Null Limits and Variance Estimation for Quadratic Tests

Why Null Calibration Requires More Than a Variance

A quadratic mean statistic measures whether the sample mean vector is far from zero without relying on an inverse covariance matrix. To turn it into a test, however, one needs both its null variance and a justified reference distribution. In high dimensions a standard normal approximation can be appropriate, but it is not automatic: it depends on how the covariance is distributed across directions.

Let $X_1,\ldots,X_n$ be independent and identically distributed $p$-dimensional random vectors, with mean vector $\mu=E[X_i]$ and covariance matrix $\Sigma=\operatorname{Cov}(X_i)$. The null hypothesis considered here is $H_0:\mu=0$. Coordinate variances may differ: the diagonal entries of $\Sigma$ need not be equal. Coordinates within a vector may also be correlated.

Definition: For $n\geq 2$, define the quadratic U-statistic $$ Q_n=\frac{1}{n(n-1)}\sum_{i\ne j}X_i^\top X_j =\frac{2}{n(n-1)}\sum_{1\leq i<j\leq n}X_i^\top X_j. $$ Write $\tau=\operatorname{tr}(\Sigma^2)$, where $\operatorname{tr}$ denotes the matrix trace. When $H_0$ holds and $\tau>0$, the null variance of $Q_n$ is $2\tau/[n(n-1)]$.

The statistic excludes the terms $X_i^\top X_i$. Under the null, each cross-product has mean zero, and cross-products from distinct pairs are uncorrelated. This degeneracy is why the variance has a factor $n(n-1)$ in the denominator and why the limiting distribution needs particular care.

Null Mean and Variance

Assume $E\|X_i\|^2<\infty$. Under $H_0$, independence gives $E[X_i^\top X_j]=E[X_i]^\top E[X_j]=0$ for $i\ne j$, so $E[Q_n]=0$. For the variance, put $h_{ij}=X_i^\top X_j$ for $i<j$. If two pairs have no index in common, their cross-products are independent and have mean zero. If they share one index, for example $(i,j)$ and $(i,k)$, then conditioning on $X_i$ gives $$ E[h_{ij}h_{ik}\mid X_i] = X_i^\top E[X_j]\,E[X_k]^\top X_i=0. $$ Thus distinct pair terms are uncorrelated.

For a single pair, independence and the covariance definition yield $$ E[h_{ij}^2] =E[X_i^\top X_jX_j^\top X_i] =E[X_i^\top\Sigma X_i] =\operatorname{tr}(\Sigma^2)=\tau. $$ There are $n(n-1)/2$ unordered pairs in $Q_n$, each with coefficient $2/[n(n-1)]$. Consequently,

\operatorname{Var}(Q_n) =\left(\frac{2}{n(n-1)}\right)^2 \frac{n(n-1)}{2}\tau =\frac{2\tau}{n(n-1)}.

This variance is exact under the null, not a large-sample approximation. It accounts for unequal coordinate variances and any within-vector covariance through $\tau$. For example, if $\Sigma$ is diagonal with entries $\sigma_1^2,\ldots,\sigma_p^2$, then $\tau=\sum_{k=1}^p\sigma_k^4$, not a multiple of $p$ unless the coordinate variances happen to be equal.

Worked Example: Unequal Coordinate Variances

Suppose the observations are $N(0,\Sigma)$ vectors with $\Sigma=\operatorname{diag}(1,4)$, and $n=10$. The two coordinate variances are 1 and 4, so $$ \tau=\operatorname{tr}(\Sigma^2)=1^2+4^2=17. $$

The exact null variance and standard deviation are $$ \operatorname{Var}(Q_n)=\frac{2(17)}{10(9)}=\frac{17}{45}\approx 0.3778, \qquad \operatorname{SD}(Q_n)=\sqrt{\frac{17}{45}}\approx 0.615. $$

The contribution of the second coordinate to $\tau$ is $16/17$, even though it is only one of two coordinates. Treating coordinates as if they had a common variance would miss this substantial difference in the null scale.

