Why Treatment Model Misspecification Matters
Inverse-probability weighting is intended to construct a pseudo-population in which measured confounders no longer predict treatment. That purpose depends on the treatment probabilities used to calculate the weights. If the probability model misses important features of treatment assignment, weighting can leave treatment groups different in their confounder distributions. The resulting weighted comparison may therefore retain confounding, even when the outcome model is correctly specified for the intended marginal structural model.
The key issue is not simply whether the weights are large or whether the treatment model predicts treatment well overall. The issue is whether the fitted probabilities reproduce treatment assignment conditional on the measured covariates and histories needed to control confounding. The calculations below make the connection between probability errors, imbalance, and bias explicit.
Setup and the Role of the Treatment Model
Begin with one binary treatment \(A\), measured covariates \(X\), and an outcome \(Y\). Let \(Y^1\) and \(Y^0\) denote the potential outcomes under treatment and no treatment. The causal contrast considered here is the average treatment effect \(E[Y^1-Y^0]\). For this contrast to be identified by treatment weighting, assume consistency, conditional exchangeability \( (Y^1,Y^0)\perp A\mid X\), and positivity: every covariate pattern in the target population has a positive probability of receiving either treatment.
Define the true propensity score \(e(X)=P(A=1\mid X)\), and let \(\hat e(X)\) denote the probability supplied by the fitted treatment model. The unstabilized inverse-probability weight for an individual is \(1/\hat e(X)\) if \(A=1\), and \(1/\{1-\hat e(X)\}\) if \(A=0\). In a weighted comparison, each treatment group’s outcome mean is calculated using its corresponding weights.
The following identity shows why correct conditional probabilities matter. Conditional on \(X\), the expected weighted contribution to the treated group is \(E[A/\hat e(X)\mid X]=e(X)/\hat e(X)\). The corresponding contribution to the untreated group is \(E[(1-A)/\{1-\hat e(X)\}\mid X]=\{1-e(X)\}/\{1-\hat e(X)\}\). If \(\hat e(X)=e(X)\), both expressions equal 1, so weighting restores the same relative representation of each covariate pattern in the two groups. If the probabilities are wrong, those expressions generally vary with \(X\), and the groups need not have the same covariate distribution.
Deriving the Residual Imbalance
For a discrete covariate pattern \(x\) with population proportion \(f(x)\), the treated group’s expected weighted mass at that pattern is proportional to \(f(x)e(x)/\hat e(x)\). The untreated group’s expected weighted mass is proportional to \(f(x)\{1-e(x)\}/\{1-\hat e(x)\}\). These are then normalized separately to give the covariate distributions within the two weighted groups.
When the fitted probabilities are correct, both unnormalized distributions are proportional to \(f(x)\), so normalization makes them equal. Under misspecification, their relative weights across covariate patterns differ. This argument concerns the treatment-weighting step; it does not require the outcome to be continuous or the treatment effect to be constant. The numerical examples use simple outcomes so the consequences can be calculated directly.
Worked Example: Omitting a Prognostic Confounder
Consider a hypothetical population with 100 people in each of two covariate strata, low risk and high risk. The true treatment probabilities are 0.80 in the low-risk stratum and 0.20 in the high-risk stratum. Suppose the outcome under no treatment is 10 units in the low-risk stratum and 20 units in the high-risk stratum, and treatment adds exactly 2 units for everyone. The true average treatment effect is therefore 2 outcome units. A misspecified intercept-only model assigns probability 0.50 to treatment in both strata.
The observed counts implied by the stated probabilities are 80 treated and 20 untreated in the low-risk stratum, and 20 treated and 80 untreated in the high-risk stratum. Treated outcomes are 12 and 22 units in the two strata; untreated outcomes are 10 and 20 units.
Using the misspecified probabilities. Every person receives weight \(1/0.50=2\), regardless of treatment group or risk stratum. The common weight cancels when each group’s weighted mean is calculated. Thus the treated mean is \((80\times12+20\times22)/100=14\) units, and the untreated mean is \((20\times10+80\times20)/100=18\) units. The weighted contrast is \(14-18=-4\) units, not the true effect of 2 units.
