Why Let the Threshold Grow?
An upper clipping threshold limits how much any one observation can contribute to a weighted estimating equation. A fixed threshold offers that protection at every sample size, but it can continue to change the population equation being solved. A threshold that grows with \(n\) can reduce that asymptotic distortion: eventually, each fixed finite weight is left unchanged. The challenge is to let the cap grow slowly enough that extreme observations still cannot dominate the empirical equation.
The result in this tutorial concerns consistency, not asymptotic normality. It uses the true weights and independent sampling units; in a longitudinal study, one unit can contain the full treatment and covariate history. For causal interpretation, the underlying weighted equation must already target the desired marginal parameter under the relevant identification assumptions. Increasing the cap does not fix a misspecified weight model or a failure of positivity.
Definitions and a Sufficient Condition
Here is a convenient sufficient condition for a bounded estimating function: \(E[W]<\infty\), \(c_n\to\infty\), and \(c_n/n\to 0\). The first condition controls the population contribution of the clipped tail. The last controls the sampling variation of the empirical clipped equation. Neither condition says that all observed weights will be below the cap; rather, the cap grows while remaining small relative to the sample size.
The separation condition is an identification requirement: parameters a fixed positive distance from \(\theta_0\) cannot have population estimating equations arbitrarily close to zero. The theorem also covers approximate roots, not just exact numerical solutions, provided their residual tends to zero in probability.
Why the Conditions Work
Write \(P_n f=n^{-1}\sum_{i=1}^n f(Z_i)\), and first consider the clipped population equation \(\Psi_n(\theta)=E[W^{(n)}\psi(\theta,Z)]\). Since \(W^{(n)}\leq W\), $$ \sup_{\theta\in\Theta}\|\Psi_n(\theta)-\Psi(\theta)\| \leq M E[W-W^{(n)}] \leq M E[W\,1\{W>c_n\}]. $$ For each fixed \(W<\infty\), \(W1\{W>c_n\}\to 0\). It is bounded by the integrable random variable \(W\). The dominated convergence theorem therefore gives \(E[W1\{W>c_n\}]\to 0\). Thus clipping bias in the population equation vanishes uniformly over \(\Theta\).
For a fixed \(\theta\), each coordinate of \(W^{(n)}\psi(\theta,Z)\) has variance at most its second moment. Using the stated bound on \(\psi\) and \( (W^{(n)})^2\leq c_n W\), $$ \operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n W_i^{(n)}\psi_j(\theta,Z_i)\right) \leq \frac{M^2 E[(W^{(n)})^2]}{n} \leq \frac{M^2 c_n E[W]}{n}. $$ This tends to zero because \(c_n/n\to0\). Chebyshev's inequality gives convergence in probability at every fixed \(\theta\).
To obtain uniform convergence, fix a positive mesh width \(\delta\) and cover the compact set \(\Theta\) with a finite \(\delta\)-net. At the finitely many net points, the fixed-\(\theta\) convergence holds jointly, by a union bound. Between a parameter and its nearest net point, the Lipschitz condition bounds the difference between the empirical and population equations, after centering, by $$ L\delta\{P_n W^{(n)}+E[W^{(n)}]\}. $$ Here \(P_n W^{(n)}\leq P_nW\), and \(P_nW\) is bounded in probability because \(E[W]<\infty\) and Markov's inequality applies. Also \(E[W^{(n)}]\leq E[W]\). Choosing \(\delta\) arbitrarily small therefore proves $$ \sup_{\theta\in\Theta}\left\|P_n\{W^{(n)}\psi(\theta,Z)\}-\Psi_n(\theta)\right\|\xrightarrow{p}0. $$ Together with the uniform vanishing of \(\Psi_n-\Psi\), this gives uniform convergence of the empirical clipped equation to \(\Psi\).
Finally, if \(\|\hat\theta_n-\theta_0\|\geq\epsilon\), the separation assumption makes \(\|\Psi(\hat\theta_n)\|\) bounded away from zero. Uniform convergence and the assumed approximate-root property make that event have probability tending to zero. This proves \(\hat\theta_n\xrightarrow{p}\theta_0\). \(\square\)
Worked Examples with Growing Caps
Worked Example: A Pareto Weight with a Finite Mean
Suppose \(W\) has a Pareto distribution with tail probability \(P(W>w)=w^{-2}\) for \(w\geq1\), and use \(c_n=n^{0.4}\). The density is \(f(w)=2w^{-3}\) for \(w\geq1\), so $$ E[W]=\int_1^\infty 2w^{-2}\,dw=2. $$ The cap diverges and \(c_n/n=n^{-0.6}\to0\), meeting the theorem’s two threshold conditions.
For \(c\geq1\), the expected weight removed by clipping is $$ E[(W-c)_+]=\int_c^\infty (w-c)2w^{-3}\,dw =2\int_c^\infty w^{-2}\,dw-2c\int_c^\infty w^{-3}\,dw =\frac{2}{c}-\frac{1}{c}=\frac{1}{c}. $$ Thus the absolute population-equation clipping bias is at most \(M/c_n=M n^{-0.4}\). The variance bound for each score coordinate is at most $$ \frac{M^2 c_n E[W]}{n}=2M^2 n^{-0.6}. $$ Both bounds vanish, although the second is a bound rather than an exact variance.
