Baseline Covariates in Stabilized Numerators

Why Condition a Stabilized Numerator on Baseline Covariates?

A stabilized treatment weight does not merely make weights smaller. Its numerator helps determine the treatment-assignment distribution in the weighted population. When that numerator conditions on baseline covariates, the weighted population retains the baseline covariate–treatment association, while removing the associations between treatment and measured time-varying confounders that the denominator adjusts for.

This choice is useful when the marginal structural model (MSM) describes the outcome conditional on baseline covariates. It also changes what one should expect from balance diagnostics: baseline covariates need not be balanced across treatment groups after weighting. The introductory material on numerator models and the marginal structural model establishes the basic role of stabilized numerators; here the focus is the population induced by including baseline covariates.

Definition: Let $V$ denote baseline covariates, $A_t$ the treatment at decision time $t$, and $L_t$ time-varying covariates measured before that decision. Let $H_t=(V,\bar L_t,\bar A_{t-1})$ be the measured history available just before $A_t$. A stabilized weight through time $K$, with a baseline-conditioned numerator, is $$ W^s=\prod_{t=0}^{K} \frac{P(A_t\mid \bar A_{t-1},V)} {P(A_t\mid H_t)}. $$ Each probability is evaluated at the treatment actually received. At $t=0$, the past-treatment history is empty. For binary treatment, the probability in each factor is the probability of the observed value of $A_t$, not necessarily the probability of treatment $1$.

The denominator adjusts for the measured treatment history, including time-varying covariates. The numerator preserves a treatment law that can depend on baseline $V$ and past treatment, but not on the current time-varying covariate history. Thus, baseline covariates are deliberately treated differently from time-varying confounders.

What Population Does the Weight Create?

The interpretation is clearest when the probabilities in both parts of the weight are correctly specified and the relevant denominator probabilities are positive. Suppose the observed-data law factors as $$ p(v)\prod_{t=0}^{K} p(l_t\mid v,\bar l_{t-1},\bar a_{t-1}) p(a_t\mid v,\bar l_t,\bar a_{t-1}) \,p(y\mid v,\bar l_K,\bar a_K). $$ Write $q_t(a_t\mid\bar a_{t-1},v)=P(A_t=a_t\mid\bar A_{t-1}=\bar a_{t-1},V=v)$ for the numerator treatment law. Multiplying the observed-data density by the weight cancels each denominator treatment probability.

Result: Under the stated factorization, the density induced by weighting is $$ p(v)\prod_{t=0}^{K} p(l_t\mid v,\bar l_{t-1},\bar a_{t-1}) q_t(a_t\mid\bar a_{t-1},v) \,p(y\mid v,\bar l_K,\bar a_K). $$ This is a probability density when each $q_t$ is a normalized conditional treatment distribution. In this weighted law, treatment at time $t$ depends on baseline covariates and past treatment, but not on the time-varying covariate history available at time $t$, conditional on $V$ and past treatment.

To see the cancellation, take one factor of the original density involving treatment and multiply it by the corresponding weight factor:

p(a_t\mid v,\bar l_t,\bar a_{t-1}) \frac{q_t(a_t\mid\bar a_{t-1},v)} {p(a_t\mid v,\bar l_t,\bar a_{t-1})} =q_t(a_t\mid\bar a_{t-1},v).

Doing this at every time gives the displayed weighted density. Its integral is one: integrate over $y$, then sum over $a_K$ and integrate over $l_K$, proceeding backward through time. Each treatment distribution $q_t$ sums to one, and each covariate or outcome conditional density integrates to one. In particular, integrating over all treatment sequences conditional on $V$ leaves $p(v)$: the weighted law preserves the baseline covariate distribution.

“Preserves $V$” does not mean “balances $V$ across treatment groups.” The numerator allows treatment to remain associated with $V$, so treatment groups can have different baseline covariate distributions in the weighted population. What the factorization removes is the dependence of each treatment decision on the measured time-varying covariate history, conditional on $V$ and past treatment.

