STK4900/ Lecture 4 Program. Counterfactuals and causal effects. Example (cf. practical exercise 10)

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1 STK4900/ Leture 4 Program 1. Counterfatuals and ausal effets 2. Confoundng 3. Interaton 4. More on ANOVA Setons 4.1, 4.4, 4.6 Supplementary materal on ANOVA Example (f. pratal exerse 10) How does exerse affet blood gluose level? Use the HERS data, dsregardng women wth dabetes Smple lnear regresson: 0 1 Estmate Std. Error t value Pr(> t ) (Interept) < 2e-16 exerse Resdual standard error: on 2030 degrees of freedom Multple R-squared: , Adjusted R-squared: F-statst: on 1 and 2030 DF, p-value: Can we onlude that exerse on average dereases the blood gluose level wth 1.7 mg/dl? 2 Problem: The women who exerse are not a random sample of all women n the ohort (as they would have been n a lnal tral), but dffer from the women who don't exerse, e.g. wth respet to age, alohol use, and body mass ndex (BMI) Gluose Further age, alohol use, and BMI may nfluene the gluose level Illustraton for BMI: Gluose Counterfatuals and ausal effets For the general dsusson we onsder some outome (e.g. gluose level) and we want to see how ths s affeted by a bnary predtor, or "exposure", X 1 (e.g. exerse) wth X 1 =1 orrespondng to "exposed" and X 1 =0 orrespondng to "unexposed" Suppose (ounter to the fat) that we ould run an experment n whh frst every ndvdual s exposed (.e. X 1 =1) and the outome Y 1 s observed then, turnng bak the lok, every ndvdual s unexposed (.e. X 1 =0) and the outome Y 0 s observed BMI Exerse=0 Exerse=1 Consderng ths problem, an anythng be sad about the "ausal effet" of exerse on blood gluose level? 3 All other haratersts of the ndvduals are assumed to be the same n the two parts of the hypothetal experment 4

2 In real lfe, we an not turn bak the lok, so one of the two expermental outomes for every ndvdual s an unobserved ounterfatual The ausal effet (n a statstal sense ) of the exposure s defned as the dfferene n populaton means under the two parts of the ounterfatual experment: Causal effet = E( Y ) E( Y ) 1 0 If the means dffer, we say that the exposure s a ausal determnant of the outome A smple model for the ounterfatual experment To make the argument smple, we assume that all other haratersts of the ndvduals are aptured by a bnary ovarate X 2 whh also has a ausal effet on the outome Further we assume that the (ounterfatual) outome for ndvdual when exposed take the form y = β + β + β x + ε whle when unexposed t beomes y = β + β x + ε Then the populaton means for the two parts of the ounterfatual experment beome exposed: E( Y ) = E( β + β + β X + ε ) = β + β + β E( X ) unexposed: E( Y ) = E( β + β X + ε ) = β + β E( X ) In the ounterfatual experment the dstrbuton of X 2 s the same n both parts of the experment, and hene ts mean s the same Hene the ausal effet of the exposure beomes Causal effet = E( Y ) E( Y ) 1 0 { E X } = β + β + β E( X ) β + β ( ) = β Confoundng In realty we annot observe the ounterfatuals We an only observe the outome for an ndvdual under one of the two ondtons (exposed/unexposed) In prate we therefore have to ompare the mean values of the outome n two dstnt populatons, one exposed and one unexposed But then there s no guarantee that the mean value of X 2 wll be the same n the exposed and unexposed populatons Let E1( X 2) denote the mean of X 2 among the exposed, and let E ( X ) denote the mean of X 2 among the unexposed 0 2 8

3 For the exposed populaton: Thus E( Y ) = β + β + β E ( X ) For the unexposed populaton: E( Y ) = β + β E ( X ) { } E( Y ) E( Y ) = β + β + β E ( X ) β + β E ( X ) { E ( X ) E ( X )} = β + β If we perform a study where we sample from the exposed and unexposed populatons, and estmate the dfferene based on the exposed and unexposed samples, we wll estmate { E ( X ) E ( X )} β + β If the mean value of X 2 dffers between the exposed and unexposed populatons, we wll get a based estmate of the ausal effet β We say that the (ausal) effet of X 1 s onfounded by X No onfoundng If the dstrbuton of X 2 s ndependent of the level of exposure (.e. X 1 = 0,1), then E1( X 2) = E0( X 2) and there wll be no onfoundng In partular ths wll be the ase n an experment where ndvduals are randomly alloated to exposure/no exposure Condtons for onfoundng A ovarate X 2 s a onfounder for the ausal effet of X 1 provded that X 2 s a ausal determnant of the outome Y (or a proxy for suh determnants) X 2 s a ausal determnant of X 1 (or they share a ommon ausal determnant) 10 Confoundng patterns Examples of onfoundng patterns when X 2 s a numeral ovarate Control of onfoundng Consder the stuaton where all ausal determnants other than X 1 are aptured by the bnary ovarate X 2 Complete onfoundng Then, gven the level of X 2 (= 0,1), there s no more onfoundng and the ausal effet of X 1 may estmated by omparng the means of exposed and unexposed wthn levels of X 2 In prate ths s obtaned by fttng the lnear model y = β + β x + β x + ε Negatve onfoundng β 1 sne here s the effet of one unt's nrease n X 1 the value of X 2 onstant keepng Fg. 4.1 n the book 11 In general we may use multple lnear regresson to orret for a number of onfounders by nludng them as ovarates n the model (assumng that all relevant onfounders are reorded n the data) 12

