From Stabilized Weights to an Estimating Equation
A stabilized weight is not merely a rescaled inverse probability. Its numerator specifies a treatment distribution that replaces the observed treatment-assignment mechanism in a weighted population. The weighted estimating equation therefore targets a parameter defined under that numerator distribution. To interpret a fitted marginal structural model (MSM), it matters both how the weight is calculated and what the numerator law represents.
Let \(X\) denote baseline covariates, \(A_t\) treatment at decision time \(t\), and \(H_t\) the measured history immediately before that treatment. Let \(Y\) be the outcome after the treatment history \(\bar A_T=(A_1,\ldots,A_T)\). Write \(g_t(a\mid h)\) for the observed treatment mechanism and \(q_t(a\mid h)\) for a chosen numerator mechanism. The stabilized weight for one individual is
The numerator mechanisms may depend on baseline variables or on permitted past treatment history. Their product is not an arbitrary adjustment: it defines the treatment assignment laws in the target distribution. The essential consequence is that a weighted score is an empirical version of a score averaged under that target distribution.
Conditions Behind the Change of Measure
The ratio in the weight must define a valid change of measure. In particular, at histories generated under the target law, require \(q_t(\cdot\mid h)\ll g_t(\cdot\mid h)\): whenever \(g_t(a\mid h)=0\), also \(q_t(a\mid h)=0\). Equivalently, the numerator law must not assign treatment mass where the observed treatment mechanism assigns none. For continuous treatments, the same condition applies to probability measures and their densities. The denominator must be positive for treatment values that occur under \(q_t\), so the ratio is defined \(P^q\)-almost surely. These support conditions apply at each time along histories that can arise under the sequential target law.
For the expectation identities below, also require that the weighted function is integrable, for example \(E_P[W^S|\psi(O;\beta)|]<\infty\). The sequential factorization assumes that the treatment mechanisms are correctly specified as the relevant conditional laws, and that the other conditional mechanisms in the data law are retained when defining \(P^q\). These are distributional requirements for the score identity. A causal interpretation additionally requires consistency and sequential exchangeability, and adequate treatment positivity for the target interventions.
Proof. Factor the observed-data law into its baseline, covariate-transition, treatment, and outcome mechanisms. In schematic density notation, with the covariate and outcome factors collected as \(r\), $$ p(o)=r(o)\prod_{t=1}^T g_t(a_t\mid h_t), \qquad p^q(o)=r(o)\prod_{t=1}^T q_t(a_t\mid h_t). $$ The support condition ensures that \(p^q\) is absolutely continuous with respect to \(p\) and that the ratio is defined \(P^q\)-almost surely. Wherever \(p(o)>0\), the ratio of the displayed factorizations is $$ \frac{p^q(o)}{p(o)} =\prod_{t=1}^T\frac{q_t(a_t\mid h_t)}{g_t(a_t\mid h_t)} =W^S(o). $$ Thus integration under \(P\) gives $$ E_P[W^S h(O)] =\int h(o)\frac{p^q(o)}{p(o)}p(o)\,do =\int h(o)p^q(o)\,do =E_{P^q}[h(O)]. $$ The same argument uses sums for discrete variables and integrals with respect to dominating measures for continuous variables. The stated integrability condition makes these expectations well-defined. \(\square\)
What the Weighted Score Estimates
Suppose an MSM specifies a conditional mean under \(P^q\), such as \(E_{P^q}[Y\mid \bar A_T]=m(\bar A_T;\beta)\). For a differentiable mean model, let \(D(\bar A_T;\beta)\) be the derivative of \(m\) with respect to \(\beta\). A familiar least-squares estimating function is $$ \psi(O;\beta) = D(\bar A_T;\beta)^{\mathsf T} \{Y-m(\bar A_T;\beta)\}. $$ If the conditional mean model is correct and the conditional expectation exists, then \(E_{P^q}[\psi(O;\beta_0)]=0\) at its true parameter \(\beta_0\). The theorem therefore yields the population weighted score equation \(E_P[W^S\psi(O;\beta_0)]=0\), and its empirical counterpart is the weighted estimating equation above.
