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**|**Int J Biostat**|**PMC2854089

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- Abstract
- Introduction
- Proof of Claims in ORR-I
- Confidence Set for xopt (z) when is Finite and h· (z, x; β) is Linear in β
- Positivity Assumption: Example
- References

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Int J Biostat. 2010 January 6; 6(2): Article 9.

Published online 2010 March 3. doi: 10.2202/1557-4679.1242

PMCID: PMC2854089

Copyright © 2010 The Berkeley Electronic Press. All rights reserved

This article has been cited by other articles in PMC.

In this companion article to “Dynamic Regime Marginal Structural Mean Models for Estimation of Optimal Dynamic Treatment Regimes, Part I: Main Content” [Orellana, Rotnitzky and Robins (2010), IJB, Vol. 6, Iss. 2, Art. 7] we present (i) proofs of the claims in that paper, (ii) a proposal for the computation of a confidence set for the optimal index when this lies in a finite set, and (iii) an example to aid the interpretation of the positivity assumption.

In this companion article to “Dynamic regime marginal structural mean models for estimation of optimal dynamic treatment regimes. Part I: Main Content” (Orellana, Rotnitzky and Robins, 2010) we present (i) proofs of the claims in that paper, (ii) a proposal for the computation of a confidence set for the optimal index when this lies in a finite set, and (iii) an example to aid the interpretation of the positivity assumption.

The notation, definitions and acronyms are the same as in the companion paper. Througout, we refer to the companion article as ORR-I.

First note that the consistency assumption C implies that the event

is the same as the event

So, with the definitions

we obtain

Next, note that the fact that unless , implies that

Then, it follows from the second to last displayed equality that

So, part 1 of the Lemma is proved if we show that

(1)

Define for any *k* = 0, ..., *K*,

To prove equality (1) first note that,

where the second to last equality follows because given
and
,
is a fixed, i.e. non-random function of
and consequently by the sequential randomization assumption,
is conditionally independent of *A _{K}* given
and
. The last equality follows by the definition of λ

We thus arrive at

This proves the result for the case *k* = *K.* If *k* < *K* – 1, we analyze the conditional expectation of the last equality in a similar fashion. Specifically, following the same steps as in the long sequence of equalities in the second to last display we arrive at

the last equality follows once again from the sequential randomization assumption. This is so because given
and
,
and
are fixed, i.e. deterministic, functions of
and the SR assumption ensures then that
and
are conditionally independent of *A _{K}*

Equality (1) is thus shown by continuing in this fashion recursively for *K* – 2, *K* – 3, ..., *K* – *l* until *l* such that *K* – *l* = *k* – 1.

To show Part 2 of the Lemma, note that specializing part 1 to the case *k* = 0, we obtain

Thus, taking expectations on both sides of the equality in the last display we obtain

This shows part 2 because *B* is an arbitrary Borel set.

Lemma 1, part 2 implies that the densities factors as

In particular, the event has probability 1. Consequently,

Therefore,

(2)

Lemma 1, part 1 implies that

The left hand side of this equality is equal to

and this coincides with the right hand side of (2) which, as we have just argued, is equal to _{k+}_{1} (*ō _{k}*).

Let *X* be the identity random element on
and let *E*_{Pmarg × PX} (·) stand for the expectation operation computed under the product law *P*^{marg} × *P _{X}* for the random vector (

(3)

and the restriction stated in 3) is equivalent to

(4)

To show 2) let *d* (*O*, *A*, *X*) ω* _{K}* (

(ORR-I, (14)) (3).

where the last equality follows because *E*_{Pmarg × PX} [*d* (*O*, *A*, *X*) |*X* = *x*, *Z*] = *E*_{Pmarg} [*d* (*O*, *A*, *x*) |*Z*] by independence of (*O*, *A*) with *X* under the law *P*^{marg} × *P _{X}* and, by assumption,

(3) (ORR-I, (14)). Define *b ^{*}* (

consequently, *E*_{Pmarg × PX} [*d* (*O*, *A*, *X*) |*X*, *Z*] = 0 with *P*^{marg} × *P _{X}* prob. 1 which is equivalent to (ORR-I, (14)) because

To show 3) redefine *d* (*O*, *A*, *X*) as ω* _{K}* (

(ORR-1, (15)) (4)

where the third equality follows because *E*_{Pmarg × PX} {*d* (*O*, *A*, *X*) |*X* = *x*, *Z*} = *E*_{Pmarg} {*d* (*O*, *A*, *x*) |*Z*} and *E*_{Pmarg} {*d* (*O*, *A*, *x*) |*Z*}= *q* (*Z*) μ-a.e.(*x*) and hence *P _{X}*-a.e.(

(4) (ORR-I, (15)). Define *b*^{*} (*X; Z*) = *E _{P × PX}* [

Consequently, *b*^{*} (*X*, *Z*) = *E*_{Pmarg × PX} [*b*^{*} (*X*, *Z*) |*Z*] *q* (*Z*) *P _{X}* –