Worked Example: Calculating Weights Within Baseline Strata

Let $V$ and a binary time-varying confounder $L$ each have two values, 0 and 1. Suppose each of the four $(V,L)$ cells has probability $0.25$. Treatment probabilities are $0.1$ and $0.5$ for $(V,L)=(0,0)$ and $(0,1)$, respectively, and $0.5$ and $0.9$ for $(1,0)$ and $(1,1)$. Find the numerator treatment probabilities and the weights for treatment $A=1$ when $V=0,L=0$, and treatment $A=0$ when $V=1,L=1$.

Within $V=0$, the two $L$ values are equally prevalent, so $P(A=1\mid V=0)=0.5(0.1)+0.5(0.5)=0.3$. Within $V=1$, they are also equally prevalent, giving $P(A=1\mid V=1)=0.5(0.5)+0.5(0.9)=0.7$.

For $V=0,L=0,A=1$, the numerator is 0.3 and the denominator is 0.1, so the weight is $0.3/0.1=3$. For $V=1,L=1,A=0$, use the probabilities of the observed value $A=0$: the numerator is $1-0.7=0.3$, and the denominator is $1-0.9=0.1$. The weight is $0.3/0.1=3$.

The cancellation can be checked directly. In the first cell, the weighted mass for $A=1$ is $0.25(0.1)(3)=0.075$; for $V=0,L=1,A=1$, it is $0.25(0.5)(0.3/0.5)=0.075$. Thus, among weighted $V=0,A=1$ observations, the two $L$ values have equal mass, as they did in the full $V=0$ stratum.

Connection to a Conditional Marginal Structural Model

A conditional MSM specifies a potential-outcome mean as a function of the treatment history and baseline covariates. For a fixed treatment history $\bar a$, write this mean as $m(\bar a,v;\beta)=E[Y^{\bar a}\mid V=v]$. A treatment effect may vary with $v$; for example, the model can include a treatment-by-$V$ interaction. The weight with a baseline-conditioned numerator makes treatment assignment in the weighted law depend on $V$, so the outcome model should account for $V$ when estimating conditional effects.

The identification argument requires consistency, sequential exchangeability, and positivity. Consistency says that when a person follows $\bar a$, the observed outcome equals $Y^{\bar a}$. Sequential exchangeability says that, at each decision time, the treatment received is conditionally independent of the relevant future potential outcomes given $H_t$. Positivity requires positive probability of every treatment choice needed for the target histories at histories that occur in the population. These assumptions concern the causal interpretation; specifying the numerator does not make them true.

Result: Under consistency, sequential exchangeability, positivity, and correctly specified treatment probabilities, the conditional weighted mean for a treatment history $\bar a$ identifies $E[Y^{\bar a}\mid V=v]$. In the weighted law, conditional on $V=v$ and $\bar A=\bar a$, the distribution of the measured covariate history and outcome follows the corresponding intervention law.

For justification, the weighted density derived above has treatment-sequence probability $Q(\bar a\mid v)=\prod_{t=0}^{K}q_t(a_t\mid\bar a_{t-1},v)$, which depends on the fixed $(\bar a,v)$, not on the covariate history or outcome. Conditional on that sequence and $V=v$, this factor cancels from the conditional distribution. The remaining covariate and outcome factors are those obtained by setting treatment to $\bar a$ at each time. Sequential exchangeability and consistency equate the observed outcome distribution under that intervention with the distribution of $Y^{\bar a}$ conditional on $V=v$. Hence $$ E_W[Y\mid \bar A=\bar a,V=v]=E[Y^{\bar a}\mid V=v], $$ where $E_W$ denotes expectation under the weighted law. This is a population identification statement; estimated weights and outcome models introduce sampling and modeling error.

Worked Example: Why the Outcome Model Must Retain Baseline Covariates

Suppose $P(V=0)=P(V=1)=0.5$. Under the baseline-conditioned numerator, let the weighted treatment probabilities be $P_W(A=1\mid V=0)=0.3$ and $P_W(A=1\mid V=1)=0.7$. Suppose the conditional potential-outcome means are $E[Y^0\mid V=0]=10$, $E[Y^0\mid V=1]=20$, and the treatment effects are 2 for $V=0$ and 8 for $V=1$. Compare the treatment-only weighted mean difference with the marginal causal effect in the original population.

