Appendix for Causal Interaction in Factorial Experiments: Application to Conjoint Analysis

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1 A Appendx for Causal Interacton n Factoral Experments: Applcaton to Conjont Analyss Mathematcal Appendx: Proofs of Theorems A. Lemmas Below, we descrbe all the lemmas, whch are used to prove the man theorems of ths paper. For completeness, ther proofs appear n the supplementary appendx. Lemma (An Alternatve Defnton of the K-way Average Interacton Effect) The K-way average nteracton e ect (AIE) of treatment combnaton T :K t :K (t,...,t K ) relatve to baselne condton T :K t :K 0 (t 0,...,t 0K ), gven n Defnton 3, can be rewrtten as, :(K ) :(K ) :K (t :K ; t :K :(K ) (t :(K ) ; t0 T K t K ) :(K ) (t :(K ) ; t0 T K t 0K ). Lemma 2 Under Assumpton 2, for any k,...,k, the followng equalty holds, KK ( T K k df ( T K k ) F K k k ( )` F K k \K` t K` df ( T K k\k`). Lemma 3 (Decomposton of the K-way AIE) The K-way Average Treatment Interacton E ect (AIE) (Defnton 3), can be decomposed nto the sum of the K- way condtonal Average Treatment Combnaton E ects (ACEs). Formally, let K k K K {,...,K} wth K k k where k,...,k. Then, the K-way AIE can be wrtten as follows, KK (t K K K ( ) K k Kk (t Kk ; t Kk, where the second summaton s taken over the set of all possble K k and the k-way condtonal ACE s defned as, Kk (t K k E apple { (t K k K 0,T K ) (t K k 0 K 0,T K )}df (T K K ). F K K Lemma 4 (Decomposton of the K-way AMIE) The K-way Average Margnal Treatment Interacton E ect (AMIE), defned n Defnton 2, can be decomposed nto the sum of the K-way Average Treatment Combnaton E ects (ACEs). Formally, let {,...,K} wth K k k where k,...,k. Then, the K-way AMIE can be wrtten as follows, KK (t K K K ( ) K k Kk (t Kk ; t Kk, where the second summaton s taken over the set of all possble K k.

2 A.2 Proof of Theorem We use proof by nducton. Under Assumpton 2, we frst show for K 2. Tosmplfy the notaton, we do not wrte out the J 2factorsthatwemargnalzeout. Webegn by decomposng the AME as follows, A(a l,a E{ (a`,b ) (a 0,B )} df (B ) B E{ (a`,b (a 0,b } E{ (a`,b ) (a 0,B ) (a`,b (a 0,b } df (B ) B E{ (a`,b (a 0,b } AB (a`,b ; a 0,b df (B ). B Smlarly, we have B (b m,b E{ (a 0,b m ) (a 0,b } R A AB(A,b m ; a 0,b df (A ). Gven the defnton of the AMIE n equaton (5), we have, AB (a`,b m,a 0,b E{ (a`,b m ) (a 0,b } A (a`,a B (b m,b AB (a`,b m ; a 0,b AB (a`,b ; a 0,b df (B ) AB (A,b m ; a 0,b df (A ). B Ths proves that the AMIE s a lnear functon of the AIEs. We next show that the AIE s also a lnear functon of the AMIEs. AB (a`,b m ; a 0,b E{ (a`,b m ) (a 0,b } A (a`,a A (b m,b A E{ (a`,b (a 0,b } A (a`,a E{ (a 0,b m ) (a 0,b } A (b m,b AB (a`,b m ; a 0,b AB (a`,b 0 ; a 0,b AB (a 0,b m ; a 0,b. Thus, we obtan the desred results for K 2. Now we show that f the theorem holds for any K wth K 2, t also holds for K. Frst, usng Lemma 2, we rewrte the equaton of nterest as follows, K KK (t K K KK (t K K ( ) k k ( )` F K k \K` t K` df ( T K k\k`). Utlzng the the defnton of the K-way AMIE gven n Defnton 2 and the assumpton that the theorem holds for K, wehave, KK (t K K KK (t K K KK (t K K " K k Kk (t K k ( ) m m ( )` K` K m F Km\K` m K m K k K Kk (t Kk ; t Kk Kk \K m (t K k\k m, t K k\k m Kk \K m (t K k\k m, t K k\k m 0 T Km\K`, T K` t K` df ( T Km\K`) (7) #. 2

