Why the Timing of Truncation Matters
A fixed clipping rule is applied to estimated weights in practice: first estimate the treatment probabilities, then transform the resulting inverse probabilities. The resulting weights are random for two reasons. The observed treatment and outcome records vary across samples, and the fitted propensity model varies across samples. If the clipping rule is locally flat at a fitted probability, the second source can have no first-order effect for that observation. Where the rule is not flat, it generally contributes to the estimator’s sampling variability.
It is useful to compare this procedure with an oracle analysis that uses the true treatment probabilities. The oracle makes a clean reference point, but it is not ordinarily available. Its variance is not automatically the variance of the estimator that uses fitted weights, and fitting the propensity model does not necessarily increase variance: the additional term can be positively or negatively associated with the outcome term.
Under consistency, conditional exchangeability, and positivity, the unclipped quantity \(E[A Y/e_0(X)]\) identifies the mean of the potential outcome under treatment. Clipping can change that target: \(\mu_\tau\) need not equal the causal mean when some true weights exceed \(\tau\). The results here concern the sampling variability around the clipped target. They do not remove the clipping bias relative to an unclipped causal target.
For a fixed \(\tau\), clipping the inverse weight at \(\tau\) is equivalent to replacing \(e\) by \(\max(e,1/\tau)\) before taking its reciprocal. The kink occurs at \(e=1/\tau\). Derivatives below apply away from that boundary; a positive probability mass exactly at the boundary can make the plug-in functional non-differentiable in the ordinary first-order sense.
Derivative and Influence of the Fitted Propensity
For clarity, suppose the propensity model is logistic: \(e_\beta(X)=\operatorname{expit}\{\beta^\mathsf{T}Z(X)\}\), where \(Z(X)\) is a fixed-dimensional covariate vector. At the true parameter \(\beta_0\), let \(e_0=e_{\beta_0}\). The derivative of the unclipped inverse weight follows by the chain rule: \[ \frac{\partial(1/e_\beta)}{\partial\beta} =-\frac{1}{e_\beta^2}\frac{\partial e_\beta}{\partial\beta} =-\frac{1-e_\beta}{e_\beta}Z(X). \] Consequently, \[ \dot g_\tau(e_{\beta_0}) =-\mathbf{1}\{e_0>1/\tau\}\frac{1-e_0}{e_0}Z(X), \] provided \(P\{e_0(X)=1/\tau\}=0\). When \(e_0<1/\tau\), the weight is capped and its local derivative is zero. When \(e_0>1/\tau\), it is uncapped and the usual inverse-weight derivative applies.
The logistic maximum-likelihood score for one observation and its expected information matrix are \[ s_{\beta_0}=(A-e_0)Z(X),\qquad I=E[e_0(1-e_0)Z(X)Z(X)^\mathsf{T}]. \] Assuming a correctly specified regular model, nonsingular \(I\), and the usual likelihood regularity conditions, the estimator satisfies \(\hat\beta-\beta_0=I^{-1}P_n s_{\beta_0}+o_p(n^{-1/2})\). The following result shows how that fitted-parameter variation enters the clipped estimator.
Proof. Write \(f(O,\beta)=A Y g_\tau\{e_\beta(X)\}\), with \(O=(X,A,Y)\). A first-order expansion of the sample average around \(\beta_0\) gives \[ P_n f(O,\hat\beta) =P_n f(O,\beta_0) +\left\{P_n\frac{\partial f(O,\beta_0)}{\partial\beta}\right\}^{\mathsf T} (\hat\beta-\beta_0)+o_p(n^{-1/2}). \] By the law of large numbers, the derivative average converges in probability to \(G\). Substituting the likelihood expansion for \(\hat\beta-\beta_0\), and using that the product of a term converging to \(G\) and an \(O_p(n^{-1/2})\) term differs from \(G^\mathsf{T}(\hat\beta-\beta_0)\) by \(o_p(n^{-1/2})\), yields the fitted-parameter contribution \(G^\mathsf{T}I^{-1}P_n s_{\beta_0}\). Also, \[ P_n f(O,\beta_0)-\mu_\tau =P_n\{f(O,\beta_0)-\mu_\tau\}. \] Combining these expressions gives the stated expansion. The central limit theorem applies to the displayed influence function under the assumed finite second moments. \(\square\)
The oracle influence function, which treats \(e_0\) as known, is only \(A Y g_\tau(e_0)-\mu_\tau\). If \(V_{\rm oracle}\) is its variance and \(V_{\rm fitted}\) is the variance of the theorem’s full influence function, then \[ V_{\rm fitted}=V_{\rm oracle} +2\operatorname{Cov}\!\left(A Y g_\tau(e_0),G^\mathsf{T}I^{-1}s_{\beta_0}\right) +\operatorname{Var}\!\left(G^\mathsf{T}I^{-1}s_{\beta_0}\right). \] The last term is nonnegative, but the covariance term can have either sign. Therefore estimating the propensity can increase or decrease the variance relative to the oracle calculation; simply adding a positive “nuisance variance” to the oracle variance is not correct.
