Why Weights Change a Normal Limit
A sum of many independent observations often has an approximately normal distribution, but that conclusion depends on how much each observation contributes. If one term carries most of the variance, adding many small terms may not make the sum behave like an average of many comparable contributions. For weighted sums of iid variables, the key diagnostic is the largest absolute weight relative to the square root of the sum of squared weights.
The result below applies to deterministic weights and iid variables with finite, positive variance. It includes equal weights as a special case, but also covers unequal and signed weights. The normalization must reflect the actual variance of the weighted sum.
The absolute value matters because a large negative weight can dominate just as a large positive one can. The weights may vary with both \(i\) and \(n\), but they are not random in this result.
Variance Normalization and the Limit Theorem
Independence and centering give the variance directly: $$ \operatorname{Var}\left(\sum_{i=1}^n a_{ni}(X_i-\mu)\right) =\sum_{i=1}^n a_{ni}^2\operatorname{Var}(X_i) =\sigma^2 Q_n. $$ Thus \(T_n\) has mean zero and variance one for every \(n\). This exact variance calculation explains why dividing by \(\sigma\sqrt{Q_n}\), rather than automatically by \(\sqrt n\), is essential.
The condition says that no single summand accounts for a substantial fraction of the total variance: the variance fraction of the largest-weight term is at most \(\max_i a_{ni}^2/Q_n\), which tends to zero. Here is a proof using the Lindeberg–Feller central limit theorem for triangular arrays. In the form needed, that theorem says that for each row of independent, mean-zero variables with total variance one, the row sum converges in distribution to \(N(0,1)\) if its Lindeberg sum tends to zero for every \(\varepsilon>0\).
Write \(Y_i=(X_i-\mu)/\sigma\) and \(c_{ni}=a_{ni}/\sqrt{Q_n}\). The row variables \(c_{ni}Y_i\) are independent and centered, and \(\sum_i c_{ni}^2=1\). Their row sum is \(T_n\).
Let \(m_n=\max_i|c_{ni}|\). For every \(\varepsilon>0\), using \(E[Y_i^2]=1\) and identical distributions, $$ \sum_{i=1}^n E\left[c_{ni}^2Y_i^2\mathbf{1}\{|c_{ni}Y_i|>\varepsilon\}\right] \le E\left[Y_1^2\mathbf{1}\{|Y_1|>\varepsilon/m_n\}\right]. $$ The inequality holds because \(|c_{ni}|\le m_n\), so the event \(|c_{ni}Y_i|>\varepsilon\) implies \(|Y_i|>\varepsilon/m_n\); the coefficients' squared values sum to one.
The maximal-weight condition gives \(m_n\to0\). Since \(Y_1^2\) is integrable, dominated convergence applied to \(Y_1^2\mathbf{1}\{|Y_1|>\varepsilon/m_n\}\) shows that its expectation tends to zero. The Lindeberg condition therefore holds for every \(\varepsilon>0\).
The Lindeberg–Feller theorem applies to these independent triangular-array rows, whose total variance is one. Therefore \(T_n\xrightarrow{d}N(0,1)\). \(\square\)
Worked Examples
Worked Example: Equal Weights Recover the Usual IID CLT
Let \(X_1,X_2,\ldots\) be iid exponential random variables with rate \(1\). Then \(\mu=1\) and \(\sigma^2=1\). Give each of the first \(n\) observations weight \(a_{ni}=1\).
Here \(Q_n=\sum_{i=1}^n 1^2=n\), and the maximal-weight ratio is $$ \frac{\max_i |a_{ni}|}{\sqrt{Q_n}}=\frac{1}{\sqrt n}\longrightarrow 0. $$ The weighted statistic is $$ T_n=\frac{\sum_{i=1}^n(X_i-1)}{\sqrt n} =\frac{\sum_{i=1}^nX_i-n}{\sqrt n}. $$ The theorem gives \(T_n\xrightarrow{d}N(0,1)\). The variance check agrees: the numerator has variance \(n\), so the denominator \(\sqrt n\) makes the variance one. This is the usual iid central limit theorem expressed as a weighted-sum result.