4 Example (ontd) We ft a multple regresson model wth blood gluose level as response and exerse, age, alohol use, and body mass ndex (BMI) as ovarates Multple lnear regresson: Estmate Std. Error t value Pr(> t ) (Interept) <2e-16 exerse age drnkany BMI <2e-16 Interaton for bnary ovarates We have onsdered the stuaton where two bnary predtors X 1 and X 2 have a ausal effet on the outome We ould then estmate the (ausal) effets by fttng the lnear model y = β + β x + β x + ε Note that we assume that the effet of X 1 s the same for both levels of X 2 (and ve versa): Resdual standard error: on 2023 degrees of freedom (4 observatons deleted due to mssngness) Multple R-squared: 0.072, Adjusted R-squared: F-statst: on 4 and 2023 DF, p-value: < 2.2e-16 We now fnd that exerse on average dereases the blood gluose level wth 1.0 mg/dl Ths should be loser to the ausal effet of exerse 13 X X E( y x) β 0 β + β 0 1 β + β 0 2 β + β + β If the effet of X 1 depends on the level of X 2 we have an nteraton We may then ft a model of the form y = β + β x + β x + β x x + ε The effet for dfferent values of the ovarates are then gven by: X X X X E( y x) β 0 β + β 0 1 β + β 0 2 β + β + β + β Example Use the HERS data to study how low-densty lpoproten holesterol after one year (LDL1) depends on hormone therapy (HT) and statn use (both bnary) R ommands: ht.ft=lm(ldl1~ht+statns+ht:statns, data=hers) summary(ht.ft) R output (edted): Estmate Std. Error t value Pr(> t ) (Interept) < 2e-16 HT < 2e-16 statns e-10 HT:statns (In the model formula HT:statn spefes the nteraton term "HT*statn") The effet of HT seems to be lower among statn users 15 16

5 Estmate Std. Error t value Pr(> t ) (Interept) < 2e-16 HT < 2e-16 statns e-10 HT:statns HT redues LDL holesterol for non-users of statns by 17.7 mg/dl For users of statns the estmated reduton s = 11.5 mg/dl To obtan the unertanty, we use the "ontrast" lbrary R ommands: lbrary(ontrast) par1= lst(ht=1,statns=1) # spefy one set of values of the ovarates par2= lst(ht=0,statns=1) # spefy another set of values of the ovarates ontrast(ht.ft, par1,par2) # ompute the dfferene between the two sets R output (edted): Contrast S.E. Lower Upper t df Pr(> t ) Interaton for one bnary and one numeral ovarate We now onsder the stuaton where X 1 s a bnary predtor and X 2 s numeral As an llustraton we onsder the HERS data, and we wll see how baselne LDL holesterol depends on statn use ( X 1 ) and BMI ( X 2 ) The model y = β + β x + β x + ε assumes that the effet of BMI s the same for statn users and those who don't use statns It may be of nterest to onsder a model where the effet of BMI may dffer between statn users and those who don't use statns,.e. where there s an nteraton 18 We then onsder the model y = β + β x + β x + β x x + ε Note that the model may be wrtten β0 + β2x2 + ε when x1 = 0 y β0 + β1 + ( β2 + β3) x2 + ε when x1 = 1 Ths s a model wth dfferent nterepts and dfferent slopes for the numeral ovarate dependng on the value of the bnary ovarate When onsderng suh a model, t s useful to enter the numer ovarate (by subtratng ts mean) to ease nterpretaton In the example, we let X 2 orrespond to the entered BMI-values, denoted BMI R ommands: hers$bmi=hers$bmi - mean(hers$bmi[!s.na(hers$bmi)]) stat.ft=lm(ldl~statns+bmi+statns:bmi,data=hers) summary(stat.ft) par1=lst(statns=1,bmi=1) par2=lst(statns=1,bmi=0) ontrast(stat.ft,par1,par2) R output (edted): Estmate Std. Error t value Pr(> t ) (Interept) < 2e-16 statns < 2e-16 BMI e-05 statns:bmi Contrast S.E. Lower Upper t df Pr(> t )