Any element

of the set Λ is the sum of *K* + 1 uncorrelated terms because for any *l*, *j* such that 0 ≤ *l* < *l* + *j* ≤ *K* + 1,

Thus, Λ is equal to Λ_{0} Λ_{1} . . . Λ* _{K}* where

and stands for the direct sum operator. Then,

and it can be easily checked that Π [*Q*|Λ* _{k}*] =

Applying formula (26, in ORR-I) we obtain

So, for *k* = 0, ..., *K*,

But,

So formula ((27), ORR-I) is proved if we show that

(5)

This follows immediately from the preceding proof of Result (b) of Section 3.2. Specifically, it was shown there that

Consequently, the left hand side of (5) is equal to

where the last equality follows by the definition of
and the fact that
(as this is just the function
resulting from applying the integration to the utility *u* (*O*, *A*) = 1).

It suffices to show that where

But by definition

where the last equality follows because

Write for short, _{·} (*b*) _{·} (*b*, * _{·, opt}*),

We will show that *J*_{·} (*b*) = *E* {*Q*_{·} (*b*) *Q*_{·} (*b _{·, opt}*)

For · = *sem* and with the definitions (*x*, *Z*) *b* (*x*, *Z*) – (*Z*) and * _{sem}* (

Now, with *var _{A}* (

and consequently

Thus, 0 ≤ *var _{A}* (

We first prove the assertion that the computation of the confidence set *B _{b}* entails an algorithm for determining if the intersection of
half spaces in

(6)

Define the *p* × 1 vector
whose *j ^{th}* entry is equal to

we conclude that the condition in the display (6) is equivalent to

The set
is a hyper-plane in * ^{p}* which divides the Euclidean space

Turn now to the construction of a confidence set
that includes *B _{b}*. Our construction relies on the following Lemma.

**Lemma.** Let

where **u**_{0} is a fixed *p* × 1 real valued vector and Σ is a fixed non-singular *p* × *p* matrix.

Let **α** be a fixed, non-null, *p*×1 real valued vector. Let τ_{0} **α′ u**_{0} and α^{*} = Σ^{1/2}**α**. Assume that α_{1} ≠ 0. Let,
be the *p*×1 vector
. Let be the linear space generated by the *p*×1 vectors
,
and define

where

Then there exists satisfying

if and only if

**Proof**

Then, with τ_{0} −**α′ u _{0}** and

Now, by the assumption
we have −**α* ^{′} u*** = τ

where and are defined in the statement of the lemma. The vector is the residual from the (Euclidean) projection of into the space .

Thus, −**α* ^{′} u*** τ

Therefore, since is unrestricted,

if and only if

(7)

This concludes the proof of the Lemma.

To construct the set we note that the condition in the display (6) implies the negation, for every subset of , of the statement

(8)

Thus, suppose that for a given *x _{l}* we find that (8) holds for some subset
of
, then we know that

(9)

holds. If (9) does not hold, then we conclude that (8) is false for this choice of
. If (9) holds, then we conclude that (8) is true and we then exclude *x _{l}* from the set
.

Suppose that *K* = 1 and that
with probability 1 for *k* = 0, 1, so that no subject dies in neither the actual world nor in the hypothetical world in which *g* is enforced in the population. Thus, for *k* = 0, 1, *O _{k}* =

Assume that

(10)

Assumption PO imposes two requirements,

(11)

(12)

Because by definition of regime *g*,
, then requirement (11) can be re-expressed as

Since indicators can only take the values 0 or 1 and
, *l*_{0} = 0, 1 (by assumption (10)), the preceding equality is equivalent to

that is to say,

By the definition of λ_{0} (·|·) (see (3) in ORR-I), the last display is equivalent to

(13)

Likewise, because , and because by the fact that , requirement (12) can be re-expressed as

or equivalently, (again because the events and have the same probability by ,

Under the assumption (10), the last display is equivalent to

which, by the definition of λ_{0} (·|·, ·, ·) in ((3), ORR-I), is, in turn, the same as

(14)

We conclude that in this example, the assumption PO is equivalent to the conditions (13) and (14). We will now analyze what these conditions encode.

Condition (13) encodes two requirements:

- i) the requirement that in the actual world there exist subjects with
*L*_{0}= 1 and*L*_{0}= 0 (i.e. that the conditioning events*L*_{0}= 1 and*L*_{0}= 0 have positive probabilities), for otherwise at least one of the conditional probabilities in (13) would not be defined, and - ii) the requirement that in the actual world there be subjects with
*L*_{0}= 0 that take treatment*A*_{0}= 1 and subjects with*L*_{0}= 1 that take treatment*A*_{0}= 0, for otherwise at least one of the conditional probabilities in (13) would be 0.

