Bartlett Correction For A Specified-Covariance Test

Why a Bartlett Correction Is Needed

The likelihood-ratio test for a specified covariance matrix has a simple asymptotic reference distribution: a chi-square distribution with degrees of freedom equal to the number of distinct covariance parameters. But a chi-square approximation is an approximation, and its finite-sample calibration can be imperfect. The Bartlett correction adjusts the statistic by a dimension- and sample-size-dependent factor to improve that calibration to first order.

Here the mean vector is unknown, so it must be estimated under both the null and the unrestricted model. That detail affects the finite-sample correction: after centering the observations, the relevant scatter matrix has \(n-1\), not \(n\), Wishart degrees of freedom. We derive the correction for independent multivariate normal observations with a specified positive-definite covariance matrix.

Model and Likelihood-Ratio Statistic

Definition: Let \(X_1,\ldots,X_n\) be independent \(p\)-dimensional observations with distribution \(N_p(\mu,\Sigma)\), where \(\mu\) is unrestricted and \(\Sigma\) is positive definite. The specified-covariance null hypothesis is \(H_0:\Sigma=\Sigma_0\), for a fixed positive-definite matrix \(\Sigma_0\). The alternative allows \(\Sigma\) to be any positive-definite matrix. Write

\(\bar X=n^{-1}\sum_{i=1}^n X_i\), \(\quad A=\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)^\mathsf{T}\).

The likelihood-ratio statistic is \(T=-2\log\Lambda\), where \(\Lambda\) is the maximized likelihood under \(H_0\) divided by the maximized likelihood under the unrestricted alternative.

Under either model, the maximum-likelihood estimate of \(\mu\) is \(\bar X\). Under the alternative, the maximum-likelihood estimate of \(\Sigma\) is \(A/n\). Substituting these estimates in the Gaussian likelihood gives

T=n\left\{\operatorname{tr}(\Sigma_0^{-1}A/n)-\log|\Sigma_0^{-1}A/n|-p\right\}.

The statistic is defined when \(A\) is positive definite, which holds with probability one under the model if \(n>p\). The degrees of freedom in the Wilks chi-square calibration are

q=\frac{p(p+1)}{2},

because a symmetric \(p\times p\) covariance matrix has \(q\) distinct entries, all fixed by the null hypothesis.

Deriving the First-Order Mean Correction

Under \(H_0\), define the whitened scatter matrix \(W=\Sigma_0^{-1/2}A\Sigma_0^{-1/2}\). Since the mean is estimated from the same observations, \(W\) has a Wishart distribution with identity scale and \(n-1\) degrees of freedom. The trace and determinant in the statistic are unchanged by this whitening, so

T=\operatorname{tr}(W)-n\log|W|+np\log n-np.

For \(W\sim W_p(I,n-1)\), the trace has expectation \(p(n-1)\). Bartlett decomposition expresses \(|W|\) as the product of independent chi-square random variables with degrees of freedom \(n-1,n-2,\ldots,n-p\). A chi-square variable with \(k\) degrees of freedom has expected logarithm \(\psi(k/2)+\log 2\), where \(\psi\) is the digamma function. Therefore,

E[\log|W|]=\sum_{j=1}^p\left\{\psi\left(\frac{n-j}{2}\right)+\log 2\right\}.

Combining these expectations yields an exact expression for the mean of the likelihood-ratio statistic under the null:

E[T]=-p+np\log n-n\sum_{j=1}^p\left\{\psi\left(\frac{n-j}{2}\right)+\log 2\right\}.

To find its first-order departure from \(q\), keep \(p\) fixed and let \(n\) grow. The digamma expansion gives

\psi\left(\frac{n-j}{2}\right)+\log 2 =\log(n-j)-\frac{1}{n-j}-\frac{1}{3(n-j)^2}+O(n^{-4}).

For each fixed \(j\), expand the terms around \(n\):

-\log\left(1-\frac{j}{n}\right)+\frac{1}{n-j}+\frac{1}{3(n-j)^2} =\frac{j+1}{n}+\frac{j^2/2+j+1/3}{n^2}+O(n^{-3}).

Substitution into the exact mean and multiplication by \(n\) give

\begin{aligned} E[T] &=-p+\sum_{j=1}^p(j+1) +\frac{1}{n}\sum_{j=1}^p\left(\frac{j^2}{2}+j+\frac13\right)+O(n^{-2})\\ &=q+\frac{p(2p^2+9p+11)}{12n}+O(n^{-2}). \end{aligned}

In the last equality, \(\sum_{j=1}^p(j+1)=q+p\), and the remaining sum equals \(p(2p^2+9p+11)/12\). This shows that the uncorrected statistic has a mean slightly greater than the mean \(q\) of its limiting chi-square distribution.

The Bartlett-Corrected Calibration

Theorem (first-order Bartlett correction): In a regular, fixed-dimensional likelihood problem with a smooth null hypothesis and an interior true parameter, the first-order null distribution expansion of a likelihood-ratio statistic \(T\) with \(q\) restrictions has the form

\(P(T\le x)=F_q(x)-\dfrac{b}{qn}x f_q(x)+O(n^{-2})\),

where \(F_q\) and \(f_q\) are the chi-square distribution function and density with \(q\) degrees of freedom, and \(E[T]=q+b/n+O(n^{-2})\). Thus multiplying \(T\) by \(1-b/(qn)\) removes the order-\(n^{-1}\) term in this expansion. The fixed-dimensional Gaussian covariance model above satisfies the regularity conditions when \(\Sigma_0\) is positive definite and \(n>p\).