When the Standard Normal Calibration Holds

The exact variance alone does not imply that $Q_n$, divided by its standard deviation, is approximately normal. A general route to a central limit theorem uses a martingale decomposition. Under $H_0$, let $\mathcal F_j$ be the information in $X_1,\ldots,X_j$, and define $$ D_j=\left(\sum_{i<j}X_i\right)^\top X_j,\qquad S_n=\sum_{j=2}^nD_j. $$ Then $Q_n=2S_n/[n(n-1)]$. Independence and the zero mean imply $E[D_j\mid\mathcal F_{j-1}]=0$, so the $D_j$ are martingale differences. Their total unconditional variance is $$ s_n^2=\sum_{j=2}^n E[D_j^2] =\frac{n(n-1)}{2}\tau. $$

Theorem: Suppose $H_0$ holds, $0<\tau<\infty$, and $E\|X_i\|^4<\infty$. If, as the sample size and possibly the dimension vary, $$ \frac{\sum_{j=2}^n E[D_j^2\mid\mathcal F_{j-1}]}{s_n^2}\xrightarrow{p}1 \quad\text{and}\quad \frac{\sum_{j=2}^n E[D_j^4]}{s_n^4}\longrightarrow 0, $$ then $$ \frac{Q_n}{\sqrt{2\tau/[n(n-1)]}}\xrightarrow{d}N(0,1). $$

These are sufficient martingale central limit conditions. The first requires the conditional variance, given the observations so far, to be close in probability to its nonrandom target. The second is a Lyapunov-type condition that prevents a small number of unusually large increments from dominating the sum. To see why the theorem applies, the martingale central limit theorem gives $S_n/s_n\xrightarrow{d}N(0,1)$ under these two conditions. Since $Q_n=2S_n/[n(n-1)]$ and $2s_n/[n(n-1)]=\sqrt{2\tau/[n(n-1)]}$, the displayed conclusion follows.

The conditions are meaningful restrictions, not formalities. One useful Gaussian case makes their role visible. If $X_i\sim N(0,\Sigma)$, then, writing $r_4=\operatorname{tr}(\Sigma^4)$, direct Gaussian moment calculations bound the relative conditional-variance fluctuation by a constant times $1/n+r_4/\tau^2$, and give $$ \frac{\sum_{j=2}^nE[D_j^4]}{s_n^4} \leq \frac{C}{n}\left(1+\frac{r_4}{\tau^2}\right) $$ for an absolute constant $C$. Thus, when $n\to\infty$ and $r_4/\tau^2\to0$, both conditions hold. Since $r_4/\tau^2$ measures how concentrated the squared covariance eigenvalues are, this condition is satisfied when no single covariance direction dominates the total quadratic scale.

Worked Example: A Gaussian Spectrum That Supports Normal Calibration

Suppose $\Sigma=I_p$ and the observations are independent $N(0,I_p)$ vectors. Then $\tau=p$ and $\operatorname{tr}(\Sigma^4)=p$, so $$ \frac{\operatorname{tr}(\Sigma^4)}{\tau^2}=\frac{p}{p^2}=\frac{1}{p}. $$ If both $n\to\infty$ and $p\to\infty$, the Gaussian bounds above tend to zero: the conditional-variance fluctuation is bounded by a constant times $1/n+1/p$, and the fourth-moment ratio by a constant times $(1+1/p)/n$. Hence the theorem gives $$ \frac{Q_n}{\sqrt{2p/[n(n-1)]}}\xrightarrow{d}N(0,1). $$

This calculation illustrates a high-dimensional calibration regime. It does not say that every quadratic statistic is normal merely because the number of coordinates is large; the covariance spectrum and the sampling conditions matter.

Estimating the Null Variance

The null variance depends on $\tau=\operatorname{tr}(\Sigma^2)$, which is usually unknown. Under $H_0$, a natural estimator uses squared cross-products: $$ \widehat{\tau}=\frac{1}{n(n-1)}\sum_{i\ne j}(X_i^\top X_j)^2 =\frac{1}{\binom n2}\sum_{i<j}(X_i^\top X_j)^2. $$ For independent observations with mean zero and finite second moments, each squared cross-product has expectation $\tau$, by the calculation above. Therefore $E[\widehat{\tau}]=\tau$. An unbiased variance estimator is $$ \widehat{\operatorname{Var}}_0(Q_n)=\frac{2\widehat{\tau}}{n(n-1)}. $$