More generally, any estimating function designed to have expectation zero under \(P^q\) can be used; the particular score need not be a likelihood score. The weight multiplies the entire individual score contribution. Consequently, each observed treatment history receives the numerator probabilities evaluated at that individual’s observed treatments, not a single common numerator constant.
Proof. Apply the change-of-measure identity component by component to \(\psi(O;\beta_0)\): $$ E_P[W^S\psi(O;\beta_0)] =E_{P^q}[\psi(O;\beta_0)] =0. $$ For the sample score, each summand \(W_i^S\psi(O_i;\beta_0)\) has finite first moment and mean zero. The weak law of large numbers therefore gives $$ \frac{1}{n}\sum_{i=1}^n W_i^S\psi(O_i;\beta_0) \xrightarrow{p}0. $$ This establishes the population equation and convergence of the score at the target parameter; consistency of a chosen root additionally requires the usual identification and regularity conditions for that estimating equation. \(\square\)
Worked Example: A Weighted Linear MSM Score
Consider one binary treatment \(A\), with a constant numerator probability \(q(A=1)=0.25\). Four observed individuals have the following outcomes and fitted denominator probabilities \(g_i=P(A_i=1\mid H_i)\). Their observed treatment values are shown in the first column.
| \(A_i\) | \(Y_i\) | \(g_i\) | \(W_i^S\) |
|---|---|---|---|
| 1 | 4 | 0.80 | 0.25 / 0.80 = 0.3125 |
| 1 | 8 | 0.40 | 0.25 / 0.40 = 0.625 |
| 0 | 1 | 0.20 | 0.75 / 0.80 = 0.9375 |
| 0 | 3 | 0.60 | 0.75 / 0.40 = 1.875 |
The last two rows require care: for \(A=0\), the denominator probability is \(1-g_i\). Thus their correct weights are \(0.75/0.80=0.9375\) and \(0.75/0.40=1.875\). Fit the saturated linear mean \(m(A;\beta)=\beta_0+\beta_1A\). Its weighted score equations are $$ \sum_i W_i^S(Y_i-\beta_0-\beta_1 A_i)=0, \qquad \sum_i W_i^S A_i(Y_i-\beta_0-\beta_1 A_i)=0. $$
The second equation involves only treated individuals, so their fitted weighted mean is $$ \hat\beta_0+\hat\beta_1 =\frac{0.3125(4)+0.625(8)}{0.3125+0.625} =\frac{6.25}{0.9375} =\frac{20}{3}\approx 6.667. $$ The control weighted mean is $$ \hat\beta_0 =\frac{0.9375(1)+1.875(3)}{0.9375+1.875} =\frac{6.5625}{2.8125} =\frac{7}{3}\approx 2.333. $$ Therefore \(\hat\beta_1=20/3-7/3=13/3\approx4.333\). The fitted contrast is the treated-minus-control difference in this weighted sample. Its causal interpretation depends on the identification assumptions and on the target law, not on the arithmetic alone.
Worked Example: A Baseline-Dependent Numerator Changes the Target
Let \(X\) be binary with \(P(X=0)=P(X=1)=0.5\). At a single treatment time, suppose \(g(1\mid X=0)=0.2\) and \(g(1\mid X=1)=0.6\), while the numerator uses \(q(1\mid X=0)=0.1\) and \(q(1\mid X=1)=0.7\). Both treatment values have positive probability under both mechanisms in each stratum, so the numerator law is absolutely continuous with respect to the denominator law.
Under the target law, the joint treatment-baseline probabilities are $$ P^q(X=0,A=1)=0.5(0.1)=0.05,\qquad P^q(X=1,A=1)=0.5(0.7)=0.35. $$ Hence \(P^q(A=1)=0.05+0.35=0.40\), and \(E_{P^q}[XA]=0.35\). For example, the weighted estimating equation for the target quantity \(\theta=E_{P^q}[XA]\) is $$ \frac{1}{n}\sum_{i=1}^n W_i^S(X_iA_i-\theta)=0, \quad\text{so}\quad \hat\theta=\frac{\sum_iW_i^S X_iA_i}{\sum_iW_i^S}. $$
If instead the numerator were constant with \(q(1)=0.4\), the marginal target treatment probability would still be \(0.4\), but \(P^q(X=1,A=1)\) would be \(0.5(0.4)=0.20\), not \(0.35\). Thus equal marginal treatment probabilities do not imply equal target distributions or equal scores involving baseline-treatment terms. The numerator’s dependence on \(X\) matters whenever the estimating function distinguishes those strata.