The scaling claim follows directly from the expansion. Set \(a=b/q\), so the multiplier is \(1-a/n\). For fixed \(x>0\),

\begin{aligned} P\left((1-a/n)T\le x\right) &=P\left(T\le \frac{x}{1-a/n}\right)\\ &=P\left(T\le x+\frac{ax}{n}+O(n^{-2})\right)\\ &=F_q(x)+\frac{ax}{n}f_q(x)-\frac{ax}{n}f_q(x)+O(n^{-2})\\ &=F_q(x)+O(n^{-2}). \end{aligned}

The two order-\(n^{-1}\) terms cancel. For the specified-covariance test, the mean expansion gives \(b=p(2p^2+9p+11)/12\), and hence the first-order multiplier is

T_B=\left(1-\frac{2p^2+9p+11}{6n(p+1)}\right)T.

Equivalently, to first order this multiplier is \(q/E[T]\). The theorem explains why matching the mean is useful here: for a regular likelihood-ratio statistic, the first-order distribution correction has the particular form that the rescaling cancels. Mean matching alone would not establish that distributional improvement for an arbitrary statistic.

Worked Applications

Worked Example: A One-Variable Covariance Test

Suppose \(p=1\) and \(n=20\). The null concerns one variance, so \(q=1\). The correction coefficient is

\(\dfrac{2p^2+9p+11}{6(p+1)}=\dfrac{2+9+11}{12}=\dfrac{11}{6}\).

The multiplier is therefore \(1-11/(6\cdot20)=1-11/120=109/120\), or approximately \(0.9083\). If the observed likelihood-ratio statistic is \(T=4.20\), then

\(T_B=(109/120)(4.20)=3.815\).

The 5% upper critical value for \(\chi^2_1\) is approximately \(3.841\). The uncorrected statistic exceeds that value, while the corrected statistic does not. This comparison illustrates the calibration adjustment; it does not establish that the corrected test is exact in a sample of 20.

Worked Example: A Two-Dimensional Test Near a Critical Value

Let \(p=2\) and \(n=30\). There are \(q=2(3)/2=3\) covariance restrictions. The correction coefficient and multiplier are

\(\dfrac{2(2)^2+9(2)+11}{6(2+1)}=\dfrac{37}{18}\),
\(1-\dfrac{37}{18\cdot30}=1-\dfrac{37}{540}=\dfrac{503}{540}\approx0.9315\).

For an observed value \(T=8.00\), the corrected statistic is

\(T_B=(503/540)(8.00)=4024/540\approx7.452\).

The 5% upper critical value for \(\chi^2_3\) is approximately \(7.815\). The uncorrected value is above the critical value, but the corrected value is below it. The change occurs because the multiplier is less than one: the uncorrected statistic has a positive first-order mean bias relative to its chi-square limit.

Worked Example: Applying the Correction in Dimension Three

Suppose \(p=3\) and \(n=40\), so \(q=3(4)/2=6\). The coefficient is

\(\dfrac{2(3)^2+9(3)+11}{6(3+1)}=\dfrac{56}{24}=\dfrac{7}{3}\).

Thus the multiplier is \(1-(7/3)/40=1-7/120=113/120\approx0.9417\). If \(T=13.00\), then

\(T_B=(113/120)(13.00)=1469/120\approx12.242\).

The 5% upper critical value for \(\chi^2_6\) is approximately \(12.592\). The uncorrected statistic exceeds that reference value, whereas the corrected statistic does not. Again, the stated comparison uses a first-order corrected chi-square calibration, not an exact finite-sample null distribution.

Interpretation and Common Pitfalls

The correction is useful when the dimension is not negligible relative to the sample size and the chi-square approximation is being used for inference. Its coefficient increases with \(p\), reflecting the number of covariance parameters being estimated. For fixed \(p\), the adjustment vanishes as \(n\) grows, and the corrected and uncorrected statistics become asymptotically equivalent.

Using \(n\) Wishart degrees of freedom. That is wrong when the mean is estimated. Centering the observations leaves \(n-1\) degrees of freedom, which is why the log-determinant expectation contains \(\psi((n-j)/2)\).
Using the correction for a known mean. A known mean gives a different scatter-matrix distribution and a different finite-sample expansion. The multiplier derived here is specifically for an unknown mean profiled out under both hypotheses.
Changing the degrees of freedom. The correction rescales \(T\); it does not change \(q=p(p+1)/2\), the number of covariance restrictions.
Treating the adjusted test as exact. The correction removes a first-order term in an asymptotic distribution expansion. It does not turn the likelihood-ratio statistic into an exact chi-square variable or guarantee improved performance in every small sample.

Key takeaway: With an unknown Gaussian mean, the first-order Bartlett-corrected statistic for testing \(\Sigma=\Sigma_0\) is \(\left(1-\dfrac{2p^2+9p+11}{6n(p+1)}\right)T\). Compare this adjusted statistic with \(\chi^2_{p(p+1)/2}\), while retaining the distinction between improved asymptotic calibration and an exact finite-sample test.

Check Your Understanding

Use the derivation and examples to check that you can distinguish the statistic, its correction, and its reference distribution.

Why does the centered scatter matrix have \(n-1\) Wishart degrees of freedom when the mean is unknown?
For \(p=2\), how many degrees of freedom does the chi-square calibration use, and what is the first-order multiplier as a function of \(n\)?
What feature of the Wishart distribution is used to derive the exact expectation of \(\log|W|\)?
Does multiplying by the Bartlett factor change the number of restrictions? Explain.
Why does matching the expected statistic to \(q\) support the correction here, but not prove that any arbitrary rescaled statistic has a better null distribution?