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S d 6@ #" `^`  F*  d 6F #" `^   H*  d 6J #" `  H* V d s *h޽h ?#" @@ MMM̙80___PPT10.ݿ 8,  Clouds  0 1)0h(  h h BX ?#" `  T Click to edit Master title style! ! h 08 " `    W#Click to edit Master subtitle style$ $ h 6 #" `^`  F*  h 6  #" `^   H*  h 6P #" `^   H* V h s *h޽h ?#" @@ MMM̙80___PPT10.ݿ 8, 0 TLD(   D D N Ƙ|@|@ -   x*   e44ee D N ǘ|@|@ 2 -  *   e44eed D c $ ?$   D Nܘ|@|@ H ]  8________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________   D TT|@|@ s   x*   e44ee D T8|@|@ s2   z*   e44eeH D 0ηo~ ? ̙3380___PPT10.bUp,Q  ,( t; 4  C x|@|@ Ԕ?-   t*    e44ee>  C xD!|@|@ Ԕ?2 -  ~*   e44ee4  C x!|@|@ Ԕ?s   t*    e44ee6  C x!|@|@ Ԕ?s2  ! v*    e44eeH  0ηo~ ? ̙3380___PPT10.bUQ  0  (    0Tb(h R CONFOUNDING EQUIVALENCE ,  0}   JJudea Pearl  UCLA, USA Azaria Paz  Technion, Israel (www.cs.ucla.edu/~judea/).P8nKZP$Z               H  0޽h ? f̙f___PPT10i.:U+D=' = @B +$  0  p!33$(  $ %$ s *"B uCONFOUNDING EQUIVALENCE ,     B &$ N D1?"jnjfl = 2$=,$D 0`B $ 0DԔ999 n2 $ 0"`Y\`B $ 0DԔ?w n2 $ 0"`< [ `B $B 0DԔ n2 $ 0"`  ? n2 $ 0"` Y `B  $B 0DԔ9 n2  $B 0"`YA`B  $B 0DԔ w n2  $B 0"`< B n2  $B 0"`G ? `B $ 0DԔ S n2 $B 0"`G . n2 $ 0"`  . `B $ 0DԔ?] ] `B $B 0DԔE E `B $ 0DԔ  E `B $ 0DԔ Y  $ <"KB PW120   $ <Ș" K  PW420   $ <|"9 | OX20   $ <j9  OY20   $ <o 7 @  PV220   $ <oo @  PV120   $ <@oV `  PW220   $ <[0QBl  3$,$D 0 0$ T ?". BBias = 0 1$ c $A '??"`x 'dH $ 0޽h ? 330(___PPT10.{+ D' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*2$%(D' =-o6Bdissolve*<3<*2$D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*3$%(D' =-o6Bdissolve*<3<*3$+   0 !2(4(  ( ( s *` ` 2CONFOUNDING EQUIVALENCE DEFINITION AND SOLUTION,32   3  B ( N D1?" ( 0_%  \TL___PPT9.& >DEFINITION: T and Z are c-equivalent if DEFINITION (Markov boundary): Markov boundary Sm of S (relative to X) is the minimal subset of S that d-separates X from all other members of S. THEOREM: Z and T are c-equivalent iff Zm=Tm, or Z and T are admissible (i.e., satisfy the back-door condition))0 0 0 J1 0        1>XV@h` ( c $A '??"`x 'dH ( 0޽h ? MMM̙___PPT10i.LP+D=' = @B +!  0 z!*\,(  , , s *47` &ARE Z AND T CONFOUNDING EQUIVALET?^'      '  B 0, N D1?" 