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Stat 516 Homework 5 Solutions 1. Consider conducting m hypothesis tests. Let V denote the number of type I errors. Let R denote the number of rejected null hypotheses. Let Q = V /R if R > 0, and let Q = 0 if R = 0. Let α ∈ (0, 1) be fixed. By definition, a method that strongly controls FDR at level α has the property that E(Q) ≤ α no matter which or how many of the m null hypotheses are true or false. Note that if all null hypotheses are true, Q = 1 if R > 0, Q = 0 if R = 0, and V = R. Thus, E(Q) = E(Q|R > 0)P (R > 0) + E(Q|R = 0)P (R = 0) = 1 ∗ P (R > 0) + 0 ∗ P (R = 0) = P (R > 0) = P (V > 0) = FWER, when all null hypotheses are true. Thus, when all null hypotheses are true, FDR control (E(Q) ≤ α) implies FWER control (FWER≤ α). 2. (a) By definition of conditional probability, P (H2 |H1 ) = P (H1 H2 ) . P (H1 ) Now note that P (H1 H2 ) = P (H1 H2 |A1 A2 )P (A1 A2 ) + P (H1 H2 |A1 B2 )P (A1 B2 ) +P (H1 H2 |B1 A2 )P (B1 A2 ) + P (H1 H2 |B1 B2 )P (B1 B2 ) = 0.52 ∗ 0.7 ∗ 0.9 + 0.5 ∗ 1 ∗ 0.7 ∗ 0.1 + 1 ∗ .5 ∗ 0.3 ∗ 0.6 + 1 ∗ 1 ∗ 0.3 ∗ 0.4 = 0.4025 and P (H1 ) = P (H1 |A1 )P (A1 ) + P (H1 |B1 )P (B1 ) = 0.5 ∗ 0.7 + 1 ∗ 0.3 = 0.65. Thus, P (H2 |H1 ) = 0.4025/0.65 = 0.6192308. (b) It is most likely that coin A was used for both flips. See the relevant calculations below. P (A1 A2 |H1 H2 ) = P (H1 H2 |A1 A2 )P (A1 A2 )/0.4025 = 0.52 ∗ 0.7 ∗ 0.9/0.4025 = 0.3913043 P (B1 A2 |H1 H2 ) = P (H1 H2 |B1 A2 )P (B1 A2 )/0.4025 = 1 ∗ .5 ∗ 0.3 ∗ 0.6/0.4025 = 0.2236025 P (A1 B2 |H1 H2 ) = P (H1 H2 |A1 B2 )P (A1 B2 )/0.4025 = 0.5 ∗ 1 ∗ 0.7 ∗ 0.1/0.4025 = 0.08695652 P (B1 B2 |H1 H2 ) = P (H1 H2 |B1 B2 )P (B1 B2 )/0.4025 = 1 ∗ 1 ∗ 0.3 ∗ 0.4/0.4025 = 0.2981366 3. #Density of the absolute value of a noncentral #t random variable with df degrees of freedom #and noncentrality parameter ncp. f.abs.t=function(t,df,ncp=0){ 1 (t>0)*(dt(t,df=df,ncp=ncp)+dt(-t,df=df,ncp=ncp)) } #Density of the p-value from a t-test #when the noncentrality parameter is ncp #and the degrees of freedom are df. f=function(p,df,ncp=0) { tt=qt(1-p/2,df=df) f.abs.t(tt,df=df,ncp=ncp)/f.abs.t(tt,df=df) } f(.1,18,1.6) [1] 2.072947 4. (a) NR = 2.12 + 2.23 + 4.42 + 2.14 = 10.91 N − NH = 10 − 4 = 6 i Ordered Stat Phit (S, i) Pmiss (S, i) Phit (S, i) − Pmiss (S, i) 1 -4.42 4.42/10.91 0 4.42/10.91 2 -2.14 6.56/10.91 0 6.56/10.91 3 -2.12 8.68/10.91 0 8.68/10.91 4 -0.25 8.68/10.91 1/6 8.68/10.91-1/6 5 -0.17 8.68/10.91 1/3 8.68/10.91-1/3 6 -0.14 8.68/10.91 1/2 8.68/10.91-1/2 7 0.26 8.68/10.91 2/3 8.68/10.91-2/3 8 0.36 8.68/10.91 5/6 8.68/10.91-5/6 9 0.60 8.68/10.91 1 8.68/10.91-1 10 2.23 1 1 0 [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] [10,] Phit 0.4051329 0.6012832 0.7956004 0.7956004 0.7956004 0.7956004 0.7956004 0.7956004 0.7956004 1.0000000 Pmiss Phit-Pmiss 0.0000000 0.40513291 0.0000000 0.60128323 0.0000000 0.79560037 0.1666667 0.62893370 0.3333333 0.46226703 0.5000000 0.29560037 0.6666667 0.12893370 0.8333333 -0.03773297 1.0000000 -0.20439963 1.0000000 0.00000000 (b) M (S) = 0.7956, m(S) = −0.2044 (c) ES(S) = 0.7956 (d) getES=function(r,S,p=1,plt=F) { N=length(r) NH=length(S) inS=(1:N)%in%S 2 Pmiss=(!inS)/(N-NH) Phit=inS*abs(r)ˆp/sum(abs(r[inS])ˆp) ordr=order(r) Pmiss=cumsum(Pmiss[ordr]) Phit=cumsum(Phit[ordr]) Phit.Pmiss=Phit-Pmiss mM=range(Phit.Pmiss) Mgenm=(mM[2]>=-mM[1]) ES=mM[2]*Mgenm+mM[1]*(1-Mgenm) if(plt){ plot(1:N,Phit.Pmiss) lines(1:N,Phit.Pmiss) } ES } (e) n=10000 es=rep(0,n) for(i in 1:n){ es[i]=getES(rnorm(1000),1:100) } hist(es) 3 1500 1000 0 500 Frequency 2000 Histogram of es −0.4 −0.2 0.0 0.2 0.4 es (f) No simulation is actually necessary here. It is not difficult to see that M (S) and m(S) are guaranteed to be 0.5 and -0.5, respectively. (g) Yes. An enrichment score of 0.5 or -0.5 is more extreme (farther from 0) than the statistics in part (e). 4