The reason is visible in the weighted covariate composition. Among treated people, 20 of 100 are high risk; among untreated people, 80 of 100 are high risk. Multiplying everyone by the same weight changes neither proportion. Since risk predicts the outcome, this imbalance remains confounding in the weighted comparison.
Check with the correct probabilities. If the weights use the true probabilities, the treated weighted mean is \((80\times12/0.80+20\times22/0.20)/(80/0.80+20/0.20)=3400/200=17\) units. The untreated weighted mean is \((20\times10/0.20+80\times20/0.80)/(20/0.20+80/0.80)=3000/200=15\) units. Their difference is \(17-15=2\) units, as stipulated. In this example, correct weighting makes the risk-stratum representation equal in the two groups.
Partial Probability Errors Still Leave Bias
A model does not have to omit a covariate entirely to cause consequential imbalance. It may capture part of the variation in treatment probabilities but still get their conditional values wrong. The next calculation holds the population and causal effect fixed while replacing the constant fitted probability with probabilities that are closer to the truth.
Worked Example: A Closer but Still Incorrect Model
Use the same hypothetical population: 100 low-risk and 100 high-risk people; true treatment probabilities 0.80 and 0.20; outcome means 12 and 22 under treatment and 10 and 20 without treatment. Now suppose the fitted model assigns probabilities 0.60 in the low-risk stratum and 0.40 in the high-risk stratum. These values distinguish the strata, but they are still not the true conditional probabilities.
For treated people, the low-risk weighted mass is \(80/0.60=400/3\), and the high-risk weighted mass is \(20/0.40=50\). The treated weighted outcome sum is \((400/3)\times12+50\times22=2700\), and the total treated weighted mass is \(400/3+50=550/3\). The weighted treated mean is therefore \(2700/(550/3)=162/11\approx14.73\) outcome units.
For untreated people, the low-risk weighted mass is \(20/(1-0.60)=50\), and the high-risk weighted mass is \(80/(1-0.40)=400/3\). The untreated weighted outcome sum is \(50\times10+(400/3)\times20=9500/3\), and its total weighted mass is \(550/3\). The weighted untreated mean is \((9500/3)/(550/3)=190/11\approx17.27\) outcome units.
The resulting weighted contrast is \(162/11-190/11=-28/11\approx-2.545\) outcome units. It remains far from the stipulated causal effect of 2 units. The weighted high-risk proportions are \(3/11\) among treated people and \(8/11\) among untreated people: the model reduces the original imbalance but does not remove it. A probability model can improve balance without achieving the balance required for an unbiased contrast.
Misspecification Across Covariate Patterns
The conditional-probability identity applies to every covariate pattern, not just to a single omitted binary confounder. An overall treatment proportion can agree with its fitted counterpart while conditional probabilities are wrong. Such marginal agreement does not ensure that the weighted groups have matching distributions of prognostic covariates.
Worked Example: Correct Marginal Prevalence, Incorrect Conditional Probabilities
Consider four equally sized covariate strata, each with 100 people. Their true treatment probabilities are 0.80, 0.60, 0.40, and 0.20, respectively. Suppose the fitted model assigns probabilities 0.70 to each of the first two strata and 0.30 to each of the last two. The average fitted probability is 0.50, equal to the actual average treatment probability of 0.50, but the conditional probabilities are misspecified.
Suppose the outcome under no treatment is 10, 15, 20, and 25 units across the four strata, and treatment adds 2 units in every stratum. For the treated group, the weighted masses, after dividing by the common stratum size of 100, are \(0.80/0.70=8/7\), \(0.60/0.70=6/7\), \(0.40/0.30=4/3\), and \(0.20/0.30=2/3\). They sum to 4. The weighted treated mean is \[ \frac{(8/7)(12)+(6/7)(17)+(4/3)(22)+(2/3)(27)}{4} =\frac{397}{21}\approx18.90 \] outcome units.
For the untreated group, the corresponding weighted masses are \(0.20/0.30=2/3\), \(0.40/0.30=4/3\), \(0.60/0.70=6/7\), and \(0.80/0.70=8/7\), also summing to 4. The weighted untreated mean is \[ \frac{(2/3)(10)+(4/3)(15)+(6/7)(20)+(8/7)(25)}{4} =\frac{380}{21}\approx18.10 \] outcome units. The weighted contrast is \(397/21-380/21=17/21\approx0.8095\) outcome units, rather than 2.
