OPTIMAL DESIGNS FOR TWO-LEVEL FACTORIAL EXPERIMENTS WITH BINARY RESPONSE

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1 Statistica Sinica: Sulement OPTIMAL DESIGNS FOR TWO-LEVEL FACTORIAL EXPERIMENTS WITH BINARY RESPONSE Jie Yang 1, Abhyuday Mandal, and Dibyen Majumdar 1 1 University of Illinois at Chicago and University of Georgia Sulementary Material 1 Proof of Lemma 311 Suose v 1 > v Let 1,, 3, 4 be a solution maximizing 3 If 1 >, then L = 1 v 4 3 +v v 1 +v 1 1 v 4 3 +v v v 1, S11 where 1 = = 1 + / The = in S11 is true if and only if 3 = 4 = 0 which is imossible for a solution Therefore, 1,, 3, 4 is strictly better than the solution 1,, 3, 4 This contradiction guarantees 1 Note that 1 = is ossible even if v 1 > v For examle, v 1, v, v 3, v 4 =, 1, 1, 5 leads to the solution 1,, 3, 4 = 1/3, 1/3, 1/3, 0 Suose v 1 = v We also have L 1 v v v 1 + v 1 Since 3 = 4 = 0 is imossible here, the = is true if and only if 1 = More Results for Section 31 in terms of β under Logit Link Write r = e β0, s = e β1, and t = e β Recall that v i = 1/w i, i = 1,, 3, 4 Then v rst, v t + rs, v 3 s + rt, v 4 r + st 1 Pairwise equal cases of v i s or w i s equivalently 1 v 1 = v if and only if β = 0 or β 0 + β 1 = 0; v 1 = v 3 if and only if β 1 = 0 or β 0 + β = 0; 3 v 1 = v 4 if and only if β 0 = 0 or β 1 + β = 0;

2 S JIE YANG, ABHYUDAY MANDAL AND DIBYEN MAJUMDAR 4 v = v 3 if and only if β 0 = 0 or β 1 = β ; 5 v = v 4 if and only if β 1 = 0 or β 0 = β ; 6 v 3 = v 4 if and only if β = 0 or β 0 = β 1 Trilet-wise equal cases v i s 1 v 1 = v = v 3 v 4 if and only if β 0 = β 1 = β 0; v 1 = v = v 4 v 3 if and only if β 0 = β 1 = β 0; 3 v 1 = v 3 = v 4 v if and only if β 0 = β 1 = β 0; 4 v = v 3 = v 4 v 1 if and only if β 0 = β 1 = β 0 3 v 1 = v = v 3 = v 4 if and only if at least two of β 0, β 1, β are zeros 4 Pairwise nonequal cases of v i s 1 v 1 > v if and only if β β 0 + β 1 > 0; v 1 > v 3 if and only if β 1 β 0 + β > 0; 3 v 1 > v 4 if and only if β 0 β 1 + β > 0; 4 v > v 3 if and only if β 0 β 1 β > 0; 5 v > v 4 if and only if β 1 β 0 β > 0; 6 v 3 > v 4 if and only if β β 0 β 1 > 0 For the other direction of inequalities, for examle, v 1 < v, just change the the direction of the inequality, ie β β 0 + β 1 < 0 Note that the airwise equal cases, for examle, v 1 = v if and only if β β 0 + β 1 = 0 5 β s version of conditions in Corollary v = v 3 = v 4 = v and v 1 < 3v if and only if either two of β 0, β 1, β are zeros in this case v 1 = v = v 3 = v 4 = 4 or β 0 = β 1 = β and β 0 < log ; v 1 = v 3 = v 4 = v and v < 3v if and only if either two of β 0, β 1, β are zeros in this case v 1 = v = v 3 = v 4 = 4 or β 0 = β 1 = β and β 0 < log ;