Condition i) is automatically satisfied, i.e. it does not impose a restriction on the law of *L*_{0}, by the fact that
(since baseline covariates cannot be affected by interventions taking place after baseline) and the fact that we have assumed that
, *l*_{0} = 0, 1.

Condition ii) is indeed a non-trivial requirement and coincides with the interpretation of the PO assumption given in section 3.1 for the case *k* = 0. Specifically, in the world in which *g* were to be implemented there would exist subjects with *L*_{0} = 0. In such world the subjects with *L*_{0} = 0 would take treatment
, then the PO assumption for *k* = 0 requires that in the actual world there also be subjects with *L*_{0} = 0 that at time 0 take treatment *A*_{0} = 1. Likewise the PO condition also requires that for *k* = 0 the same be true with 0 and 1 reversed in the right hand side of each of the equalities of the preceding sentence. A key point is that (11) does not require that in the observational world there be subjects with *L*_{0} = 0 that take *A*_{0} = 0, nor subjects with *L*_{0} = 1 that take *A*_{1} = 1. The intuition is clear. If we want to learn from data collected in the actual (observational) world what would happen in the hypothetical world in which everybody obeyed regime *g*, we must observe people in the study that obeyed the treatment at every level of *L*_{0} for otherwise if, say, nobody in the actual world with *L*_{0} = 0 obeyed regime *g* there would be no way to learn what the distribution of the outcomes for subjects in that stratum would be if *g* were enforced. However, we don t care that there be subjects with *L*_{0} = 0 that do not obey *g*, i.e. that take *A*_{0} = 0, because data from those subjects are not informative about the distribution of outcomes when *g* is enforced.

Condition (14) encodes two requirements:

- iii) the requirement that in the actual world there be subjects in the four strata (
*L*_{0}= 0,*L*_{1}= 0,*A*_{0}= 1), (*L*_{0}= 0,*L*_{1}= 1,*A*_{0}= 1), (*L*_{0}= 1,*L*_{1}= 0,*A*_{0}= 0) and (*L*_{0}= 1,*L*_{1}= 1,*A*_{0}= 0) (i.e. that the conditioning events in the display (14) have positive probabilities), for otherwise at least one of the conditional probabilities would not be defined, and - iv) the requirement that in the actual world there be subjects in every one of the strata (
*L*_{0}= 0,*L*_{1}= 0,*A*_{0}= 1), (*L*_{0}= 0,*L*_{1}= 1,*A*_{0}= 1), (*L*_{0}= 1,*L*_{1}= 1,*A*_{0}= 0) that have*A*_{1}= 0 at time 1 and the requirement that there be subjects in stratum (*L*_{0}= 1,*L*_{1}= 0,*A*_{0}= 0) that have*A*_{1}= 1 at time 1, for otherwise at least one of the conditional probabilities in (14) would be 0.

Given condition ii) and the sequential randomization (SR) and consistency (C) assumptions, condition iii) is automatically satisfied, i.e. it does not impose a further restriction on the joint distribution of (*L*_{0}, *L*_{1}, *A*_{0}). To see this, first note that by condition (ii) the strata (*L*_{0} = 0, *A*_{0} = 1) and (*L*_{0} = 1, *A*_{0} = 0) are non-empty. So condition (iii) is satisfied provided

But

and
by (10). An analogous argument shows that
. Finally, condition (iv) is a formalization our interpretation of assumption PO in section 3.1 for *k* = 1. In the world in which *g* was implemented there would exist subjects that would have all four combination of values for
. However, subjects with
will only have
, so in this hypothetical world we will see at time 1 only four possible recorded histories,
,
,
and
. In this hypothetical world subjects with any of the first three possible recorded histories will take
and subjects with the last one will take
. Thus, in the actual world we must require that there be subjects in each of the first three strata (*L*_{0} = 0, *L*_{1} = 0, *A*_{0} = 1), (*L*_{0} = 0, *L*_{1} = 1, *A*_{0} = 1), (*L*_{0} = 1, *L*_{1} = 0, *A*_{0} = 0) that take *A*_{1} = 0 and subjects in the stratum (*L*_{0} = 1, *L*_{1} = 1, *A*_{0} = 0) that take *A*_{1} = 1. A point of note is that we don t make any requirement about the existence of subjects in strata other than the four mentioned in (iii) or about the treatment that subjects in these remaining strata take. The reason is that subjects that are not in the four strata of condition (iii) have already violated regime *g* at time 0 so they are uninformative about the outcome distribution under regime *g*.

^{*}This work was supported by NIH grant R01 GM48704.

- Orellana L, Rotnitzky A, Robins JM. 2010. Dynamic regime marginal structural mean models for estimation of optimal dynamic treatment regimes, Part I: Main content The International Journal of Biostatistics 62Article 7. [PubMed]

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