The fitted and actual marginal treatment proportions agree, but the strata do not receive the same relative weights in the treated and untreated groups. The discrepancy in the contrast is caused by this residual imbalance in a covariate that predicts the outcome, not by a change in the stipulated treatment effect.
Why the Consequences Can Compound Over Time
In longitudinal analyses, treatment weights involve probabilities conditional on treatment and covariate histories. For a binary treatment at visits \(t=0,\ldots,K\), an unstabilized treatment weight for an observed treatment history is a product of inverse conditional probabilities. A stabilized weight uses a numerator probability as well; in either case, the denominator model must represent treatment probabilities conditional on the measured history needed to control confounding.
At a given visit, let \(H_t\) denote the relevant observed history before treatment, and write \(e_t(H_t)=P(A_t=1\mid H_t)\) for the true probability and \(\hat e_t(H_t)\) for the fitted probability. The same conditional calculation applies at that visit: the expected treated contribution after applying the fitted inverse probability is \(e_t(H_t)/\hat e_t(H_t)\), and the untreated contribution is \(\{1-e_t(H_t)\}/\{1-\hat e_t(H_t)\}\). If these ratios vary across histories, the weighted treatment groups can remain imbalanced within the history distribution.
Because a longitudinal weight multiplies contributions from multiple visits, probability errors at more than one visit can interact. For example, a model that misses a nonlinear history effect at one visit can leave imbalance in a time-varying confounder; a later model may then also assign incorrect probabilities across histories affected by that imbalance. This does not mean every small error mechanically creates a large bias. The consequence depends on which histories are misweighted, how strongly their covariates predict the outcome, and how treatment effects vary across the target population.
What Model Checks Can and Cannot Establish
Balance checks examine whether measured covariates and histories are similarly distributed across weighted treatment groups. They address a consequence of treatment model fitting rather than proving that a model is correct. The earlier material on covariate balance and weight-model diagnosis is useful here: balance should be examined for relevant covariates and, where needed, transformations and interactions, not inferred from a single overall fit statistic.
A well-behaved weight distribution is not proof of adequate confounding control. In the first worked example, all weights were identical and finite, yet the weighted groups retained severe risk imbalance. Conversely, extreme weights may indicate limited practical overlap or a poorly estimated model, but avoiding extreme weights alone does not guarantee that the remaining weights remove confounding.
Model changes should be guided by the treatment-assignment process and subject-matter knowledge, with balance and overlap assessed after fitting. If the measured data do not support adequate adjustment for some histories, changing the functional form cannot create information that is absent from the data. Weight truncation also does not correct the underlying probability model: it may trade some variance for additional bias, rather than restore conditional balance.
Common Mistake: Treating a Good Marginal Fit as Conditional Balance
- Wrong reasoning: “The predicted treatment proportion matches the observed proportion, so the treatment weights must have removed confounding.” The four-stratum example shows why this fails: the overall proportions both equal 0.50, while conditional treatment probabilities are wrong and the weighted effect is biased.
- Correction: The required probabilities are conditional on the measured confounders or histories, not merely calibrated in the aggregate. Evaluate whether the weighted groups are balanced on relevant measured features and whether the model represents plausible treatment-assignment structure.
- Another wrong reasoning: “A robust standard error or bootstrap will fix the bias from a misspecified treatment model.” These procedures quantify sampling uncertainty under their respective assumptions; they do not make an incorrect weighting model correct. A narrow interval around a biased estimate can still be misleading.
Check Your Understanding
Answer each question using the assumptions and calculations in this tutorial.
- In the first worked example, why did constant weights leave the high-risk proportions unequal between treatment groups?
- For a covariate pattern \(x\), what are the expected weighted contributions for treated and untreated people when the fitted probability is \(\hat e(x)\)?
- Why does matching the average fitted probability to the average observed treatment proportion fail to establish conditional balance?
- If a treatment model is misspecified but the weights are finite, what additional condition must still be considered before interpreting a weighted contrast causally?
- Can bootstrap inference eliminate bias caused by an incorrect treatment probability model? Explain the distinction between bias and sampling uncertainty.