Among weighted treated observations, the proportion with $V=1$ is $0.5(0.7)/[0.5(0.3)+0.5(0.7)]=0.7$. Their mean outcome is $0.3(12)+0.7(28)=23.2$. Among weighted untreated observations, the proportion with $V=1$ is $0.5(0.3)/[0.5(0.7)+0.5(0.3)]=0.3$. Their mean outcome is $0.7(10)+0.3(20)=13$. The treatment-only weighted mean difference is therefore $23.2-13=10.2$, not the marginal causal effect.

The original-population marginal causal effect is $0.5(2)+0.5(8)=5$. A conditional MSM can recover the stratum-specific effects, 2 and 8, and standardizing them to the original $V$ distribution gives 5. The example shows why baseline-conditioned stabilization does not justify omitting $V$ from the outcome model or treating a raw weighted group difference as the target effect.

Worked Example: Changing the Target Baseline Distribution

Using the conditional treatment effects 2 for $V=0$ and 8 for $V=1$, suppose a health program wants an average effect for a target population with $P(V=1)=0.7$, rather than the study population’s $P(V=1)=0.5$. The standardized effect for the target population is $0.3(2)+0.7(8)=0.6+5.6=6.2$.

This calculation uses the conditional effects and the target population’s baseline distribution. The numerator’s inclusion of $V$ helps retain the information needed for conditional modeling; it does not by itself change the target distribution of $V$ to the program population. That change is made explicitly in standardization.

Diagnostics, Alternatives, and Limitations

Diagnostics should match the intended weighted population. With a baseline-conditioned numerator, assess whether measured time-varying covariates are balanced across treatment groups within strata of $V$ and past treatment, or use appropriate summaries of those conditional associations. Do not demand unconditional balance in $V$ across treatment groups: such balance is not implied by this numerator. Also inspect fitted probabilities and the distribution of the weights within baseline strata; very small denominator probabilities can produce influential observations. A good-looking marginal balance table cannot establish sequential exchangeability or rule out unmeasured confounding.

The main alternative is a numerator that does not condition on $V$, such as $P(A_t\mid\bar A_{t-1})$. That choice targets a weighted treatment law that does not retain the baseline treatment association in the numerator. A baseline-conditioned numerator is often convenient for a conditional MSM and may yield less variable weights than an unstabilized numerator, but it is not automatically more efficient, and it does not guarantee stable weights. The choice should be aligned with the outcome model and the estimand, rather than made solely to obtain a preferred balance table.

The result depends on correctly modeling the treatment probabilities used in the weights and having adequate positivity. If a relevant treatment choice is absent or nearly absent within histories defined by $V$ and $L_t$, the weights cannot create information that the data do not contain. Further, conditioning the outcome model on $V$ does not repair a misspecified treatment model, unmeasured confounding, or a poorly chosen MSM. In practice, compare specifications, inspect conditional balance and weight behavior, and state clearly which baseline distribution is used when reporting standardized effects.

Common pitfall: “The weights worked, so $V$ should be balanced between treated and untreated groups.” This is incorrect when the numerator conditions on $V$: the weighted treatment law is allowed to depend on $V$. Check balance of time-varying confounders conditional on baseline covariates and past treatment, and include $V$ in the conditional outcome model when that is the target MSM.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If the numerator is $P(A_t\mid\bar A_{t-1},V)$, should $V$ necessarily be balanced across treatment groups in the weighted population?
In one sentence, explain what factor cancels from the observed-data density after weighting.
In the numerical example, what is the numerator probability for $A=0$ when $V=1$?
With conditional effects 2 and 8 and target $P(V=1)=0.7$, what is the standardized effect?

Solutions. (1) No. The numerator permits treatment to depend on $V$, so baseline balance across treatment groups is not required. (2) The denominator treatment probability at each time cancels with the corresponding treatment factor in the observed-data density, leaving the numerator treatment law. (3) It is $1-0.7=0.3$. (4) It is $0.3(2)+0.7(8)=6.2$.

Baseline Covariates in Stabilized Numerators

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