Asymptotic Normality with Truncated Weights

Why a moving threshold requires two separate checks

Clipping can make unusually large weights less influential, but a fixed cutoff generally changes the population quantity estimated by a weighted estimating equation. Letting the cutoff increase with sample size can remove that asymptotic clipping bias. It does not, by itself, guarantee a normal limit: the bias must disappear quickly enough, and the random part of the estimator must satisfy a central limit theorem.

The key distinction is between consistency and root-n asymptotic normality. Consistency asks whether the clipping bias tends to zero. Root-n inference asks whether that bias is negligible after multiplication by $\sqrt n$. A bias can tend to zero and still be too large for a centered root-n normal approximation.

Setup and the moving-cutoff target

Definition: Let $(W_i,Y_i)$, $i=1,\ldots,n$, be independent observations from a common distribution, with $W_i\geq 0$, $0<E[W]<\infty$, and $|Y|\leq B<\infty$. Define the untruncated weighted mean by $$ \theta=\frac{E[WY]}{E[W]}. $$ For a deterministic sequence of cutoffs $c_n$, let $W_{i,n}=\min(W_i,c_n)$, $\mu_n=E[W_{i,n}]$, and $$ \theta_n=\frac{E[W_{i,n}Y]}{\mu_n}, \qquad \hat\theta_n=\frac{\sum_{i=1}^n W_{i,n}Y_i}{\sum_{i=1}^n W_{i,n}}. $$ Here $\theta_n$ is the population target after clipping at sample size $n$; $\hat\theta_n$ estimates that target.

These assumptions make the ratios well-defined: boundedness of $Y$ and $E[W]<\infty$ imply $E[|WY|]<\infty$, while $E[W]>0$. Assume below that $c_n\to\infty$. Then $W_{i,n}\to W_i$, and dominated convergence gives $\mu_n\to\mu=E[W]>0$ and $\theta_n\to\theta$.

The exact residual bias has a useful form. Write $x_+=\max(x,0)$. Since $W-\min(W,c_n)=(W-c_n)_+$, subtraction of the two ratios gives

\theta_n-\theta = -\frac{E[(W-c_n)_+(Y-\theta)]}{\mu_n}.

This identity shows why bias depends on both the tail of $W$ and the association between large weights and the centered outcome $Y-\theta$. It also gives the general bound

|\theta_n-\theta| \leq \frac{E[(W-c_n)_+|Y-\theta|]}{\mu_n} \leq \frac{2B\,E[(W-c_n)_+]}{\mu_n}.

The last inequality uses $|\theta|\leq B$, which follows because $\theta$ is an average of $Y$ with nonnegative weights and positive finite total mean weight.

Worked Example: A Pareto Tail Gives a Bias-Rate Requirement

Suppose $W$ has a Pareto distribution with $P(W>x)=x^{-3}$ for $x\geq1$, and $0\leq Y\leq1$. Find a sufficient cutoff rate for the clipping bias to be negligible at the root-n scale.

The mean weight is $\mu=3/2$. For $c\geq1$, the tail integral identity gives $$ E[(W-c)_+]=\int_c^\infty P(W>x)\,dx =\int_c^\infty x^{-3}\,dx =\frac{1}{2c^2}. $$ Here $B=1$, so the bias bound is $2B E[(W-c)_+]/\mu_n$. Since $\mu_n\to3/2$, it is of order $c_n^{-2}$. Therefore a sufficient condition for $\sqrt n|\theta_n-\theta|\to0$ is $$ \sqrt n\,c_n^{-2}\to0, \qquad\text{or equivalently}\qquad \frac{c_n}{n^{1/4}}\to\infty. $$ For example, $c_n=n^{0.30}$ satisfies this condition: $\sqrt n\,c_n^{-2}=n^{0.5-0.6}=n^{-0.1}\to0$. The Pareto distribution also has $E[W^2]=3<\infty$, so it permits the finite-variance central limit argument below. The bias calculation is sufficient, not necessary: if large weights are weakly associated with $Y-\theta$, the actual bias can be smaller.

A central limit theorem for the clipped estimator

A general statement can be made directly in terms of the clipped contributions. Define $$ X_{i,n}=W_{i,n}(Y_i-\theta_n), \qquad s_n^2=E[X_{i,n}^2]. $$ For every $n$, $E[X_{i,n}]=0$. The following result separates the normal approximation around the clipped target from the bias incurred by that target.

Theorem: Suppose $c_n\to\infty$, $0<E[W]<\infty$, $|Y|\leq B<\infty$, and $s_n^2\to s^2$ for some $0<s^2<\infty$. Suppose also that the triangular-array Lindeberg condition holds: for every $\varepsilon>0$, $$ E\!\left[X_{i,n}^2\,\mathbf{1}\{|X_{i,n}|>\varepsilon\sqrt n\}\right]\to0. $$ If $\sqrt n(\theta_n-\theta)\to0$, then $$ \sqrt n(\hat\theta_n-\theta)\xrightarrow{d} N\!\left(0,\frac{s^2}{\mu^2}\right), \qquad \mu=E[W]. $$

Proof. First, the sample denominator obeys $n^{-1}\sum_i W_{i,n}\xrightarrow{p}\mu$. To see this under only a finite first moment, note that $\mu_n\to\mu$. For any fixed $K>0$, the variables $\min(W_{i,n},K)$ are bounded by $K$, so their sample average minus its expectation converges to zero in probability by Chebyshev's inequality. The expected omitted tail is bounded uniformly in $n$ by $E[W\mathbf{1}\{W>K\}]$, which tends to zero as $K\to\infty$. Markov's inequality then shows that the sample average of $W_{i,n}$ minus $\mu_n$ converges to zero in probability.

