Power of Quadratic Tests Under Dense Alternatives

Why Power Depends on More Than the Mean Norm

A quadratic mean test is designed to detect a mean vector whose signal may be spread across many coordinates, with no single coordinate carrying most of the change. Its power depends on how large that vector is compared with the random fluctuation of the test statistic. In high dimensions, that fluctuation reflects the covariance matrix as well as the sample size and dimension.

We study the one-sample model $X_i=\mu_n+\varepsilon_i$, for $i=1,\ldots,n$, where $X_i$ and $\mu_n$ are $p_n$-dimensional, $E[\varepsilon_i]=0$, and $\operatorname{Cov}(\varepsilon_i)=\Sigma_n$. The null hypothesis is $\mu_n=0$. A dense alternative here means a sequence of mean vectors whose signal is distributed across coordinates; the analysis below does not require exact equality of their coordinates.

Definition: The quadratic U-statistic for testing a zero mean is $$ Q_n=\frac{1}{n(n-1)}\sum_{i\ne j}X_i'X_j. $$ Its null variance is $v_{0,n}^2=2\operatorname{tr}(\Sigma_n^2)/[n(n-1)]$, provided $\operatorname{tr}(\Sigma_n^2)>0$. The corresponding standardized statistic is $Q_n/\widehat v_{0,n}$, where $\widehat v_{0,n}$ estimates the null standard deviation $v_{0,n}$.

The statistic avoids squaring each observation’s noisy sample mean separately: its cross-products use distinct observations. Its expectation is $\|\mu_n\|^2$, so positive mean shifts move it toward the upper tail. But the relevant comparison is not simply $\|\mu_n\|^2$ against zero. It is the signal relative to $v_{0,n}$, which incorporates $\operatorname{tr}(\Sigma_n^2)$, the aggregate squared covariance eigenvalues.

Decomposing the Statistic Under an Alternative

Write $\bar\varepsilon=n^{-1}\sum_i\varepsilon_i$, and define the centered quadratic component $$ D_n=\frac{1}{n(n-1)}\sum_{i\ne j}\varepsilon_i'\varepsilon_j. $$ Expanding each cross-product gives the exact decomposition

Q_n=\|\mu_n\|^2+2\mu_n'\bar\varepsilon+D_n.

The first term is the signal. The second is a random linear term, and the third is the centered quadratic fluctuation. Since the errors are independent with mean zero and covariance $\Sigma_n$, $E[D_n]=0$, $\operatorname{Var}(D_n)=v_{0,n}^2$, and $\operatorname{Var}(2\mu_n'\bar\varepsilon)=4\mu_n'\Sigma_n\mu_n/n$. These two random terms are uncorrelated: every covariance between a linear error term and a product of two distinct centered errors has expectation zero by independence and zero means.

The statistic’s expected shift in null-standard-deviation units is

\delta_n=\frac{\|\mu_n\|^2}{v_{0,n}} =\|\mu_n\|^2\sqrt{\frac{n(n-1)}{2\operatorname{tr}(\Sigma_n^2)}}.

This ratio suggests the detection scale. A signal with $\delta_n$ tending to zero is small relative to the null fluctuation; one with a finite positive limit gives a nontrivial power limit; one with a diverging ratio is larger than the null standard deviation. A finite ratio alone does not imply that the alternative distribution is a shifted standard normal: the random linear term must also be negligible on the null-standard-deviation scale.

Limiting Power Under Local Dense Alternatives

Theorem: For each $n$, suppose $X_i=\mu_n+\varepsilon_i$, where the $\varepsilon_i$ are independent and identically distributed, have mean zero and covariance $\Sigma_n$, and have the same distribution under the null and alternative at that $n$. Suppose $\operatorname{tr}(\Sigma_n^2)>0$. Assume the centered quadratic component satisfies the null limit $D_n/v_{0,n}\xrightarrow{d}N(0,1)$, and that $\widehat v_{0,n}/v_{0,n}\xrightarrow{p}1$ under both the null and the alternatives. If $\delta_n\to\delta\in[0,\infty)$ and $n\mu_n'\Sigma_n\mu_n/\operatorname{tr}(\Sigma_n^2)\to0$, then the test that rejects when $Q_n/\widehat v_{0,n}>z_{1-\alpha}$ has asymptotic size $\alpha$ and limiting rejection probability $1-\Phi(z_{1-\alpha}-\delta)$ under the alternatives.