Worked Example: A Numerator Determines an Intercept-Only Target Mean
Suppose \(A\) is binary, \(P(A=1)=0.5\) under the observed law, and \(Y=10A\). Choose a numerator law \(q(A=1)=0.2\). The stabilized weights are \(0.2/0.5=0.4\) for treated individuals and \(0.8/0.5=1.6\) for controls. For the intercept-only model \(m(\beta)=\beta\), the score is \(Y-\beta\). Its target-law solution is $$ E_{P^q}[Y-\beta]=0 \quad\Longrightarrow\quad \beta=E_{P^q}[Y]=0.2(10)+0.8(0)=2. $$
The weighted population score gives the same answer: $$ E_P[W^S(Y-\beta)] =0.5(0.4)(10-\beta)+0.5(1.6)(0-\beta) =2-1.0\beta. $$ Setting it to zero yields \(\beta=2\). By contrast, the unweighted outcome mean is \(0.5(10)+0.5(0)=5\), which is the mean under the observed treatment distribution, not the chosen numerator distribution. The numerator changes the target score and its root.
Diagnostics, Alternatives, and Limitations
A score equation can be algebraically correct and still be unreliable in practice. Inspect the fitted denominator and numerator probabilities at observed treatment values, the resulting weight distribution, and the contributions \(W_i^S\psi(O_i;\beta)\). Very large weights, a small number of observations dominating the score, or near-zero denominator probabilities signal weak empirical support and potential instability. A score that does not approach zero at the fitted estimate can indicate an optimization or implementation error. Also check that the numerator mechanism actually represents the intended target distribution; the baseline-dependent example shows why checking only marginal treatment probabilities can miss a mismatch.
Stabilized weights and normalized weights serve different roles. Stabilization changes the numerator treatment law and therefore the target distribution. Dividing all weights by their sample mean is a common rescaling: for an equation whose entire score is multiplied by that same constant, it leaves the root unchanged, but it does not replace the numerator law or repair absent support. An unweighted outcome regression instead relies on its outcome-model assumptions and does not use this change-of-treatment-law score. These approaches answer different questions and rely on different modeling assumptions.
The weighted equation is not automatically unbiased for every proposed MSM. Its estimating function must have mean zero under \(P^q\); model misspecification can change the parameter to a score-dependent projection rather than the intended MSM parameter. Estimated treatment mechanisms add uncertainty, and treating estimated weights as fixed can give an incomplete variance calculation. Finally, no choice of numerator creates information for treatment histories that are impossible under the observed mechanism. These limitations motivate careful target specification, model checks, and appropriate uncertainty assessment.
Check Your Understanding
Answer each question, then compare with the solutions below.
- State the change-of-measure condition on \(q_t\) and \(g_t\) needed for the stabilized-weight identity.
- With \(q(A=1)=0.3\), \(g(A=1\mid H)=0.6\), and observed \(A=1\), calculate the stabilized weight.
- In the baseline-dependent example, calculate \(P^q(X=1,A=0)\).
- Why can multiplying every weight by one positive constant leave the score root unchanged without making normalization equivalent to stabilization?
Solutions. (1) Require \(q_t(\cdot\mid h)\ll g_t(\cdot\mid h)\) at relevant target histories, with the ratios defined \(P^q\)-almost surely; numerator mass cannot occur where denominator mass is zero. (2) The weight is \(0.3/0.6=0.5\). (3) It is \(P(X=1)q(A=0\mid X=1)=0.5(1-0.7)=0.15\). (4) If the estimating equation is multiplied throughout by a common positive constant \(c\), then \(c\sum_i W_i\psi_i=0\) has the same roots as \(\sum_iW_i\psi_i=0\). Stabilization instead changes the individual numerator factors and hence the target law; it is not generally a common rescaling.