1, T7?" O )ANSWER: Yes! Because both are admissible>*0!XB 2, 0DԔ 99wf2 3, 0"` \ XB 4, 0DԔ ?G f2 5, 0"`z[XB 6,@ 0DԔ  ; f2 7, 0"`8  } f2 8, 0"` XB 9,@ 0DԔ wf2 :,@ 0"` A XB ;,@ 0DԔ G f2 <,@ 0"`zBf2 =,@ 0"`8 G } XB >, 0DԔ S ; f2 ?,@ 0"`' G l f2 @, 0"`'  l XB A, 0DԔ?XB B,@ 0DԔW EXB C, 0DԔW XB D, 0DԔK Y K  E, <PB  PW120   F, <P   PW420   G, < Pw| OX20   H, <Pw OY20   I, <ԄP7 7 ~ PV220   J, <+7 o ~ PV120   K, <W V  PW220   L, <W P  PW320   z P  M, P ,$D  0f2 N, 6ԔPP  O, < ?Z"0 ZB P, s *DԔ@ z @   Q, @ ,$D  0f2 R, 6Ԕ@ P  S,B <  ?T"0 ZB T,B s *DԔ [ x5z =  Z,  = ,$D 0 [, <Bh =  | Z ~ TV0      \, <28  ? "0 H , 0޽h ? MMM̙  ___PPT10 .LI+}DD ' = @B D ' = @BA?%,( < +O%,( < +D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*M,%(D' =-o6Bwipe(up)*<3<*M,D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*Q,%(D' =-o6Bwipe(up)*<3<*Q,D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*Z,%(D' =-o6Bdissolve*<3<*Z,+n+  0 !140Y(  0 0 s *9` &ARE Z AND T CONFOUNDING EQUIVALET?^'      '  XB 0 0DԔ 99wf2 0 0"` \ XB 0 0DԔ ?G f2 0 0"`z[XB  0@ 0DԔ  ; f2  0 0"`8  } f2  0 0"` XB  0@ 0DԔ wf2  0@ 0"` A XB 0@ 0DԔ G f2 0@ 0"`zBf2 0@ 0"`8 G } XB 0 0DԔ S ; f2 0@ 0"`' G l f2 0 0"`'  l XB 0 0DԔ?XB 0@ 0DԔW EXB 0 0DԔW XB 0 0DԔK Y K  0 <XB  PW120   0 <d"   PW420   0 <D&w| OX20   0 <t+w OY20   0 <%7 7 ~ PV220   0 <7 o ~ PV120   0 < W V  PW220   0 <DW P  PW320   z P   0 P ,$D  0f2 !0 6ԔPP  "0 < ?Z"0 ZB #0 s *DԔ@ z @   $0 @ ,$D  0f2 %0 6Ԕ@ P  &0B <dg  ?T"0 ZB '0B s *DԔ [ xz 8Y (0 78 ,$@  0n2 )0 0"` 4 y *0 < Y  OL20  fB +0B 6DԔ8X fB ,0 6DԔ hB 00 N D1?" 10 TtT?"@ I ~$ANSWER: No! None is admissible and >%0 20 c $A (??"`0h @x (d2 =  -0#  = ,$D 0 .0 <h =  | Z ~ TV0      /0 <   6"0 B 30@ T Do?"4 8 ,$D  0H 0 0޽h ? MMM̙A9___PPT10.M vG^+g8PD' = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* 0%(D' =-o6Bwipe(up)*<3<* 0D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*$0%(D' =-o6Bwipe(up)*<3<*$0D' =%(D' =%(D3' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*-0%(D' =-o6Bdissolve*<3<*-0D' =%(D' =%(D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*(0%(D' =-o6Bwipe(up)*<3<*(0D3' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*30%(D' =-o6Bwipe(up)*<3<*30+   0 !4[(  4$ 4 HX?"8  The definition of c-equivalence is purely statistical. The graphical criterion we found invokes  admissibility, which is causal  why? Corollary 1: c-equivalence is an invariant property of M-equivalent graphs Corollary 2: There ought to be a graphical criterion that does not invoke  admissibility, based on d-separation only. Indeed, Theorem: Z ~ T iffn0u ) W%$%j 4 c $A '??"