The null limit and consistency of the variance estimator are stated as assumptions. In applications, the null limit is supplied by the regularity conditions for the quadratic test, while the estimator’s consistency must be checked for the estimator actually used. In particular, a consistency statement under alternatives does not by itself establish null calibration.

Proof. Under the null, $\mu_n=0$ and $Q_n=D_n$. Thus $Q_n/v_{0,n}\xrightarrow{d}N(0,1)$. Since $\widehat v_{0,n}/v_{0,n}\xrightarrow{p}1$ under the null, Slutsky’s theorem gives $Q_n/\widehat v_{0,n}\xrightarrow{d}N(0,1)$. Its upper-tail rejection probability therefore tends to $\alpha$.

Under the alternatives, the variance of the linear term divided by the null variance is

\frac{\operatorname{Var}(2\mu_n'\bar\varepsilon)}{v_{0,n}^2} =\frac{4\mu_n'\Sigma_n\mu_n/n}{2\operatorname{tr}(\Sigma_n^2)/[n(n-1)]} =\frac{2(n-1)\mu_n'\Sigma_n\mu_n}{\operatorname{tr}(\Sigma_n^2)} \longrightarrow 0.

Chebyshev’s inequality consequently gives $2\mu_n'\bar\varepsilon/v_{0,n}\xrightarrow{p}0$. Because the centered errors have the same distribution under the null and alternatives, the assumed null limit for $D_n/v_{0,n}$ also holds under the alternatives. The decomposition now yields $Q_n/v_{0,n}=\delta_n+2\mu_n'\bar\varepsilon/v_{0,n}+D_n/v_{0,n}\xrightarrow{d}N(\delta,1)$. Consistency of $\widehat v_{0,n}/v_{0,n}$ under the alternatives and Slutsky’s theorem preserve this limit after studentization. The limiting upper-tail probability is $P\{N(\delta,1)>z_{1-\alpha}\}=1-\Phi(z_{1-\alpha}-\delta)$. $\square$

Worked Examples

Worked Example: The Detection Scale With Identity Covariance

Let $p_n=n$, $\Sigma_n=I_{p_n}$, and let every coordinate of $\mu_n$ equal $\sqrt{c}\,n^{-3/4}$, where $c>0$ is fixed. Assume the theorem’s limit and variance-estimation conditions hold.

Here $\operatorname{tr}(\Sigma_n^2)=p_n=n$, and the squared signal norm is $\|\mu_n\|^2=n(cn^{-3/2})=c n^{-1/2}$. Hence $v_{0,n}^2=2n/[n(n-1)]=2/(n-1)$, and

\delta_n=\frac{c n^{-1/2}}{\sqrt{2/(n-1)}} =c\sqrt{\frac{n-1}{2n}}\longrightarrow\frac{c}{\sqrt{2}}.

The signal is dense: all $n$ coordinates are nonzero, and each coordinate tends to zero. The linear-term condition also holds, because $2(n-1)\mu_n'\Sigma_n\mu_n/\operatorname{tr}(\Sigma_n^2) =2(n-1)c n^{-1/2}/n\to0$. The limiting power is therefore $1-\Phi(z_{1-\alpha}-c/\sqrt{2})$. In this identity-covariance case, the nontrivial detection scale is $\|\mu_n\|^2$ of order $\sqrt{p_n}/n$.

Worked Example: How the Covariance Spectrum Changes the Scale

Let $p_n=n$ be even, and take $\Sigma_n$ diagonal with half its diagonal entries equal to $1$ and half equal to $4$. Distribute the mean so that all coordinates have the same squared magnitude, and choose its norm to satisfy $\|\mu_n\|^2=\delta v_{0,n}$ for a fixed $\delta>0$.

The squared-eigenvalue sum is $\operatorname{tr}(\Sigma_n^2)=(n/2)(1)+(n/2)(16)=17n/2$, so $v_{0,n}^2=17/[n-1]$. The chosen norm is thus $\|\mu_n\|^2=\delta\sqrt{17/(n-1)}$, and the standardized signal is exactly $\delta_n=\delta$.

The average diagonal entry of $\Sigma_n$ is $(1+4)/2=2.5$. Equal squared magnitudes therefore give $\mu_n'\Sigma_n\mu_n=2.5\|\mu_n\|^2$. The linear-term condition becomes

\frac{2(n-1)\mu_n'\Sigma_n\mu_n}{\operatorname{tr}(\Sigma_n^2)} =\frac{5(n-1)\delta\sqrt{17/(n-1)}}{17n/2} \longrightarrow 0.

