Chi-Square Calibration for Known-Covariance Mean Tests

Why Calibration Matters

A quadratic statistic measures how far an estimated mean vector lies from a null value, accounting for the scale and correlation of its coordinates. To turn that distance into a test, we need its distribution when the null hypothesis is true. With a known covariance matrix and a Gaussian sample, that distribution is exact: it is a central chi-square distribution, with degrees of freedom equal to the dimension of the mean vector.

This calibration is what makes a test decision interpretable. It supplies critical values for a chosen significance level and a null tail probability for an observed statistic. The result depends on the Gaussian sampling model; knowing only the mean and covariance does not, by itself, give this exact finite-sample distribution.

Model and Statistic

Let \(\mathbf{X}_1,\ldots,\mathbf{X}_n\) be independent and identically distributed \(p\)-dimensional random vectors with distribution \(N_p(\boldsymbol{\mu},\boldsymbol{\Sigma})\). Assume \(n\geq 1\), the dimension \(p\geq 1\), and that the known covariance matrix \(\boldsymbol{\Sigma}\) is positive definite. We test the point null hypothesis \(H_0:\boldsymbol{\mu}=\boldsymbol{\mu}_0\), where \(\boldsymbol{\mu}_0\) is specified.

Write \(\bar{\mathbf{X}}=n^{-1}\sum_{i=1}^n\mathbf{X}_i\). The earlier result on sampling laws for the Gaussian sample mean gives \(\bar{\mathbf{X}}\sim N_p(\boldsymbol{\mu},\boldsymbol{\Sigma}/n)\). The statistic is the squared Mahalanobis distance from \(\bar{\mathbf{X}}\) to the null mean, with the sample-size factor included:

Positive definiteness ensures that \(\boldsymbol{\Sigma}^{-1}\) exists and that \(Q\geq 0\). Large values indicate that the observed mean is far from the null value relative to the covariance structure. The calibration question is the distribution of \(Q\) when \(H_0\) holds.

Whitening Gives the Exact Null Law

The key step is to express the sample-mean error in coordinates with unit variance and no correlation. Since \(\boldsymbol{\Sigma}\) is positive definite, it has a symmetric positive-definite square root \(\boldsymbol{\Sigma}^{1/2}\), with inverse \(\boldsymbol{\Sigma}^{-1/2}\). Define

Under \(H_0\), the vector \(\mathbf{Z}\) has a \(p\)-dimensional standard normal distribution. Its coordinates are independent \(N(0,1)\) variables. Also, because the square root is symmetric,

Proof. Under the null, \(\bar{\mathbf{X}}-\boldsymbol{\mu}_0\sim N_p(\mathbf{0},\boldsymbol{\Sigma}/n)\). A linear transformation of a multivariate normal vector is multivariate normal, so the defined \(\mathbf{Z}\) is normal. Its mean is

and its covariance matrix is

Thus \(\mathbf{Z}\sim N_p(\mathbf{0},\mathbf{I}_p)\). Its joint density factors into the product of \(p\) standard normal densities, so its coordinates \(Z_1,\ldots,Z_p\) are independent standard normal variables. The quadratic identity above gives \(Q=\sum_{j=1}^p Z_j^2\). By the definition of the central chi-square distribution, \(Q\sim\chi^2_p\). \(\square\)

The dimension, not the sample size, determines the degrees of freedom. Increasing \(n\) changes the variability of the sample mean, but the factor \(n\) in \(Q\) standardizes that variability. After standardization there are still \(p\) independent coordinates.

Using the Calibration for Tests

For a significance level \(\alpha\), let \(c_{p,1-\alpha}\) denote the \((1-\alpha)\)-quantile of \(\chi^2_p\). The upper-tail test rejects \(H_0\) when \(Q>c_{p,1-\alpha}\). Equivalently, for an observed value \(q_{\mathrm{obs}}\), the p-value is \(P(\chi^2_p\geq q_{\mathrm{obs}})\). These are two ways of using the same null distribution: the quantile compares the statistic with a fixed cutoff, while the p-value reports the null probability of a value at least as large as the one observed.

The calibration is exact, not an approximation, under the stated Gaussian model. To see the size directly, the \(\chi^2_p\) distribution is continuous, with a strictly increasing distribution function on positive arguments. Thus \(c_{p,1-\alpha}\) satisfies \(P(\chi^2_p\leq c_{p,1-\alpha})=1-\alpha\), and it assigns zero probability to the cutoff itself. Under \(H_0\), the theorem therefore gives

So the test has rejection probability exactly \(\alpha\) under the null. The continuity matters: it means that using \(>\) rather than \(\geq\) at the cutoff does not change the probability.

Common Scaling Errors

A frequent mistake is to use the covariance matrix of the sample mean, \(\boldsymbol{\Sigma}/n\), and then multiply the resulting quadratic form by \(n\) again. The inverse of that covariance already contains the sample-size factor:

There is no additional factor of \(n\) to apply to this last expression. Doing so would produce \(nQ\), not the calibrated statistic. Conversely, omitting \(n\) when using \(\boldsymbol{\Sigma}^{-1}\) gives \(Q/n\), which has distribution \(\chi^2_p/n\) under the null rather than \(\chi^2_p\).

Another error is to use the sample size as the degrees of freedom. The standardization turns the mean error into a vector with \(p\) independent standard normal coordinates; the statistic sums their \(p\) squares. Hence the degrees of freedom are \(p\), regardless of how many observations were used to estimate the mean.

Why the Exact Result Is Useful

The result separates the covariance adjustment from the reference distribution. A diagonal covariance matrix scales coordinates by their variances; a nondiagonal matrix also accounts for correlations. In either case, whitening converts the null mean error into \(p\) standard normal coordinates, so the same \(\chi^2_p\) calibration applies. This makes it possible to compare statistics from different known covariance structures using a common null reference, provided the model assumptions and dimension are as stated.

The word exact should be read carefully. Here it follows from normal sampling and known positive-definite covariance. If the observations are not Gaussian, their sample mean need not be normal at a finite sample size; a chi-square calibration may then require a separate approximation or justification. The theorem does not establish that every quadratic mean statistic is chi-square merely because it uses an inverse covariance matrix.