Next, the Lindeberg–Feller central limit theorem for triangular arrays gives $$ \frac{1}{\sqrt n}\sum_{i=1}^n X_{i,n}\xrightarrow{d}N(0,s^2). $$ By the definition of $\hat\theta_n$, on the event of a positive sample denominator, $$ \sqrt n(\hat\theta_n-\theta_n) = \frac{n^{-1/2}\sum_{i=1}^n X_{i,n}}{n^{-1}\sum_{i=1}^n W_{i,n}}. $$ The denominator converges in probability to $\mu>0$, so Slutsky's theorem yields a centered normal limit with variance $s^2/\mu^2$. Finally, $$ \sqrt n(\hat\theta_n-\theta) = \sqrt n(\hat\theta_n-\theta_n)+\sqrt n(\theta_n-\theta), $$ and the second term tends to zero by assumption. Slutsky's theorem gives the stated limit. $\square$

This theorem's variance condition matters: clipping must not leave the contribution variance tending to zero or infinity if the desired limit is a nondegenerate normal distribution on the root-n scale. The Lindeberg condition rules out a small number of observations dominating the centered sum.

A practical sufficient condition and a threshold rule

Corollary: Under the setup above, suppose $E[W^2]<\infty$, $c_n\to\infty$, and $$ \sqrt n\,E[(W-c_n)_+]\to0. $$ If $E[W^2(Y-\theta)^2]>0$, then $$ \sqrt n(\hat\theta_n-\theta)\xrightarrow{d} N\!\left(0,\frac{E[W^2(Y-\theta)^2]}{\mu^2}\right). $$

Proof. Because $|Y-\theta_n|\leq2B$ and $W_{i,n}\leq W_i$, we have $|X_{i,n}|\leq2BW_i$. Also, $\theta_n\to\theta$, so $X_{i,n}\to W_i(Y_i-\theta)$ almost surely. The squared contributions are bounded by $4B^2W_i^2$, which is integrable. Dominated convergence therefore implies $$ s_n^2\to E[W^2(Y-\theta)^2]. $$ For any $\varepsilon>0$, $$ E[X_{i,n}^2\mathbf{1}\{|X_{i,n}|>\varepsilon\sqrt n\}] \leq 4B^2 E[W^2\mathbf{1}\{2BW>\varepsilon\sqrt n\}] \to0, $$ since $E[W^2]<\infty$. Thus the Lindeberg condition holds. The bias bound and $\mu_n\to\mu>0$ imply $\sqrt n|\theta_n-\theta|\to0$. All conditions of the theorem are satisfied. $\square$

A moment condition turns this bias requirement into a simple growth rate. If $E[W^q]<\infty$ for some $q>2$, then $$ E[(W-c)_+]\leq E[W\mathbf{1}\{W>c\}] \leq \frac{E[W^q]}{c^{q-1}}. $$ Consequently, a sufficient rate is $c_n/n^{1/[2(q-1)]}\to\infty$. The exponent is a sufficient bound based on the available moment, not a universal optimal cutoff. A sharper rate may follow from the actual tail or from cancellation in $E[(W-c_n)_+(Y-\theta)]$.

Worked Example: Verifying a Moment-Based Threshold

Suppose $0\leq Y\leq1$, $E[W^{5/2}]<\infty$, and choose $c_n=n^{0.34}$. Check whether the moment-based conditions give root-n normality.

Here $q=5/2$, so $q-1=3/2$ and the sufficient rate is $c_n/n^{1/3}\to\infty$. Since $0.34>1/3$, $$ \frac{c_n}{n^{1/3}}=n^{0.34-1/3}\to\infty. $$ Equivalently, the bound on the scaled bias is a constant times $$ \sqrt n\,c_n^{-3/2}=n^{0.5-(3/2)(0.34)}=n^{-0.01}\to0. $$ The $5/2$-moment implies $E[W^2]<\infty$, so the variance and Lindeberg conditions in the corollary also hold, provided $E[W^2(Y-\theta)^2]>0$. Thus this threshold is sufficient for a nondegenerate root-n normal limit. The small exponent margin also warns that finite-sample bias reduction may be gradual.

Worked Example: Exact Bias Cancellation Under Independence

Let $W$ have the Pareto distribution with $P(W>x)=x^{-3}$ for $x\geq1$, and let $Y$ be independent of $W$, with $P(Y=1)=P(Y=0)=1/2$. Take any deterministic $c_n\to\infty$.