The limiting power is $1-\Phi(z_{1-\alpha}-\delta)$. For the same dimension and sample size, the identity covariance has $v_{0,n}^2=2/(n-1)$, rather than $17/(n-1)$. This example shows how a larger aggregate covariance fluctuation raises the signal norm needed to attain the same $\delta$.

Worked Example: Translating a Signal Ratio Into Power

Suppose $n=200$, $p=10{,}000$, and $\Sigma=I_p$. Consider a mean vector with equal coordinates and choose its norm so that the standardized signal ratio is $\delta_n=1.5$. Then $\operatorname{tr}(\Sigma^2)=10{,}000$, and

v_{0,n}=\sqrt{\frac{20{,}000}{200(199)}}\approx0.7089, \qquad \|\mu_n\|^2=1.5v_{0,n}\approx1.0633.

Each coordinate is $\sqrt{1.0633/10{,}000}\approx0.01031$, so the shift is spread over all coordinates. At a nominal level $\alpha=0.05$, $z_{0.95}\approx1.645$, giving the normal-limit power approximation

1-\Phi(1.645-1.5)=1-\Phi(0.145)\approx0.442.

The linear-term variance is small relative to the null variance in this illustration: $(4/200)(1.0633)/(0.7089^2)\approx0.042$. This calculation supports the local approximation but does not make the finite-sample rejection probability exactly $0.442$; that conclusion requires the relevant asymptotic conditions.

Interpreting the Power Curve

The theorem separates the roles of signal and noise. The mean affects the leading shift through $\|\mu_n\|^2$; the covariance affects the null scale through $\operatorname{tr}(\Sigma_n^2)$; and $n$ reduces that scale through the factor $n(n-1)$. With identity covariance, the signal norm needed for a fixed nonzero $\delta$ is approximately $\sqrt{2p_n}\,z/n$ if the desired signal ratio is $z$. More generally, it is of order $\sqrt{\operatorname{tr}(\Sigma_n^2)}/n$, with the exact constant given by the definition of $\delta_n$.

When $\delta=0$, the limiting rejection probability is $\alpha$; when $\delta$ is finite and positive, the limit is strictly between $\alpha$ and $1$. These are local alternatives: they yield a nontrivial limiting power rather than power tending to one. A diverging $\delta_n$ suggests increasing separation, but the finite-$\delta$ theorem does not establish consistency for that regime. Such a conclusion requires conditions controlling the alternative variance and the tail behavior at the diverging scale.

Common Pitfall: Assuming the Null Limit Automatically Transfers

A tempting argument is: “$D_n$ is formed from centered errors, so its null distribution is unchanged under the alternative.” Centering alone does not justify that claim. The assumptions $E[\varepsilon_i]=0$ and $\operatorname{Cov}(\varepsilon_i)=\Sigma_n$ do not require the full error distribution to be the same under the null and alternative. Different centered distributions can have different higher-order behavior, so a null central limit theorem for $D_n/v_{0,n}$ need not hold under alternatives.

The correction is to impose a common centered error distribution under null and alternative, as in the theorem, or to assume the required alternative limit directly. Separately, null calibration requires $\widehat v_{0,n}/v_{0,n}\xrightarrow{p}1$ under the null, not just under alternatives. Verifying both conditions keeps the claimed power curve tied to a test that actually has asymptotic size $\alpha$.

Key takeaway: For local dense alternatives, the quadratic test’s limiting power is governed by $\|\mu_n\|^2\sqrt{n(n-1)/[2\operatorname{tr}(\Sigma_n^2)]}$, provided the linear fluctuation is negligible and the centered quadratic limit and variance calibration hold under the relevant distributions.

Check Your Understanding

Use the definitions and conditions above to check your understanding of the power calculation.

For $\Sigma_n=I_{p_n}$, what order of $\|\mu_n\|^2$ gives a finite, nonzero signal ratio when both $n$ and $p_n$ grow?
Why does the condition $n\mu_n'\Sigma_n\mu_n/\operatorname{tr}(\Sigma_n^2)\to0$ make the linear term negligible?
If $\delta_n\to0$, what is the limiting rejection probability under the theorem’s assumptions?
Why is a common centered-error distribution, or a direct alternative central limit assumption, needed in the power argument?
What additional variance-estimator condition is needed to establish asymptotic size under the null?