Worked Example: A Clipped Region with No First-Order Fit Effect
Suppose \(A\) is Bernoulli with propensity \(e_0=0.4\), and \(Y\) is Bernoulli with mean \(0.5\), independent of treatment. Use a logistic intercept-only propensity model and clip at \(\tau=2\). The clipping boundary is \(1/\tau=0.5\), so the true weight \(1/0.4=2.5\) is clipped to \(2\).
Because \(0.4<0.5\), the fitted probability remains in the clipped region with probability tending to one. There, \(g_\tau(e)=2\) is locally constant, so \(\dot g_\tau(e_0)=0\) and \(G=0\). The clipped target is \(\mu_\tau=E[2AY]=2(0.4)(0.5)=0.4\).
The oracle influence function is \(2AY-0.4\). Since \(AY\) is Bernoulli with success probability \(0.4(0.5)=0.2\), its variance is \(4\operatorname{Var}(AY)=4(0.2)(0.8)=0.64\). The fitted-propensity influence function is the same to first order, so its asymptotic variance is also \(0.64\). This equality is local and asymptotic: the fitted probability can cross the boundary in a finite sample, but consistency and the positive distance from the boundary make that event vanish asymptotically.
Worked Example: Fitting Can Reduce Variance When Weights Are Unclipped
Let \(A\) be Bernoulli with known true probability \(p=0.5\), and let \(Y\) be Bernoulli with mean \(\mu=0.4\), independent of \(A\). Fit an intercept-only logistic propensity and use \(\tau=3\). Since \(1/p=2<3\), clipping is inactive near the truth. The target is \(\mu_\tau=E[AY/p]=0.4\).
With the true probability treated as known, the influence function is \(AY/p-\mu\). As \(AY\) is Bernoulli with probability \(0.5(0.4)=0.2\), the oracle variance is \[ \operatorname{Var}(AY/p-\mu) =\frac{0.2(0.8)}{0.5^2}=0.64. \]
For the fitted intercept, take \(Z=1\). The score is \(A-p\), and the information is \(I=p(1-p)=0.25\). The derivative of the inverse propensity with respect to the logit intercept is \(-(1-p)/p=-1\), so \(G=E[AY(-1)]=-0.2\). Hence \(G I^{-1}=-0.8\), and the fitted influence function is \[ AY/p-\mu-0.8(A-0.5) =2AY-0.4-0.8A+0.4 =2A(Y-0.4). \] Its variance is \(4E[A(Y-0.4)^2]=4(0.5)(0.4)(0.6)=0.48\). Here fitting reduces the variance from \(0.64\) to \(0.48\). The reduction comes from the covariance between the score correction and the oracle term; it is not a general promise about fitted weights.
Worked Example: Only the Unclipped Stratum Contributes to the Derivative
Let \(X\) have two equally likely strata, \(X=1\) and \(X=2\). Fit a saturated logistic propensity with \(Z(X)=(\mathbf{1}\{X=1\},\mathbf{1}\{X=2\})^\mathsf{T}\). Set \(e_0(1)=0.8\), \(e_0(2)=0.4\), and \(\tau=2\). Suppose \(E[Y\mid A=1,X=1]=0.6\) and \(E[Y\mid A=1,X=2]=0.3\).
The threshold is \(0.5\). In stratum 1 the weight is \(1/0.8=1.25\), which is uncapped; in stratum 2 the weight \(1/0.4=2.5\) is capped at \(2\). The clipped target is \[ \mu_\tau =\tfrac12(0.8)(0.6)(1.25)+\tfrac12(0.4)(0.3)(2) =0.30+0.12=0.42. \]
For stratum 1, the derivative of the weight with respect to its logit is \(-(1-e)/e=-0.2/0.8=-0.25\). Thus its contribution to the first component of \(G\) is \(\tfrac12(0.8)(0.6)(-0.25)=-0.06\). In stratum 2, the weight is capped and its derivative is zero. Therefore \(G=(-0.06,0)^\mathsf{T}\).