3 After rearrangng equaton (7), the coe cent for KK \K u (t K K\K u, t K K\K u s equal to ( ) u. Smlarly, the coe cent of the followng term s equal to ( ) uv. KK \K u (t K K\K u, t K K\Ku F Ku\Kv 0 T Ku\Kv, T Kv t Kv df ( T Ku\Kv ). Therefore, we can rewrte equaton (7) as follows, KK (t K K KK (t K K k ( )` K ( ) k KK (t K K k ( )` KK (t K K F K k \K` K ( ) k F K k \K` (t K K\K k, t K K\K k (t K K\K k, t K K\K k t K` df ( T K k\k`) (t K K\K k, t K K\K k (t K K\K k, t K K\K k t K` df ( T K k\k`) K ( ) k (T K k, t K K\K k df (T K k ), where the second equalty follows from applyng Lemma to KK (t K K and the fnal equalty from Lemma 2. Ths proves that the K-way AMIE s a lnear functon of the K-way AIEs. We next prove that the K-way AIE can be wrtten as a lnear functon of the K-way AMIEs. We wll show ths by mathematcal nducton. We already show the desred result holds for K 2. ChooseanyK 2andassumethatthefollowng equalty holds, KK (t K K K ( ) K k KK (t Kk, t KK\Kk 0 0. Usng the defnton of the K-way AIE gven n Lemma, we have KK (t K K KK (t K K 0 T K t K ) KK (t K K 0 T K t K K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0,t K ) K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0 0,t K, where the second equalty follows from the assumpton. Let us consder the followng decomposton. K ( ) K k KK (t Kk, t KK\Kk 0 3

4 K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0,t K ( ) K KK (t K K 0,t K K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0 0,t K, (8) 0,t K where the frst and second terms together represent the cases wth K 2K k,whle the thrd term corresponds to the cases wth K 2K K \K k. Note that these two cases are mutually exclusve and exhaustve. Fnally, note the followng equalty, K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0,t K ) K ( ) K k KK (t Kk, t KK\Kk 0,t K 0 0,t K ( ) K KK (t K K 0,t K (9) 0,t K. Then, together wth equatons (8) and (9), we obtan, KK (t K K K ( ) K k KK (t Kk, t KK\Kk 0. Thus, the desred lnear relatonshp holds for any K 2. 2 A.3 Proof of Theorem 2 To prove the nvarance of the K-way AMIE, note that Lemma 4 mples, KK (t; t KK ( t; t KK (t; t KK ( t; t K ( ) K k Kk (t Kk ; t Kk ). (20) K ( ) K k Kk (t Kk ; t Kk ). (2) Thus, the K-way AMIE s nterval nvarant. To prove the lack of nvarance of the K-way AIE, note that accordng to Lemma 3, we can rewrte equaton () as follows. K ( ) K k K ( ) K k Kk (t K k Kk ( t K k Kk (t K k t K K\K k Kk ( t K k t K K\K k. It s clear that ths equalty does not hold n general because the K-way condtonal ACEs are condtoned on d erent treatment values. Thus, the K-way AIE s not nterval nvarant. 2 4

5 A.4 Proof of Theorem 3 We use L to denote the objectve functon n equaton (2). Snce t s a convex optmzaton problem, t has one unque soluton and the soluton should satsfy the K k t K k 0 forallj, and ` 2{0,,...,L j }, 0, for all j 6 j 0,`2{0,,...,L j } and m 2{0,,...,L j 0 }, 0 for all t K k, and K k K J such that k 3. (22) For the sake of smplcty, we ntroduce the followng notaton. n S(t K k ) {; T K k t K k }, N t K k {T K k t K k }, Ê[ T K k t K k ] N t K k 2S(t K k ). K J t K J t K J 2S(t K k ) 0forallt K J, 2 µ J j J L j k3 K k K J t K k j ` {T j `} J L j j j 0 >j K k t K k {TK k t K k } L j 0 m0 jj 0 `m {T j `, T j 0 m} 0. (23) Therefore, for all t K J, ˆµ J K k K J t K k K k t K k {tk k t K J } b E[ T K J t K J ]. For the frst-order e ect, we can use the weghted zero-sum constrants for all factors except for the j th factor. In partcular, for all j and t j` 2 t K J, j 0 6j j 0 6j L j 0 L j 0 () ˆj ` j 0 6j Pr(T j0 `) ˆµ t j 0`2t K J \j J K k K J t K k K k t K k {tk k 2 t K J } t j 0`2t K J \j Pr(T j0 `) b E[ T j `, T KJ \j t K J \j ] L j 0 In general, for all t K k, Kk K J and k 2, t j 0`2t K J \j Pr(T j0 `) b E[ T j `, T KJ \j t K J \j ] ˆµ. ˆKk t K k j 0 2K J \K k L j 0 t j 0`2t K J \K k Pr(T j 0 `) b E[ T K k t K k, T K J \K k t K J \K k ] 5