6 Interaton for two numeral ovarates We fnally onsder the stuaton where X 1 and X 2 are both numeral A model wth nteraton s then gven by y = β + β x + β x + β x x + ε For suh a model, t s useful to enter the ovarates But even then the nterpretaton of the estmates s a bt omplated Two-way ANOVA Consder the stuaton where the outome y for an ndvdual depends on two fators, A and B, eah wth two levels, denoted a 1, a 2 and b 1, b 2 One suh example s how LDL holesterol depends on HT (wth levels "plaebo" and "hormone therapy") and statn use (wth levels "no" and "yes"); f. slde 16 We may here ntrodue the ovarates: x 1 0 f ndvd has level a for fator A (referene) 1 f ndvd has level a for fator A 1 2 x 2 0 f ndvd has level b for fator B (referene) 1 f ndvd has level b for fator B Then a regresson model wth nteraton takes the form (f slde 15) y = β + β x + β x + β x x + ε If (e.g.) fator B has three levels b 1, b 2, b 3, we need to ntrodue two x's for ths fator (f slde 26 of Leture 3): x 2 x 3 1 f ndvd has level b for fator B 0 otherwse 1 f ndvd has level b for fator B 0 otherwse A model wth nteraton then takes the form y x x x x x x x = β 0 + β1 1 + β β β β ε (*) It beomes qute omplated to wrte the model lke ths, so t s ommon to use an alternatve formulaton We reaptulate: y x x x x x x x In order to rewrte model (*), we denote the outomes for level a j of fator A and level b k of fator B by y for = 1,..., n jk We may then rewrte model (*) as jk y = µ + α + β + ( αβ ) + ε (**) jk j k jk jk We have the followng relatons between the parameters n model (*) and model (**) (*) = β 0 + β1 1 + β β β β ε (*) β β β β β β (**) µ α β β ( αβ ) ( αβ ) In model (**) the parameters for the referene levels are 0 : α = β = ( αβ ) = ( αβ ) = ( αβ ) = ( αβ ) =

7 Note that the model formulaton y = µ + α + β + ( αβ ) + ε (**) jk j k jk jk works equally well when fator A has J levels and fator B has K levels, whle the formulaton (*) would beome muh more omplated In Leture 3 (f. slde 30), we onsdered a study of how the extraton rate of a ertan polymer depends on temperature and the amount of atalyst used. We there assumed a lnear effet of temperature and the amount of atalyst We wll here onsder temperature and atalyst as fators, eah wth three levels 25 R ommands: polymer=read.table(" polymer$ftemp=fator(polymer$temp) polymer$fat=fator(polymer$at) ft=lm(rate~ftemp+fat+ftemp:fat,data=polymer) summary(ft) R output: Estmate Std. Error t value Pr(> t ) (Interept) e-10 ftemp ftemp fat fat e-06 ftemp60:fat ftemp70:fat ftemp60:fat ftemp70:fat Resdual standard error: 1.73 on 9 degrees of freedom Multple R-squared: 0.986, Adjusted R-squared: F-statst: on 8 and 9 DF, p-value: 2.012e In a planned experment we an make sure that we have the same number of observatons for all the J x K ombnatons of levels of fator A and fator B We then have a balaned desgn, and the total sum of squares (TSS) may be unquely deomposed as a sum of squares for eah of the two fators (SSA, SSB), a sum of squares for nteraton (SSAB), and a resdual sum of squares (RSS): TSS = SSA + SSB + SSAB + RSS The result of a two-way ANOVA may be summarzed n the table Soure df Sum of Mean sum F statsts squares of squares Fator A J 1 SSA SSA/( J 1) SSA/( J 1) F = RSS /( n JK) Fator B K 1 SSB SSB /( K 1) SSB /( K 1) F = RSS /( n JK) Interaton ( J 1)( K 1) SSAB SSAB /( J 1)( K 1) SSAB /( J 1)( K 1) F = RSS /( n JK) Resdual n JK RSS RSS /( n JK) Total n 1 TSS To eah of these sum of squares there orrespond a degree of freedom as gven n the ANOVA table on the next slde NB! If the desgn s not balaned, the deomposton of the total sum of squares s not unque 27 The F-statsts (wth ther approprate degrees of freedom) may be used to test the followng null hypotheses: H0 : all ( αβ ) jk = 0 (no nteraton) H0 : all α j = 0 (no man effet of A) H0 : all β k = 0 (no man effet of B) 28

8 For the example: R ommands: anova(ft) Hgher level ANOVA Consder for llustraton the stuaton wth three fators, A, B, and C. R output: Analyss of Varane Table Df Sum Sq Mean Sq F value Pr(>F) ftemp e-06 fat e-08 ftemp:fat Resduals Data: y jkl = observaton number for level a j of fator A, level b of fator B, and level of fator C Model wth nteraton: k y = µ + α + β + γ + ( αβ ) + ( αγ ) + ( βγ ) + ( αβγ ) + ε jkl j k l jk jl kl jkl jkl l The result of a three-way ANOVA may be summarzed n the table Soure df * Sum of Mean sum F statsts squares of squares Fator A SSA SSA/ df F Fator B SSB SSB / df F Fator C SSC SSC / df F Interaton AB SSAB SSAB / df F Interaton AC SSAC SSAC / df F Interaton BC SSBC SSBC / df F Interaton ABC SSAB C SSABC / df F ABC Resdual RSS RSS / df Total n 1 TSS *) an be found on omputer output The deomposton of the total sum of squares s unque f the desgn s balaned A B C AB AC BC Hypothess testng s smlar to two-way ANOVA 31