Worked Example: Increasing Weights with Rademacher Variables
Let \(X_i\) be iid with \(P(X_i=1)=P(X_i=-1)=1/2\). Then \(\mu=0\) and \(\sigma^2=1\). Give observation \(i\) weight \(a_{ni}=i\), for \(i=1,\ldots,n\).
The sum of squared weights is $$ Q_n=\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}, $$ and the largest absolute weight is \(n\). Hence $$ \frac{\max_i|a_{ni}|}{\sqrt{Q_n}} =\frac{n}{\sqrt{n(n+1)(2n+1)/6}} \longrightarrow 0, $$ because its square is \(6n/((n+1)(2n+1))\), which tends to zero. Therefore $$ \frac{\sum_{i=1}^n iX_i}{\sqrt{n(n+1)(2n+1)/6}}\xrightarrow{d}N(0,1). $$ The largest weight grows with \(n\), but that alone does not prevent a normal limit: the total squared weight grows faster, so no single observation dominates on the normalized scale.
Worked Example: A Dominant Weight Can Preserve a Nonnormal Limit
Again let \(X_i\) be iid Rademacher variables, with mean zero and variance one. Now assign weight \(n\) to \(X_1\) and weight \(1\) to each of \(X_2,\ldots,X_n\). The normalized sum is $$ T_n=\frac{nX_1+\sum_{i=2}^nX_i}{\sqrt{n^2+n-1}}, $$ since \(Q_n=n^2+(n-1)\). The maximal-weight ratio is $$ \frac{n}{\sqrt{n^2+n-1}}\longrightarrow 1, $$ so the theorem's condition fails.
In fact, \(T_n\) does not converge to a standard normal distribution. The first coefficient \(n/\sqrt{n^2+n-1}\) tends to one. The remaining term has mean zero and variance $$ \operatorname{Var}\left(\frac{\sum_{i=2}^nX_i}{\sqrt{n^2+n-1}}\right) =\frac{n-1}{n^2+n-1}\longrightarrow 0. $$ Chebyshev's inequality therefore shows that this remaining term converges to zero in probability. By Slutsky's theorem, \(T_n\xrightarrow{d}X_1\), a two-point Rademacher distribution, not \(N(0,1)\). The large first weight leaves a nonvanishing contribution from a single nonnormal observation.
What the Condition Does—and Does Not—Say
The maximal-weight condition is a sufficient condition for a normal limit under the stated finite-variance assumption. It is useful because it does not require a third moment, a fourth moment, or a special form for the distribution of \(X_i\). It ensures that each normalized summand becomes small, while the total variance remains one.
The condition is not necessary for every particular distribution. For example, if the \(X_i\) are themselves normal, any nonzero deterministic weighted sum is exactly normal after centering and variance normalization, even if one weight dominates. The condition instead supplies a general guarantee that works for all iid distributions with finite, positive variance. When it fails, a normal limit may still occur for a special model, but it cannot be concluded from this theorem alone.
A frequent mistake is to normalize every weighted sum by \(\sqrt n\), as if all weights were one. The correct variance scale is \(\sigma\sqrt{\sum_i a_{ni}^2}\). Another is to inspect only the number of summands: having \(n\) terms is not enough if one term accounts for nearly all the variance. Finally, if weights are signed, use the largest absolute weight in the condition.
Check Your Understanding
Use the variance normalization and maximal-weight condition to assess each situation.
- For equal weights \(a_{ni}=1\), what is the maximal-weight ratio, and why does it tend to zero?
- If the weights are \(a_{ni}=i\), what is \(Q_n\), and how does the largest-weight ratio behave?
- Why is \(\max_i|a_{ni}|\), rather than \(\max_i a_{ni}\), used when weights may be negative?
- For weights \(n,1,\ldots,1\), explain why the normalized contribution from the terms after the first tends to zero in probability for Rademacher variables.
- Does failure of the maximal-weight condition prove that a normal limit is impossible? Explain using the case of normally distributed \(X_i\).