3 OPTIMAL DESIGN FOR BINARY RESPONSE S3 3 v 1 = v = v 4 = v and v 3 < 3v if and only if either two of β 0, β 1, β are zeros in this case v 1 = v = v 3 = v 4 = 4 or β 0 = β 1 = β and β 0 < log ; 4 v 1 = v = v 3 = v and v 4 < 3v if and only if either two of β 0, β 1, β are zeros in this case v 1 = v = v 3 = v 4 = 4 or β 0 = β 1 = β and β 0 < log ; 6 β s version of conditions in Corollary 31 1 v 1 = v = u, v 3 = v 4 = v and u > v if and only if β = 0 and β 0 β 1 > 0; v 1 = v 3 = u, v = v 4 = v and u > v if and only if β 1 = 0 and β 0 β > 0; 3 v 1 = v 4 = u, v = v 3 = v and u > v if and only if β 0 = 0 and β 1 β > 0 For the other direction of the inequalities, for examle, v 1 = v = u, v 3 = v 4 = v and u < v, just change the the direction of the corresonding inequality, ie β = 0 and β 0 β 1 < 0 7 β s version of conditions in Theorem 31 v 1 > v, v 3 = v 4, and v 1 < v + v 3 if and only if one of the following cases is satisfied: i log e β 0 +e 4β 0 1 e β 0 +e 4β 0 ii β < 0, if 1 log 1 < β 0 = β 1 < 0; < β < 0, if β 0 = β 1 1 log ; iii β > 0, if 0 < β 0 = β 1 1 log ; iv 0 < β < log e β 0 +e 4β 0 1 e β 0 +e 4β 0, if β 0 = β 1 > 1 log + 1 Note: i v 1 < v, v 3 = v 4 thus v 1 < v + v 3 if and only if β 0 β < 0 and β 0 = β 1 e ii v 1 > v, v 3 = v 4, and v 1 v + v 3 if and only if either β log β 0 +e 4β 0 1 e β 0 +e 4β 0 and β 0 = β 1 1 log 1 or β log and β 0 = β 1 > 1 log + 1 e β 0 +e 4β 0 1 e β 0 +e 4β 0 iii For cases similar to Theorem 31, we only need to ermute β 0, β 1, β to get similar results as above For examle, to get the β version of v > v 3, v 1 = v 4, and v < v 1 + v 3 + v 4, we only need to relace β 0, β 1, β with β 1, β, β 0, resectively 3 Proof of Theorem 33 Again, let r = e β 0, s = e β 1, and t = e β If β 1 = 0, then s = 1, which imlies v 1 = v 3 and v = w 4 Similarly, β = 0 imlies v 1 = v and v 3 = v 4 ; β 0 = 0 imlies v 1 = v 4 and v = v 3 The saturation condition cannot be true in those cases Under logit link, the saturation condition can be summarized into four cases as follows:

4 S4 JIE YANG, ABHYUDAY MANDAL AND DIBYEN MAJUMDAR 1 v 1 v + v 3 + v 4 if and only if s 1 + t < and 0 < r a 1 ; or 1 1 s 1 t > and r 1 a 3 v v 1 + v 3 + v 4 if and only if s 1 t > and r 1 a 5 ; or 1 s 1 + t > and 0 < r a 7 3 v 3 v 1 + v + v 4 if and only if s 1 t > and 0 < r a 5 ; or 3 1 s 1 + t > and r 1 a 7 4 v 4 v 1 + v + v 3 if and only if s 1 + t < and r 1 a 1 ; or 4 1 s 1 t > and 0 < r a 3 where a 1 = st + 1 s 4 1 t 4 1 s 1 t, a 3 = st + 1 s 4 1 t s 1 + t a 5 = st + 1 s 4 1 t 4 1 s 1 + t, a 7 = st + 1 s 4 1 t s 1 t Combining cases 1 and 4, it can be verified that β 0 loga 1 guarantees the saturation condition when 1 + s 1 + t < On the other hand, 1 s 1 t > in cases 1 and 4 imlies β > Λβ 1, where Λβ 1 = 1 e β log 1 + 1, e β1 1 while 1 + s 1 t > in cases and 3 imlies β < Λβ 1 Regarding a 3 as a function of β 1 and β, we can write a 3 = a 3 β 1, β and a 5 = a 3 β 1, β So cases 1, 1, 31, and 4 can be aggregated into β > Λβ 1 and β 0 log a 3 β 1, β Similarly, the other four cases can be aggregated into β > Λ β 1 and β 0 log a 3 β 1, β Thus the saturation condition is equivalent to β 1 0, β > Λ β 1, and β 0 log a 3 β 1, β That is, β 1 0, β > 1 e β log e β1 1 and e β 0 log e β1 + β + 4 β 1 1 e 4 β 1 e β 1 1 e β 1

