Triangular Arrays and Their Variance Normalization

Why Triangular Arrays?

The usual central limit setup studies longer and longer sums of observations from one fixed distribution. Many asymptotic problems instead involve observations whose distributions change with the sample size. In row \(n\), for example, there may be \(n\) Bernoulli variables with success probabilities that differ across positions and change as \(n\) grows. A triangular array represents this structure: each row is a finite collection of random variables, and the distributions may vary within and between rows.

Before deciding whether a normalized sum approaches a normal distribution, we need to define the row sum, its center, and its variance. The standardization is not a conclusion about the limiting distribution. It is a way to put the sums on a common scale so that a central limit theorem can be considered.

Definition and Row Notation

Definition: A triangular array is a collection of random variables \(\{X_{n,k}:1\leq k\leq k_n,\ n\geq1\}\), where row \(n\) contains \(k_n\) variables. It is an independent triangular array if, for each fixed \(n\), the variables \(X_{n,1},\ldots,X_{n,k_n}\) are independent. Write \(F_{n,k}\) for the distribution of \(X_{n,k}\). The distributions may vary with both indices. Independence between different rows is not required for results about each row sum.

The row index \(n\) indicates which sum is being considered; the column index \(k\) identifies a term within that row. Often \(k_n=n\), but the definition allows other row lengths. The variables need not have the same distribution in a row, and a variable in one row need not share a distribution with a variable in another row. This is the main structural difference from an identically distributed sample.

Assume for now that each variable has a finite mean and variance. Define the row sum, its mean, and its total variance by

S_n=\sum_{k=1}^{k_n}X_{n,k},\qquad m_n=E[S_n]=\sum_{k=1}^{k_n}E[X_{n,k}],\qquad s_n^2=\operatorname{Var}(S_n)=\sum_{k=1}^{k_n}\operatorname{Var}(X_{n,k}).

The mean and variance formulas use independence within the row. They do not require identical distributions. If \(s_n^2>0\), the centered, standardized row sum is

Z_n=\frac{S_n-m_n}{s_n}.

Here \(s_n\) is the positive square root of the total variance. The normalization gives each \(Z_n\) mean zero and variance one. If \(s_n^2=0\), the row sum is constant with probability one, and this standardization is not defined.

Why the Total Variance Adds

Proposition: If \(X_{n,1},\ldots,X_{n,k_n}\) are independent with finite second moments, then the variance of their sum is the sum of their variances.

Proof. Fix a row and abbreviate \(X_{n,k}\) by \(X_k\). Expanding the squared centered sum gives

\operatorname{Var}\left(\sum_{k=1}^{k_n}X_k\right) =E\left[\left(\sum_{k=1}^{k_n}(X_k-E[X_k])\right)^2\right] =\sum_{k=1}^{k_n}\operatorname{Var}(X_k) +2\sum_{1\leq i<j\leq k_n}\operatorname{Cov}(X_i,X_j).

For \(i\neq j\), independence implies \(E[X_iX_j]=E[X_i]E[X_j]\). These expectations exist because the variables have finite second moments and hence finite first moments. Therefore \(\operatorname{Cov}(X_i,X_j)=0\) for every distinct pair. All cross terms vanish, leaving the sum of the individual variances. \(\square\)

For a dependent row, the covariance terms generally remain. In that case the sum of individual variances alone is not the variance of the row sum, so the normalization must use the full variance, including covariances.

Centering and Scaling Each Row

Proposition: For an independent row with finite variances and \(s_n^2>0\), define \(\xi_{n,k}=(X_{n,k}-E[X_{n,k}])/s_n\). Then the \(\xi_{n,k}\) are independent, \(E[\xi_{n,k}]=0\), and \[ \sum_{k=1}^{k_n}\operatorname{Var}(\xi_{n,k})=1. \] Moreover, \(\sum_{k=1}^{k_n}\xi_{n,k}=Z_n\).

Proof. Centering and multiplying by a nonzero constant are measurable transformations applied separately to the variables in a row, so independence is preserved. Linearity of expectation gives

E[\xi_{n,k}] =\frac{E[X_{n,k}]-E[X_{n,k}]}{s_n}=0.

Scaling a variance by a constant \(c\) multiplies it by \(c^2\), so

\sum_{k=1}^{k_n}\operatorname{Var}(\xi_{n,k}) =\frac{1}{s_n^2}\sum_{k=1}^{k_n}\operatorname{Var}(X_{n,k}) =\frac{s_n^2}{s_n^2}=1.

Finally, summing the centered and scaled variables yields \(\sum_k \xi_{n,k}=(S_n-m_n)/s_n=Z_n\). Thus standardizing the sum is equivalent to centering and scaling every entry by the same row-specific standard deviation. \(\square\)

The individual normalized variances add to one, but need not be equal. In particular, variance normalization does not mean each entry has variance \(1/k_n\); that equality holds only in special cases.