`" :rx 'd 4 s *\!` X A PUZZLE:      B 4 N D1?"XPX 4 TY?"_j =and0H 4 0޽h ? MMM̙___PPT10i.N`b+D=' = @B +1  0 00"Uf8 0(  8^B ^8 6DԔOU5OO 8 s *Tz ?EXAMPLE: TESTING IF Z ~ T WITHOUT INVOKING ADMISSIBILITY@'! @  B 8 N D1?"vbv S8 c $A '??"`! x 'd8  b8SfB .8 6DԔ t2 /8 6"`   fB 08 6DԔ H  t2 18 6"` AfB 28B 6DԔc  | t2 38 6"`F r $ t2 48 6"`a  fB 58B 6DԔ t2 68B 6"`  fB 78B 6DԔ  t2 88B 6"`At2 98B 6"` $ fB :8 6DԔq  t2 ;8B 6"` 8 e t2 <8 6"`F 8 r e fB =8 6DԔ))fB >8B 6DԔ W K fB ?8 6DԔ W fB @8 6DԔc O O  A8 B/ O  LW1.0   B8 Bd / O  LW4.0   C8 Bl  KX.0   D8 B쟞  KY.0   E8 B # *0 PV220  F8 B # & 0 PV120  G8 B  $ !  LW2.0   H8 B  ^!  LW3.0  l2 I8 <Ԕ _ ? K J8 B0P  9 9  =Z 0 `B K8 0DԔ  )  l2 L8 <Ԕ ` jK M8B B<  9  =T 0 `B N8B 0DԔ 4y T   O8# ;   P8 <   t Z ~ TN0  Q8 <  ? "0 t2 W8 6"`g b  X8 BȞo N  KL.0  lB Y8B <DԔx lB Z8 <DԔ   \8 c $A '??"`yx 'd^B 8 6DԔ:EEDl2 8 6"`/\G^B  8 6DԔ.I 3l2  8 6"`.5a[^B  8@ 6DԔ: +l2  8 6"`) Vl2  8 6"` G^B 8@ 6DԔ:HHDl2 8@ 6"`1]G^B 8@ 6DԔ. D3l2 8@ 6"`) V^B 8 6DԔ: +l2 8@ 6"`j l2 8 6"`j ^B 8@ 6DԔM L^B 8 6DԔ ?K^B 8 6DԔ  8 B ўaL LW1.0   8 B֞a   LW4.0   8 BٞUZ v b PV220  8 B(ߞUV r b PV120   8 B3 p S LW2.0   !8 B,3V S LW3.0  d2 #8 <Ԕ } $8 BK k =Z 0 XB %8 0DԔ47 u d2 '8 <Ԕ% } (8@ BK0 k =T 0 XB )8@ 0DԔ/B L   *8# m@S) +8 <   t Z ~ TN0  ,8 <   ? "0 l2 ]8@ 6"`.8d[ `8 B P9 KX.0   a8 B / KY.0   d8 c $A '??"`&jx 'dAB e8@ N Do?",B f8@ N DjJ?"6 : H 8 0޽h ? MMM̙___PPT10i.O0n+D=' = @B +B+  0 Y*Q* "3A< )(  <8 ~ ~ 0< : B < Z DԔ?"B < Z DԔ?"DB < Z DԔ?"hhB  < Z DԔ?"B  < Z DԔ?"J2  < N ?"|D@  <2  < N ?"x,B  < T DԔ?"B <B T DԔ?"~(2 <B S  CɧENG̍HXJpSQԔ? `TЧ #ɧ`TЧ #ɧpS #ɧpS"zt(2 < S  CɧENG̍HXJpSQԔ? `TЧ #ɧ`TЧ #ɧpS #ɧpS"~ t.2 <B c  CENG>HXJ-TQԔ? `T#`T#-T#-T" < T ?"\| JW1,0  < T ?"L\| JW2,0  < T ?"l JZ1,0  < T ?"l JZ2,0  < T ?"^~ IX,0  < T ?"l^K ~ IY,0  < TTP% ?"\E IV,0 8  W~ ;< : B <  ` DԔ?" < bB <B  ` DԔ?" D bB <  ` DԔ?" hh2  < T ?"D2 "< T ?"B #< Z DԔ?"B$42 '<B s  CENG>HXJ-TQԔ? `T#`T#-T#-T"A  (< Zc% ?" T t JZ1,0  )< ZHg% ?"8TWt JZ2,0  *< Zj% ?"@  JW1,0  +< Z o% ?"P JW2,0  ,< Z,t% ?"& ^ ~ IX,0  -< Z x% ?"^~ IY,0  .< Z|% ?"s IV,0 B 1< Z DԔ?" PB 2<B T DԔ?" $B 3<B T DԔ?" 8 *B 5<B T DԔ?" 8 *B 6<B Z DԔ?"H<VB 7< T DԔ?"6B 8< T DԔ?"V JB 9<B T DԔ?" P :< HЃ% ?"   M(b).0  << H% ?" 0  M(a).0  =< s *P% ~ n APPLICATIONS TO MODEL TESTING!!$ !  B >< N D1?"vbvp ?< T% ?"8   Answer: No, $ a CI that holds in (b) but not in (a) Answer: No, because {Z1,W1,W2} ~ {W1,W2,Z2} holds in (b) but not in (a) Which is easier to detect? Which is easier to test?