5 OPTIMAL DESIGN FOR BINARY RESPONSE S5 It should be noted that a 1 3 = st + s s 4 1t 4 1/ 1t 1 By symmetry of β 0, β 1, β with resect to the saturation condition, the result of Theorem 33 can be obtained by ermuting β 1, β, β 0 above into β 0, β 1, β 4 Proof of Corollary 33 By Theorem 33, the robability of saturated design is 0 if and only if the = in the condition of β is true assuming β 0 = β 1 = β = logx The equivalent condition is x + 1x 4 x 3 x + 1 = 0 The only feasible solution is x = 5 Proof of Corollary 34 By Theorem 33, the condition of saturated design is determined by the = condition of β Let x = e β 1 = e β and A = e a An equivalent condition is A + 1A 1 x 3 A 13A + 4A + 3x + A x + A + 1A 1 = 0 The only feasible solution is x = 1 + A A /A 1, which leads to the conclusion 6 Proof of Theorem 344 First show that L 4 L 3 if v + v 4 v 3, while L 4 L 34 if v + v 4 > v 3 Let the solution to L 4 be x = 1, y = 3, z = = 4 If v + v 4 v 3, then 3 and L 3 L 4 v 4 v 3 xzz y x=1,y= 3,z= 0 If v + v 4 > v 3, then 3 and L 34 L 4 v 3 v xzy z x=1,y= 3,z= 0 Similarly, we can show that 1 L 13 L 3 if v 1 +v 3 v ; L 13 L 1 if v 1 +v 3 < v ; 3 L 14 L 13 if v 1 + v 4 v 3 ; and 4 L 14 L 4 if v 1 + v 4 v In conclusion, max{l 13, L 14, L 4 } max{l 1, L 3, L 34 } The original roblem is max L = max {v v v v } Its aroximation L 34 = max {v v v v }, where v 4 = v 3 = v 3 + v 4 / The difference max L L 34 v 4 v /, if 1,, 3, 4 maximizes L Therefore, max L max{l 1, L 3, L 34 } { } = min max L L 1, max L L 3, max L L 34 { v v 1 min 3 4 1, v 3 v 1 4 3, v } 4 v

6 S6 JIE YANG, ABHYUDAY MANDAL AND DIBYEN MAJUMDAR { v v 1 min 16, v 3 v 96, v } 4 v 3 54 Note that and = 1 7 Proof of Corollary 415 The results here can be obtained from Theorem 415 by letting b or θ = b/a go to Case i is straightforward For case ii, given v c4 < v c1 + v c + v c3, R c max = 1 min{q c b, a, a, a, Q c b, b, a, a, Q c b, b, b, a} 1/3 The exlicit exression of Q c b, a, a, a, Q c b, b, a, a, and Q c b, b, b, a can be found in the roof of Theorem 416 Let c1, c, c3, c4 be the D-otimal design with resect to v c1, v c, v c3, v c4 v c1 v c v c3 v c4 < v c1 + v c + v c3 imlies that c1 c c3 c4 > 0 As b goes to, Q c b, a, a, a 7 c c3 c4, Q c b, b, a, a 7 c c3 c4 + c1 c3 c4, and Q c b, b, b, a 81 c c3 c4 + c1 c3 c4 + c1 c c4 /4 The limit of Q c b, a, a, a is smaller than the other two, which leads to R c max 8 Proof of Theorem 517 Let θ = b/a 1 and ρ = θ θ + 1 Then Q 1 := Qb, a, a, a = { θ, if θ 3 Qθ, 1, 1, 1 = θ+39 θ 4, if 1 θ < 3 Q := Qb, b, a, a = 16θ + 1θ 1 Qθ, θ, 1, 1 = θ 1 ρθ + ρθ ρ 3θ + 19θ 1 Q 3 := Qb, b, b, a = Qθ, θ, θ, 1 = 4θ 3 Since θ 1, it can be verified that Q 3 Q and the = is true if and only if θ = 1, that is, a = b Thus min{qb, b, b, a, Qb, b, a, a, Qb, a, a, a} = min{qb, b, a, a, Qb, a, a, a} Now we only need to comare Q 1 = Qb, a, a, a with Q = Qb, b, a, a It can be verified that i If θ = 1, or θ = θ, Q 1 = Q ii If 1 < θ < θ, Q 1 > Q iii If θ > θ, then Q 1 < Q Here θ 13 is the 3rd root of equation θ θ + 596θ θ θ 5 + θ 6 = 0 Then R u max can be obtained accordingly