Worked Normalizations

Worked Example: Four Bernoulli Variables With Different Probabilities

Let \(X_{4,1},\ldots,X_{4,4}\) be independent Bernoulli variables with success probabilities \(0.1,0.3,0.6,0.8\), respectively. A Bernoulli variable with success probability \(p\) has mean \(p\) and variance \(p(1-p)\). Find the row mean, total variance, and standardized sum.

The row mean is

$$ m_4=0.1+0.3+0.6+0.8=1.8. $$

The four variances are \(0.1(0.9)=0.09\), \(0.3(0.7)=0.21\), \(0.6(0.4)=0.24\), and \(0.8(0.2)=0.16\). Independence gives

s_4^2=0.09+0.21+0.24+0.16=0.70,\qquad s_4=\sqrt{0.70}.

Writing \(S_4=\sum_{k=1}^4X_{4,k}\), the standardized sum is

Z_4=\frac{S_4-1.8}{\sqrt{0.70}}.

It has mean zero and variance one. The variables have different success probabilities, so treating them as four observations from one common Bernoulli distribution would misrepresent the row.

Worked Example: A Row With a Growing Number of Rare Events

For each \(n\geq3\), let \(X_{n,1},\ldots,X_{n,n}\) be independent Bernoulli variables, each with success probability \(2/n\). Find the mean and variance of the row sum and write its standardized form.

Each variable has mean \(2/n\), so \(m_n=n(2/n)=2\). Its variance is \((2/n)(1-2/n)\). Therefore,

s_n^2=n\frac{2}{n}\left(1-\frac{2}{n}\right) =2\left(1-\frac{2}{n}\right).

For \(n\geq3\), this variance is positive. Thus

Z_n= \frac{\sum_{k=1}^{n}X_{n,k}-2} {\sqrt{2(1-2/n)}}.

For example, at \(n=4\), the success probability is \(1/2\), the row mean is \(2\), and \(s_4^2=2(1-1/2)=1\). Consequently \(Z_4=S_4-2\), which indeed has mean zero and variance one. The normalization is valid for every \(n\geq3\); it does not by itself establish any limiting distribution.

Worked Example: Unequal Gaussian Variances

Suppose a row consists of three independent variables with distributions \(X_{n,1}\sim N(2,1)\), \(X_{n,2}\sim N(-1,4)\), and \(X_{n,3}\sim N(4,9)\). Find the standardized row sum and its distribution.

The row mean and variance are

m_n=2+(-1)+4=5,\qquad s_n^2=1+4+9=14.

The sum of independent normal variables is normal, so \(S_n\sim N(5,14)\). Therefore

Z_n=\frac{S_n-5}{\sqrt{14}}\sim N(0,1).

This exact normal distribution follows from the specific normal model, not merely from the fact that the sum has been centered and given variance one. The example distinguishes an algebraic normalization from a theorem asserting convergence to a normal law.

What Normalization Does—and Does Not—Tell Us

In central limit arguments for independent arrays, the sequence \(Z_n\) is the usual candidate for comparison with a standard normal random variable. The earlier material on the Lyapunov central limit theorem uses the same basic practice of centering and scaling a sum. For arrays, however, the terms can vary substantially across a row and from one row to the next. Later conditions determine whether this normalized sequence converges in distribution to a normal law.

Normalization alone gives only \(E[Z_n]=0\) and \(\operatorname{Var}(Z_n)=1\). Many distributions have those two moments without being normal. Even for a sequence of such variables, the two moment identities do not imply convergence in distribution. One must establish the relevant theorem's hypotheses separately.

Common Mistake: Using the Wrong Variance

A common error is to divide by the square root of the number of terms, as if every row had unit variance per term. The correct denominator is \(s_n\), the square root of the sum of the row's individual variances, provided the row variables are independent. In the first example, \(\sqrt{4}=2\) is not the correct scale: the total variance is \(0.70\), so the correct scale is \(\sqrt{0.70}\).

Another error is to infer a normal limit from variance one. The correction is to treat centering and variance normalization as the setup for a limit theorem, not as proof of its conclusion. First calculate the row-specific mean and variance; then check the additional conditions required by the theorem being used.

Key takeaway: For each independent row, compute its own mean \(m_n\) and total variance \(s_n^2\). When \(s_n^2>0\), standardize with \(Z_n=(S_n-m_n)/s_n\). This puts the row sums on a common mean-zero, variance-one scale, but does not alone guarantee a normal limit.

Check Your Understanding

Use the row-specific definitions and formulas to answer these questions.

In an independent triangular array, must variables within the same row have the same distribution? Must variables from different rows be independent?
A row contains independent variables with variances \(2\), \(3\), and \(5\). What is the variance-normalizing denominator for its sum?
Why is the sum of individual variances not generally the variance of a dependent row sum?
For a row with positive total variance, define normalized individual terms so that their sum equals the standardized row sum.
Does having mean zero and variance one prove that a random variable has a standard normal distribution? Explain.