Unbiasedness is not the same as consistency. A useful setting in which consistency can be checked explicitly is the Gaussian null. Put $W_{ij}=(X_i^\top X_j)^2$ for $i<j$, and $r_4=\operatorname{tr}(\Sigma^4)$. Gaussian moment calculations give $$ \operatorname{Var}(W_{12})=2\tau^2+6r_4, \qquad \operatorname{Cov}(W_{12},W_{13})=2r_4. $$ Disjoint pairs are independent. There are $M=\binom n2$ pairs, and each pair overlaps with $2(n-2)$ other pairs in the ordered sum for the variance of their average. Therefore $$ \operatorname{Var}(\widehat{\tau}) =\frac{2\tau^2+6r_4}{M} +\frac{4(n-2)r_4}{M}. $$ Because $r_4\leq\tau^2$, it follows that $\operatorname{Var}(\widehat{\tau})/\tau^2\to0$ as $n\to\infty$. Chebyshev’s inequality then gives $\widehat{\tau}/\tau\xrightarrow{p}1$. In this Gaussian setting, replacing $\tau$ by $\widehat{\tau}$ in the standard error is justified by this consistency and Slutsky’s theorem.

Worked Example: Calculating the Cross-Product Estimate

For four observed vectors in two dimensions, let $X_1=(1,0)^\top$, $X_2=(0,2)^\top$, $X_3=(-1,1)^\top$, and $X_4=(2,-1)^\top$. The six unordered inner products are $$ 0,\quad -1,\quad 2,\quad 2,\quad -2,\quad -3. $$ Their squared values sum to $0+1+4+4+4+9=22$. Thus $$ \widehat{\tau}=\frac{22}{\binom42}=\frac{11}{3}, \qquad \widehat{\operatorname{Var}}_0(Q_n) =\frac{2(11/3)}{4(3)}=\frac{11}{18}. $$

The inner products themselves sum to $-2$, so $Q_n=2(-2)/[4(3)]=-1/3$. The resulting standardized value is $$ \frac{Q_n}{\sqrt{\widehat{\operatorname{Var}}_0(Q_n)}} =\frac{-1/3}{\sqrt{11/18}}\approx -0.426. $$ This is a calculation of the statistic and estimated null scale; interpreting it with a normal reference distribution still requires suitable asymptotic conditions.

A Fixed-Dimension Pitfall

A common mistake is to infer normality from the central limit theorem for the sample mean alone. A quadratic function of a sample mean does not automatically have a normal limit. For a concrete check, take $p=1$ and $X_i\sim N(0,\sigma^2)$. Here $\tau=\sigma^4$, and the covariance-concentration ratio is $\operatorname{tr}(\Sigma^4)/\tau^2=1$, not a quantity tending to zero. In fact, with $Z_i=X_i/\sigma$, $$ Q_n=\frac{\sigma^2}{n(n-1)} \left\{\left(\sum_{i=1}^n Z_i\right)^2-\sum_{i=1}^n Z_i^2\right\}. $$ The first squared sum divided by $n$ has a chi-squared-one distribution, while $n^{-1}\sum_i Z_i^2\xrightarrow{p}1$. Consequently, $$ \frac{Q_n}{\sigma^2\sqrt{2/[n(n-1)]}} \xrightarrow{d}\frac{Z^2-1}{\sqrt{2}}, \qquad Z\sim N(0,1), $$ which is not standard normal. The correction is to check the limit conditions for the quadratic statistic itself, rather than assume that a normal limit for a linear statistic carries over to its square.

Key takeaway: Under the null, the exact variance is $2\operatorname{tr}(\Sigma^2)/[n(n-1)]$, with $\operatorname{tr}(\Sigma^2)$ accommodating unequal coordinate variances. Normal calibration requires additional conditions, and a cross-product estimator provides an unbiased estimate of the trace term; consistency needs its own justification.

Check Your Understanding

Use the definitions and conditions above to answer the following questions.

Why are two cross-products sharing one observation uncorrelated under the null?
If $\Sigma=\operatorname{diag}(2,3)$, what are $\tau$ and the null variance of $Q_n$ for a given $n$?
What does the conditional-variance condition in the martingale theorem require?
Why does unbiasedness of $\widehat{\tau}$ alone not establish that it can replace $\tau$ in a standard error?
In the univariate Gaussian example, which covariance-spectrum condition fails, and what is the limiting standardized distribution?