6 K pk k t Kp In addton, ˆµ s gven as follows. {t Kp t Kk } ˆKp t Kp ˆµ. (24) ˆµ K j L j t j`2t K J Pr(T j `) b E[ T K J t K J ]. Therefore, (ˆµ, ˆ) sunquelydetermned. Toconfrmthssolutonsthemnmzer of the optmzaton problem, we check all the equalty condtons. For all t K k, Kk K J,j 2K k and k, 0, L j L j Pr(T j `){t j `} ˆKk t K k Pr(T j `){t j `} j 0 2{j,K J \K k } L j 0 L j j 0 2K J \K k L j 0 t j 0`2t K J \K k Pr(T j `){t j `} K pk k t Kp Pr(T j 0 `) b E[ T K k {t Kp t Kk } ˆKp t Kp ˆµ t K k, T K J \K k t K J \K k ] Pr(T j0 `) E[ b T Kk\j t Kk\j, T {j,k J \K k } t {j,k J \K k } ] t j 0`2t {j,k J \K k } {t Kp t Kk\j } ˆKp ˆµ t Kp K p K k \j t Kp where the fnal equalty comes from equaton (24) for ˆKk\j. t K k \j Furthermore, equaton (23) mples all other equaltes n equaton (22). Therefore, the soluton (equatons (24) and (25)) satsfes all the equalty condtons. Fnally, we show that these estmators are unbased for the AMEs and the AMIEs. Snce be[ T K J t K J ]sanunbasedestmatorofe[ (t K J )], E( ˆKk t K k E( ˆKk t K k ) ˆKk t K k 0 ) j 0 2K J \K k j 0 2K J \K k L j 0 L j 0 Kk (t K k. t j 0`2t K J \K k K pk k t Kp t j 0`2t K J \K k Pr(T j 0 `) E[ (t K k, t K J \K k )] K pk k t Kp {t Kp t Kk }E[ ˆKp t Kp ] ˆµ, Pr(T j 0 `) E[ (t K k, t K J \K k ) {t Kp t Kk }E[ ˆKp t Kp ˆKp t Kp 0 ] (t K k 0, t K J \K k )] 2 6

7 B Supplementary Appendx: Proofs of Lemmas For the sake of completeness, we prove all the lemmas used n the mathematcal appendx above. B. Proof of Lemma To smplfy the proof, we start from Lemma and prove t s equvalent to Defnton 3. We prove t by nducton. Equaton (7) shows ths correspondence holds for K 2. Next, choose any K 2andassumethatthsrelatonshpholds. Thats,weassume the followng equalty, KK (t; t KK (t; t K, (25) where the second summaton s taken over all possble {,...,K} wth K k k. Usng the defnton of the K-way AIE n Lemma, we have, K K {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k 0 T,K t K, T K K\K k K 0 T,K t 0,K, T K K\K k, (26) where {Kk,K}(t K k,tk 0,t 0,K T K K\K k denotethecondtonal(k )-way AIE that ncludes the set of k treatments, K k, as well as the (K )th treatment whle fxng T K K\K k to t K K\K k 0.Therefore,wehave, KK (t K K KK (t K K 0 T,K t K ) KK (t K K 0 T,K t 0,K ) KK (t K K,t K 0,t K ) KK (t K K,t 0,K 0,t 0,K ) K {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k KK (t K K,t K 0,t K ) K K {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k t K K\K k, (27) where the second equalty follows from equaton (26), and the thrd equalty s based on the applcaton of the assumpton gven n equaton (25) whle condtonng on T,K t 0,K. 7

8 Next, consder the followng decomposton, K K K t K K\K k t K K\K k {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k (K) (t K ; t 0,K T K K t K K, (28) where the frst term corresponds to the cases wth K 2K K \K k,whlethe second and thrd terms together represent the cases wth K 2K k. Note that these two cases are mutually exclusve and exhaustve. Fnally, note the followng equalty, KK (t K K,t K 0,t K ) KK (t K K (K) (t K ; t 0,K T K K Then, together wth equatons (27) and (28), we obtan, the desred result, t K K. KK (t K K KK (t K K K t K K\K k. Thus, the lemma holds for any K 2. 2 B.2 Proof of Lemma 2 To begn, we prove the followng equalty by mathematcal nducton. KK ( T K k 0 T K k ) k ( )` (t KK\Kk, t KK\Kk Frst, t s clear that ths equalty holds when k. Thats,foragvenK,wehave, KK ( T K, t K K\K KK \K (t K K\K \K 0 T K t K ) KK \K (t K K\K \K 0 T K t K. (30) Now, assume that the equalty holds for k. Wthout loss of generalty, we suppose K k {, 2,...,k} and K k {, 2,...,k,k }. By the defnton of the K-way AIE, KK ( T K k KK \(k)( T K k \(k) 0 \K k 0 T K k) (29) t K`. T k ) KK \(k)( T K k \(k) 0 T k t k 8

