Or write me at: jmui@igr.poznan.pl Work funded by Marie Curie RTN: “Transistor” (“Trans-cis elements regulating key switches in plant development”)