Independence gives $E[W_{i,n}Y_i]=E[W_{i,n}]E[Y_i]=\mu_n/2$, so $\theta_n=1/2=\theta$ for every cutoff. There is no clipping bias, even though the cutoff may grow more slowly than the sufficient rate in the previous example. The centered contribution is $X_{i,n}=W_{i,n}(Y_i-1/2)$. By independence, $$ s_n^2=E[W_{i,n}^2]E[(Y_i-1/2)^2] =\frac14 E[W_{i,n}^2]\to\frac14 E[W^2]=\frac34. $$ Moreover, $E[W^2]=3$, and the finite-second-moment argument above verifies Lindeberg. With $\mu=3/2$, the limiting variance is $$ \frac{3/4}{(3/2)^2}=\frac{1}{3}. $$ Thus $\sqrt n(\hat\theta_n-1/2)\xrightarrow{d}N(0,1/3)$. This example shows why the tail-based bias bound is sufficient rather than necessary: the weighted outcome's independence makes the bias exactly zero.

When a root-n normal approximation fails

Finite mean weight alone is not enough for the corollary. Consider $P(W>x)=x^{-3/2}$ for $x\geq1$ and an independent Bernoulli outcome $Y$ with success probability $1/2$. Here $E[W]=3$, but $E[W^2]=\infty$. Independence still gives $\theta_n=1/2$ for every cutoff, so bias is absent.

For this Pareto tail, the clipped second moment is $$ E[\min(W,c)^2] =2\int_0^c xP(W>x)\,dx =1+2\int_1^c x^{-1/2}\,dx =4c^{1/2}-3. $$ Hence $s_n^2=\tfrac14(4c_n^{1/2}-3)$, which diverges. For $c_n=n^{0.1}$, the variance of the centered sum's root-n scaling grows on the order of $n^{0.05}$; the theorem's finite-$s^2$ condition fails. In particular, the finite-variance root-n normal conclusion cannot be inferred merely from zero clipping bias or from $c_n\to\infty$. A different normalization or stronger tail assumptions may be needed.

Diagnostics, alternatives, and limitations

A sound asymptotic argument checks the bias and random fluctuation separately. Useful diagnostics include the following:

Bias check: bound or estimate $E[(W-c_n)_+|Y-\theta|]$, not just $P(W>c_n)$. Rare exceedances can still carry substantial total weight.
Variance check: inspect whether large clipped contributions dominate. A growing cutoff can make the clipped second moment grow when the untruncated second moment is infinite.
Rate check: verify $\sqrt n|\theta_n-\theta|\to0$, not only $\theta_n\to\theta$. A slowly disappearing bias can shift a root-n confidence interval away from its nominal target.
Numerical check: track the largest clipped weights and the fraction of observations at the cap across sample sizes. A concentration of weighted contribution in a few observations signals that the normal approximation may be fragile.

A fixed cutoff is often useful for variance control, but it generally targets a clipped population quantity rather than $\theta$. An increasing cutoff can recover $\theta$ asymptotically when its residual bias is sufficiently small and the variance conditions hold. Using untruncated weights avoids clipping bias, but a root-n normal limit for the ordinary weighted estimator still requires suitable moment and central limit conditions. These are alternative trade-offs, not interchangeable guarantees.

The displayed results concern known weights and a bounded outcome with nonnegative weights. They do not establish the behavior of estimated weights, nor do they guarantee that a data-driven cutoff satisfies the deterministic rate conditions. Even when the asymptotic conditions hold, a threshold that is theoretically adequate may leave appreciable finite-sample bias or unstable uncertainty estimates. The cutoff rule and sensitivity to reasonable alternatives should therefore be examined in the application.

Common pitfall: “Because $c_n\to\infty$, clipping bias vanishes, so the clipped estimator is asymptotically normal around $\theta$.” This reasoning checks only consistency of the clipped target. The correction is to verify both a triangular-array central limit condition for the centered contributions and the stronger bias requirement $\sqrt n(\theta_n-\theta)\to0$. In addition, moment assumptions must ensure the relevant expectations and variance are finite; a finite mean weight does not imply a finite second moment.

Check Your Understanding

Answer each question, then compare with the solutions below.

Why does $c_n\to\infty$ alone not establish a root-n normal limit centered at $\theta$?
If $E[W^3]<\infty$, what sufficient threshold-rate condition follows from the moment bound?
For independent $W$ and $Y$, why can clipping bias be zero even when $E[(W-c_n)_+]$ is not small at the root-n rate?
What condition on the clipped contribution variance is violated in the Pareto example with tail exponent $3/2$?

Solutions. (1) The clipped target may approach $\theta$ too slowly: root-n inference requires $\sqrt n(\theta_n-\theta)\to0$, as well as a central limit condition for the centered contributions. (2) Set $q=3$ in $c_n/n^{1/[2(q-1)]}\to\infty$, giving $c_n/n^{1/4}\to\infty$. (3) Independence yields $E[W_{i,n}Y_i]=E[W_{i,n}]E[Y_i]$, so the clipped and untruncated ratios both equal $E[Y]$. (4) The clipped contribution variance $s_n^2$ diverges rather than converging to a finite positive limit.