Estimating 0 Estimating the proportion of true null

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

The objective n Objective n To obtain some information ( 0 and moments) to help in the construction of the critical region in a multiple hypotheses problem n The situation: n Low sample size n The distribution under the null hypothesis is unknown n But the expectation of the null distribution is known

Definitions Let Ti, 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 Ti 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 t

The idea Define: Then:

The estimators Assumed known

Any moment Because: Then:

Estimators Test value level Sample level

Example: The mean value as test statistic n The properties of the estimators can be studied by taking Taylor series. n The properties will be illustrated with the example of the mean value as test statistic n Testing m hypotheses regarding m observed samples xi, j, i=1, …m, j=1, …n, using as test statistic the mean of the observations

Properties Assuming independence

Properties Assuming independence

Properties Assuming independence

Numerical Simulations m 0=450, m 1=50, H 0 ->N(0, 1), H 1 ->N(1, 1) (2000 simulations) 5000 simulations

Numerical Simulations 5000 simulations

From moments to quantiles n A family of distributions (eg: Pearson family) can be used to calculate the quantiles error type I n=3 n=4 n=5 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, 056 0, 049 0, 052 0, 049 0, 051 0, 049 0, 01 0, 022 0, 009 0, 014 0, 012 0, 009 0, 001 0, 003 0, 001 0, 002 0, 0009 100 simulations

From moments to quantiles error type I n=3 n=4 n=5 MM Classical 0, 5 0, 476 0, 499 0, 489 0, 5 0, 494 0, 5 0, 1 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, 015 0, 01 0, 029 0, 0009 0, 007 0, 001 0, 003 0, 001 error type I n=3 n=4 n=5 MM Classical 0, 5 0, 489 0, 5 0, 495 0, 496 0, 5 0, 1 0, 097 0, 1 0, 098 0, 099 0, 098 0, 1 0, 054 0, 052 0, 049 0, 051 0, 05 0, 018 0, 013 0, 012 0, 01 0, 006 0, 0009 0, 002 0, 001

Advantages n Combine information from sample and test level. n No assumptions about the shape of the distribution (finite moments required) n Analytical solution Properties can be obtained n Estimator can be improved n

Disadvantages n Different estimator for different test statistic n Estimators of the central moments of the test statistics are required n The estimation can be outside of the parameter space

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”)