9 k ( )` (t KK\Kk, t KK\Kk \K k K k,t k 0 T t k k ( )` (t KK\Kk, t KK\Kk where the second equalty follows from the assumpton. Next, consder the followng decomposton. t K` t K` 0,T k t k, (3) k ( )` k ( )` (t KK\Kk, t KK\Kk (t KK\Kk, t KK\Kk \K k K k,t k 0 T t k k ( )` (t KK\Kk, t KK\Kk t K` t K` t K` 0,T k t k, (32) where the frst term corresponds to the case n whch K` K k n the left sde of the equaton does not nclude the (k )thtreatment, andthesecondandthrdterms jontly express the case n whch K` K k n the left sde of the equaton does nclude the (k )thtreatment. Puttng together equatons (3) and (32), we have, KK ( T K k 0 T K k ) k ( )` (t KK\Kk, t KK\Kk t K`. Therefore, equaton (29) holds n general. Fnally, under Assumpton 2, KK ( T K k F K df ( T K k ) k 0 T K k )df ( T K k ) F K k k ( )` F K k k ( )` F K k \K` F K` t K` df ( T K k ) t K` df ( T K` T K k\k`)df ( T K k\k`) 9

10 k ( )` F K k \K` t K` df ( T K k\k`). Ths completes the proof of Lemma 2. 2 B.3 Proof of Lemma 3 We prove the lemma by nducton. For K 2,equaton(7)showsthatthelemma holds. Choose any K 2andassumethatthelemmaholdsforallk wth apple k apple K. Then, KK (t K K 0 T,K t K ) K KK (t K K,t K 0,t K ) ( ) K k Kk (t Kk,t K ; t Kk 0,t K T K K\K k KK (t K K,t K 0,t K ) K apple ( ) K k {Kk,K}(t K k,t K 0,t 0,K T K K\K k K (t K ; t 0,K T K K\ (K) KK (t K K,t K 0,t K ) K ( ) K k {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k K K ( ) K k k Next, note the followng decomposton, t K K\ (K) K (t K ; t 0,K T K K\ (K) t K K\ (K). (33) KK (t K K KK (t K K 0 T,K t K ) KK (t K K ; t K L 0 T,K t 0,K ) KK (t K K 0 T,K t K ) K ( ) K k Kk (t Kk ; t Kk t K K\K k. Substtutng equaton (33) nto ths equaton, we obtan KK (t K K KK (t K K,t K 0,t K ) K ( ) K k Kk (t Kk ; t Kk K ( ) K k {Kk,K}(t Kk,t K ; t Kk 0,t 0,K T K K\K k t K K\K k 0

11 K ( ) K k k K K K ( ) K k k K (t K ; t 0,K T K K\ (K) t K K\ (K) K KK (t K K,t K 0,t K ) ( ) K k Kk (t Kk ; t Kk k2 ( ) K K (t K,t K 0 T K K\ K t K L\ K K K K K (t K ; t 0,K T K K\ (K) t K K\ (K) KK (t K K K (t K ; t 0,K T K K\ (K) K ( ) K k Kk (t Kk ; t Kk k2 ( ) K K (t K,t K 0 T K K\ K t K K\ K K K K K K ( ) K k k t K Kk\K k t K K\ (K) K (t K ; t 0,K T K K\ (K) t K K\ (K) K ( ) K k Kk (t Kk ; t Kk t K K\K k, t K K\K k where the fnal equalty follows because K K ( ) K k k ( ) K. Thus, by nducton, the theorem holds for any K 2. 2 B.4 Proof of Lemma 4 We prove the lemma by nducton. For K 2,equaton(7)showsthstheoremholds. Choose any K 2andassumethatthelemmaholdsforallk wth apple k apple K. That s, let {,...,K} wth K k k where k,...,k,andassumethe followng equalty, Kk (t K k k ( ) k ` K`(t K`; t K`. Usng ths assumpton as well as the defnton of the K-way AMIE gven n Defnton 2, we have, KK (t K K KK (t K K KK (t K K K K Kk (t Kk ; t Kk k ( ) k ` K`(t K`; t K`. (34)

12 Next, we determne the coe cent for Km (t Km ; t Km nthesecondtermofequaton (34) for each m wth apple m apple K. Note that Km (t Km ; t Km would not appear n ths term f m>k.thats,foragvenm, weonlyneedtoconsderthecaseswhere the ndex for the frst summaton satsfes m apple k apple K. Furthermore, for any gven K m such k, thereexst ways to choose K k m k n the second summaton such that K m K k. Once such K k s selected, K m appears only once n the thrd and fourth summatons together and s multpled by ( ) k m. Therefore, the coe cent for Km (t Km ; t Km sequalto, K K m ( ) k m k m km Puttng all of these together, KK (t K K KK (t K K ( ) K m. K ( ) K k K ( ) K k Kk (t Kk ; t Kk. Kk (t Kk ; t Kk Snce the theorem holds for K,wehaveshownthattholdsforanyK