<0" P A< Tp% ?"  z- a'Are these two models Markov-equivalent?(0(H < 0޽h ?O@<<< '< MMM̙___PPT10i.V+D=' = @B +-  0 --@":A@ F,(  @/F ~ ~ @ V, B @ Z DԔ?"B @ Z DԔ?"DB @ Z DԔ?"hhB @ Z DԔ?"B  @ Z DԔ?"J2  @ N ?"|DN   @ 2  @ N ?"x,B  @ T DԔ?"B @B T DԔ?"~(2 @B S  CɧENG̍HXJpSQԔ? `TЧ #ɧ`TЧ #ɧpS #ɧpS"zt(2 @ S  CɧENG̍HXJpSQԔ? `TЧ #ɧ`TЧ #ɧpS #ɧpS"~ t.2 @B c  CENG>HXJ-TQԔ? `T#`T#-T#-T" @ T(% ?"\| JW1,0  @ T1% ?"L\| JW2,0  @ Tl% ?"l JZ1,0  @ T(% ?"l JZ2,0  @ T% ?"^~ IX,0  @ T% ?"l^K ~ IY,0  @ T% ?"\E IV,0 sF  W~ @ H B @ Z DԔ?" < bB @B Z DԔ?" D bB @ Z DԔ?" hh2 @ N ?"D2 @ N ?"B @ T DԔ?"B$.2  @B c  CENG>HXJ-TQԔ? `T#`T#-T#-T"A  !@ T% ?" T t JZ1,0  "@ T`% ?"8TWt JZ2,0  #@ Td% ?"@  JW1,0  $@ T% ?"P JW2,0  %@ T% ?"& ^ ~ IX,0  &@ Tp% ?"^~ IY,0  '@ T|% ?"s IV,0 B (@ Z DԔ?" PB )@B T DԔ?" $B *@B T DԔ?" 8 *B +@B T DԔ?" 8 *B ,@B Z DԔ?"H<VB -@ T DԔ?"6B .@ T DԔ?"V JB /@B T DԔ?" P 0@ H7?"ZWz  M(b).0  1@ Hp7?"Zz  M(a).0  2@ s *T7 =WHICH IS EASIER TO TEST WHEN Z IS HIGH DIMENSIONAL?>>$    >  B 3@ N D1?"J 5@ TT7?"I x eConditional independence? or c-equivalence? c-equivalence can invoke propensity scores (Scalars): And we can ignore the outcome process05& 6@ c $A '??"`h ^.p x 'd8 \   ;@t \  8@ s *A '??"`\  x 'd 9@ Z7?"@ g  F ~&08 $ A@$ >@ s *A (??"`$x (d ?@ Zϵ?" F ~&0 @@ Tӵ?"D k  F ~&0H @ 0޽h ?O@@@@ @ MMM̙___PPT10i.[+D=' = @B +  0 |P"D(  D D s *lĵ Y CONCLUSIONS  $   B D N D1?"J D T?"vnf___PPT9H@ 3c-equivalence provides: A simple, polynomial time test for deciding whether one set of measurements has the same bias-reducing potential as another. A powerful tool for: Testing from sampled data, the validity of a given graphical model and, eventually, and systematic search for graph structures.0/1S01.1+0(1 /S.+  ( hh\<, H D 0޽h ? MMM̙80___PPT10.I  0 |p"H(  H H s *PW Y CONCLUSIONS  $   B H N D1?"J H Txq?"vnf___PPT9H@ 3c-equivalence provides: A simple, polynomial time test for deciding whether one set of measurements has the same bias-reducing potential as another. A powerful tool for: Testing from sampled data, the validity of a given graphical model and, eventually, and systematic search for graph structures.0/1S01.1+0(1 /S.+  ( hh\<, H H 0޽h ? MMM̙80___PPT10.9$ 0 4(  d  c $D$     s *DH ]    H  0ηo~ ? ̙3380___PPT10. 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