Web Appendix for:
A principal stratification approach for evaluating natural direct and indirect effects in
the presence of treatment-induced intermediate confounding
Short title: Evaluating NDEs and NIEs in the presence of intermediate confounding
Masataka Taguri1* and Yasutaka Chiba2
1
Department of Biostatistics and Epidemiology, Graduate School of Medicine, Yokohama
City University, Yokohama, Japan
2
Division of Biostatistics, Clinical Research Center, Kinki University School of Medicine,
Osaka, Japan
*Corresponding Author.
Tel: +81-45-253-9903; Fax: +81-45-253-9902; E-mail: [email protected]
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Web Appendices
In Web Appendices 1–3, we mainly discuss the case of NIE(0|c). A similar
discussion holds for NIE(1|c).
Web Appendix 1. Proofs of Result 1 and 3
Appendix 1.1. Proof of Result 1
By A1, there are relationships of potential outcome expectations between subgroups
defined by observed (A, M) and the principal strata as follows [15]:
E[Y (a, m) | A  0, M  0, c] 
 1|c
 3|c
E[Y (a, m) | S  1, c] 
E[Y (a, m) | S  3, c], (A.1)
 1|c   3|c
 1|c   3|c
E[Y (a, m) | A  1, M  0, c] 
 3|c
 4|c
E[Y (a, m) | S  3, c] 
E[Y (a, m) | S  4, c], (A.2)
 3|c   4|c
 3|c   4|c
Taking the differences between (π1|c + π3|c) × (A.1) and (π3|c + π4|c) × (A.2) yields
1|c E[Y (a, m) | S  1, c]   4|c E[Y (a, m) | S  4, c]
 (1|c   3|c ) E[Y (a, m) | A  0, M  0, c]  ( 3|c   4|c ) E[Y (a, m) | A  1, M  0, c].
(A.3)
Using (A.3) with (a,m) = (0, 1) and (0, 0), we obtain
NIE(0 | c)
  1|c E[Y (0,1)  Y (0, 0) | S  1, c]   4|c E[Y (0,1)  Y (0, 0) | S  4, c]
 ( 1|c   3|c ) E[Y (0,1)  Y (0, 0) | A  0, M  0, c]
 ( 3|c   4|c ) E[Y (0,1)  Y (0, 0) | A  1, M  0, c].
Therefore, under A2, the bias formula NIE(0|c) – Q(0|c) can be expressed as
NIE(0 | c)  Q(0 | c)
 NIE(0 | c)  ( 1|c   4|c ) E[Y (0,1)  Y (0, 0) | A  0, M  0, c]
 ( 3|c   4|c ) (0 | c)
 (1  p1c ) PIE (0 | c).
This completes the proof.
Appendix 1.2. Proof of Result 3
We first note that Q (0|c) in (6) with (a,m) = (0, 0) and (p1c –p0c)E[Y(0,1) – Y(0,0) | A = 0, M =
0, c] are no longer identical if A2 is violated. In this case, E[Y(0,1) | A = 0, M = 0, c] is
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expressed as follows:
E[Y (0,1) | A  0, M  0, c]
  l E[Y (0,1) | A  0, M  0, l , c]p(l | A  0, M  0, c)
  l E[Y (0,1) | A  0, M  1, l , c]  η(0 | l, c)} p(l | A  0, M  0, c)
  l E[Y | A  0, M  1, l , c]  η(0 | l , c)} p(l | A  0, M  0, c).
Therefore, (p1c –p0c)E[Y(0,1) – Y(0,0) | A = 0, M = 0, c] is given by
( p1c  p0c ) E[Y (0,1)  Y (0, 0) | A  0, M  0, c]
 Q(0 | c)  ( p1c  p0c )l ηPIE (0 | l , c) p(l | A  0, M  0, c).
(A.4)
From the proof of Result 1, the following equation holds under A1:
NIE(0 | c)  ( p1c  p0c ) E[Y (0,1)  Y (0,0) | A  0, M  0, c]  (1  p1c )δ(0 | c).
(A.5)
Substituting the right-hand side of (A.4) into (A.5) yields the result.
Web Appendix 2. Bias formula for the usual identification formula (8)
For the derivation, we first derive another expression of E[Y(a, m)| c] as follows:
E[Y (a, m) | c]
  m* 0 Pr[ M (0)  m* | c] E[Y ( a, m) | M (0)  m* , c]
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  m* 0 Pr[ M  m* | A  0, c] E[Y (a, m) | A  0, M  m* , c]
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(A.6)
 (1  p0 c ) E[Y (a, m) | A  0, M  0, c]  p0 c E[Y (a, m) | A  0, M  1, c].
Here, the second equality follows from the consistency assumption and A1. Using (A.3),
(A.6), and (8), the bias formula of NIE(0|c) for the usual identification formula, namely
NIE(0|c) – {(8) with a = 0}, is given as follows:
(1  p0c )(1  p1c  p0c ) A00c  p0c ( p1c  p0c ) A01c  (1  p1c ) A10c ,
(A.7)
where Aamc = E[Y(0,1) – Y(0,0)| A = a, M = m, c]. (A.7) is equal to zero if (i) (1 – p1c) = 0 and
p0c = 0, or (ii) the mediator effects among the three subgroups defined by (A, M) are the same
(A00c = A01c = A10c). Conditions (i) and (ii) are also sufficient for NIE(0|c) – Q (0|c) = 0. If (1 –
p1c) = 0 but p0c ≠ 0, then by substituting (1 – p1c) = 0 into (A.7), we obtain (A.7) = p0c(1 –
p0c)(A00c – A01c). Since (1 – p1c) = 0 implies that π3|c = π4|c = 0, “A00c = A01c” is equivalent to
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“B1c = B2c,” where Bsc = E[Y(0,1) – Y(0,0)| S = s, c]. Thus, unless A00c – A01c = 0, in other
words mediator effects with a = 0 between S = 1 (compliant intermediates) and S = 2 (always
intermediates) are the same, (A.7) is still not equal to zero.
Likewise, NIE(1|c) – {(8) with a = 1} is given as follows:
p1c D11c  p0c D10c  (1  p0c )( p1c  p0c )D00c  p0c ( p1c  p0c )D01c ,
(A.8)
where Damc = E[Y(1,1) – Y(1,0)| A = a, M = m, c]. If p0c = 0 but (1 – p1c) ≠ 0, then substituting
p0c = 0 into (A.8) yields p1c(D11c – D00c). Since p0c = 0 implies that π2|c = π4|c = 0, “D11c = D00c”
is equivalent to “E1c = E3c,” where Esc = E[Y(1,1) – Y(1,0)| S = s, c]. Thus, unless D00c – D11c
= 0, in other words mediator effects with a = 1 between S = 1 (compliant intermediates) and S
= 3 (never intermediates) are the same, (A.8) is still not equal to zero.
Web Appendix 3. Proof of Result 2
For derivation of the bounds of (1 – p1c)δ(0|c) under A1 and A2, we first derive those
of E[Y(0,m) | A = 1, M = 0, c] for m =(0,1). Note that under A1,
E[Y (0, m) | c]  (1  p1c ) E[Y (0, m) | A  1, M  0, c]  p1c E[Y (0, m) | A  1, M  1, c],
which implies
E[Y (0, m) | A  1, M  0, c] 
E[Y (0, m) | c]  p1c E[Y (0, m) | A  1, M  1, c]
.
1  p1c
(A.9)
By substituting 0 or 1 into E[Y(0,m) | A = 1, M = 1, c] in (A.9), we obtain
 E[Y (0, m) | c]  p1c 
 E[Y (0, m) | c] 
max 0,
  E[Y (0, m) | A  1, M  0, c]  min 1,
.
1  p1c
1  p1c