Worked Example: A Heavier Pareto Tail
Now let \(P(W>w)=w^{-1.2}\) for \(w\geq1\). Its density is \(1.2w^{-2.2}\), and $$ E[W]=\int_1^\infty 1.2w^{-1.2}\,dw=6. $$ Choose \(c_n=n^{0.7}\). Then \(c_n\to\infty\) and \(c_n/n=n^{-0.3}\to0\), despite the relatively heavy tail.
For \(c\geq1\), direct integration gives $$ E[(W-c)_+] =1.2\int_c^\infty w^{-1.2}\,dw-1.2c\int_c^\infty w^{-2.2}\,dw =6c^{-0.2}-c^{-0.2}=5c^{-0.2}. $$ The bias bound is consequently \(5M n^{-0.14}\). The variance bound is \(6M^2 n^{-0.3}\). Both go to zero, but the tail-bias bound decays slowly. This example shows why a valid growing threshold does not guarantee that clipping distortion is negligible at a practically modest sample size.
Worked Example: What a Growing Cap Does in a Small Sample
Take the rule \(c_n=2\sqrt n\), and consider four observed weights \(1,2,4,40\), with corresponding binary outcomes \(0,0,1,1\). At \(n=4\), the cap is \(c_4=4\), so the clipped weights are \(1,2,4,4\). The untruncated weighted mean is $$ \frac{1(0)+2(0)+4(1)+40(1)}{1+2+4+40}=\frac{44}{47}\approx0.936. $$ The clipped weighted mean is $$ \frac{1(0)+2(0)+4(1)+4(1)}{1+2+4+4}=\frac{8}{11}\approx0.727. $$ The numerical difference is substantial: consistency is an asymptotic property, not a promise of negligible finite-sample distortion. For this rule, \(c_n/n=2/\sqrt n\to0\); the maximum normalized weight contribution is at most \(c_n/n=2/\sqrt n\). At \(n=4\), that bound is \(1\), so it offers little finite-sample protection by itself.
Choosing a Rate and Checking the Analysis
For any fixed exponent \(0<\gamma<1\), the rule \(c_n=n^\gamma\) satisfies both threshold conditions. The consistency argument does not select a uniquely best exponent: faster growth reduces clipping sooner but weakens the variance bound, while slower growth controls contributions more tightly but can leave more observations clipped. The actual tail of the weights determines how quickly the clipping bias disappears.
A useful diagnostic is to examine the analysis across several increasing caps. Report the cap, the fraction of weights clipped, the largest unclipped and clipped weights, and the resulting parameter estimates. If the estimate changes markedly across reasonable caps, that is evidence that the finite-sample result depends on the tail, not proof that a particular cap is correct. Check also that the fitted weights are nonnegative and that their empirical distribution is compatible with the assumed finite first moment.
The theorem is for independent units and bounded, Lipschitz estimating functions. If the outcome or score is unbounded, the displayed moment bounds need replacement by suitable weighted-moment assumptions. If \(E[W]=\infty\), the tail argument and variance bound used here do not establish consistency. Longitudinal dependence within a person is not itself a problem when the complete person-level history is treated as one sampling unit, but dependence across units requires an appropriate argument. Estimated weights require additional conditions beyond this result.
Alternatives and a Common Pitfall
| Approach | What happens to extreme contributions? | Asymptotic implication |
|---|---|---|
| Fixed upper cap | Contributions remain bounded by the same constant. | The limiting equation may retain clipping bias. |
| Diverging cap with \(c_n/n\to0\) | The bound increases, but more slowly than the sample size. | Under the theorem’s assumptions, clipping bias vanishes and the estimator is consistent. |
| No upper cap | No clipping-induced change to the equation; individual contributions are unrestricted. | Consistency may follow under other moment conditions, but it is not established by the clipped variance bound here. |
Common mistake: “Because \(c_n\to\infty\), clipping eventually disappears, so any diverging cap preserves consistency.” Pointwise, each fixed finite weight is eventually below the cap; that fact alone does not control the random empirical equation. A cap growing too quickly need not make \(c_n/n\) small, so the variance bound in the proof need not vanish. The correction is to check both conditions: \(c_n\to\infty\) removes asymptotic clipping bias under integrability, while \(c_n/n\to0\) controls sampling variation for the bounded-score result. These are sufficient conditions, not universal necessary conditions.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Does \(c_n=3n^{0.6}\) satisfy the theorem’s two threshold conditions? State the limits.
- If \(E[W]<\infty\), why does \(E[W1\{W>c_n\}]\to0\) when \(c_n\to\infty\)?
- For a bounded score, what upper bound does the theorem give for the variance of one coordinate of the empirical clipped equation?
- Why does a fixed cap generally fail to remove clipping bias asymptotically?
Solutions: (1) Yes. \(3n^{0.6}\to\infty\), and \(c_n/n=3n^{-0.4}\to0\). (2) The integrand tends pointwise to zero and is bounded by the integrable random variable \(W\), so the dominated convergence theorem applies. (3) The bound is \(M^2c_nE[W]/n\), using \((W^{(n)})^2\leq c_nW\) and independence across units. (4) With a fixed cap \(c\), observations with \(W>c\) remain clipped at every sample size; the resulting population equation can therefore differ from the untruncated one.