The information matrix is diagonal because the stratum indicators do not overlap: \[ I= \begin{pmatrix} \tfrac12(0.8)(0.2)&0\\ 0&\tfrac12(0.4)(0.6) \end{pmatrix} = \begin{pmatrix}0.08&0\\0&0.12\end{pmatrix}. \] Consequently \(I^{-1}G=(-0.75,0)^\mathsf{T}\). The fitted-weight correction is therefore \(-0.75\mathbf{1}\{X=1\}(A-0.8)\). Stratum 2 contributes no first-order propensity-estimation correction, while stratum 1 does. The full fitted influence function still includes the ordinary weighted-outcome term from both strata.
What the Comparison Does—and Does Not—Say
| Procedure | First-order variability | Interpretation |
|---|---|---|
| Use the true propensity and clip its weight | Oracle weighted-outcome influence function only | Reference calculation; usually infeasible because the true propensity is unknown |
| Fit the propensity, then clip the fitted weight | Weighted-outcome term plus the fitted-parameter correction | Accounts for both outcome sampling and propensity estimation |
| Fit and clip, but calculate variance as if fitted weights were fixed | Omits the correction and its covariance with the outcome term | Generally not a valid first-order variance calculation |
The clipping derivative also explains why the effect is not uniform across observations. Capped weights are locally insensitive to small propensity changes; uncapped weights are not. If a positive fraction of observations lies near the threshold, even small estimation errors can change which side of the kink an observation occupies. The condition \(P\{e_0=1/\tau\}=0\) rules out probability mass exactly at the kink, but does not by itself provide a uniform margin from it. Standard differentiability arguments additionally require enough regularity to control observations near the boundary. If that control fails, the ordinary linear expansion may not apply.
A practical variance calculation should therefore reflect how the weights were obtained. A sandwich calculation based on the joint estimating equations for the propensity and weighted estimator can include the fitted-propensity contribution. A resampling procedure can also do so if it refits the propensity model and reapplies the same fixed clipping rule in each resample. Holding fitted weights fixed in every resample reproduces the oracle-style calculation, not the full fitted-weight procedure.
Diagnostics and Common Pitfalls
- Ignoring the fit. The incorrect reasoning is, “The weights are already computed, so their estimation adds no uncertainty.” The correction is that the fitted weights change with \(\hat\beta\); where \(e_0>1/\tau\), the derivative is generally nonzero.
- Assuming fitting must increase variance. The score correction has its own variance, but it is correlated with the weighted outcome term. The covariance can offset some or all of that added variance.
- Applying the uncapped derivative to every observation. Below the clipping boundary, the clipped weight is constant and its derivative is zero. Using the uncapped derivative there gives the wrong \(G\).
- Forgetting the boundary condition. At \(e=1/\tau\), the clipped function has a kink. If observations can have positive probability exactly there, the usual smooth Taylor argument needs replacement or additional analysis.
- Confusing variability with clipping bias. The expansion is centered at \(\mu_\tau\), not necessarily at the unclipped causal mean. A correct standard error does not undo the change in target caused by clipping.
A useful diagnostic is to report how much of the sample is capped and to inspect fitted propensities near \(1/\tau\). If nearly all weights are capped, the propensity correction may be small locally, but this does not establish that the clipped estimator targets the desired causal quantity. If many weights are uncapped, treating fitted weights as fixed is especially difficult to justify.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For \(\tau=4\), at which true propensity values is the clipped inverse weight locally flat?
- Why can the fitted-propensity estimator have lower variance than the oracle estimator?
- In a saturated two-stratum logistic model, what happens to a stratum’s contribution to \(G\) if its true propensity is strictly below \(1/\tau\)?
- What is omitted when a variance calculation holds estimated clipped weights fixed?
Solutions. (1) The boundary is \(1/\tau=0.25\). For \(e_0<0.25\), the inverse weight exceeds 4 and is clipped to the constant 4, so the derivative is zero. At the boundary the function has a kink; above it the weight is uncapped. (2) The score correction can be negatively correlated with the oracle weighted-outcome term. The total variance includes that covariance, which can be large enough to outweigh the correction’s own variance. (3) Its derivative is zero in that stratum, so it contributes zero to \(G\), provided the propensity is strictly away from the boundary and the regularity conditions hold. (4) It omits the first-order influence of estimating the propensity and the covariance between that influence and the weighted-outcome term.