(A.10)
Using (A.10), it follows that
max{0, E[Y (0,1) | c]  p1c }  min{1  p1c , E[Y (0, 0) | c]}
 (1  p1c ){E[Y (0,1) | A  1, M  0, c]  E[Y (0, 0) | A  1, M  0, c]}
 min{1  p1c , E[Y (0,1) | c]}  max{0, E[Y (0, 0) | c]  p1c } ,
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(A.11)
By subtracting (1 – p1c) E[Y(0,1) – Y(0,0) | A = 0, M = 0, c] = (1 – p1c) Q(0|c)/ (p1c – p0c) from
both sides of (A.11), we prove the result.
Next, we will prove that these bounds can be improved if we also assume A3. Note
that under A3,
p1c E[Y (0, m) | A  1, M  1, c]
 2|c
  1|c

 p1c 
E[Y (0, m) | S  1, c] 
E[Y (0, m) | S  2, c]
 1|c   2|c
  1|c   2|c

 ( p1c  p0 c ) E[Y (0, m) | S  1, c]  p0 c E[Y (0, m) | S  2, c],
(A.12)
and
E[Y (0, m) | S  2, c]  E[Y (0, m) | A  0, M  1, c].
(A.13)
Using (A.12) and (A.13), (A.9) can be expressed as follows:
E[Y (0, m) | A  1, M  0, c]
E[Y (0, m) | c]  p0 c E[Y (0, m) | A  0, M  1, c]  ( p1c  p0 c ) E[Y (0, m) | S  1, c]
1  p1c
(1  p0 c ) E[Y (0, m) | A  0, M  0, c]  ( p1c  p0 c ) E[Y (0, m) | S  1, c]

,
1  p1c

(A.14)
where the second equality follows from (A.6) with a = 0. Note that the only quantity we
cannot identify in (A.14) is E[Y(0,m) | S = 1, c]. By using almost the same arguments on
bounds that we used under A1 and A2 above, we can obtain the result.
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