From a Gaussian Vector to a Wishart Ratio
A quadratic form such as \(Z'W^{-1}Z\) measures the size of a Gaussian vector \(Z\) relative to a random covariance-like matrix \(W\). When the vector and matrix are independent and have the matching Gaussian and Wishart structures, this statistic has an exact distribution. That result is the key step behind the classical finite-sample law for Hotelling’s \(T^2\).
The degrees of freedom in the result are not simply those of \(W\): inversion and the dimension of the matrix both matter. Keeping the scaling explicit prevents a common error when translating the ratio into an \(F\) statistic.
For a positive integer \(\nu\), the central Wishart matrix has the Gaussian representation \(W=X'X\), where \(X\) is a \(\nu\times p\) matrix of independent \(N(0,1)\) entries. This representation will let us identify the distribution of a directional quadratic form in \(W^{-1}\).
The Directional Inverse-Wishart Factor
First fix a unit vector \(u\in\mathbb R^p\). Orthogonal invariance of the central Wishart law means that \(u'W^{-1}u\) has the same distribution as the first diagonal entry of \(W^{-1}\). Write \(X=[x_1\ X_2]\), with \(x_1\) the first column and \(X_2\) the remaining \(p-1\) columns. When \(p=1\), take \(X_2\) to have no columns and its projection matrix to be zero.
To prove the result, \(X_2\) has column rank \(p-1\) with probability one. Let \(P_{X_2}\) be the orthogonal projection onto its column space. The Schur complement of the lower-right block \(X_2'X_2\) in \(W\) is \(s=x_1'(I_\nu-P_{X_2})x_1\). The block inverse formula gives \((W^{-1})_{11}=1/s\).
Conditional on \(X_2\), the matrix \(I_\nu-P_{X_2}\) is an orthogonal projection of rank \(\nu-p+1\). The vector \(x_1\) is independent of \(X_2\) and has distribution \(N_\nu(0,I_\nu)\). An orthogonal change of coordinates therefore expresses \(s\), conditional on \(X_2\), as the sum of squares of \(\nu-p+1\) independent standard normal variables. Its conditional distribution does not depend on \(X_2\), so \(s\sim\chi^2_{\nu-p+1}\). Thus \((W^{-1})_{11}=1/s\). Finally, for any fixed unit \(u\), choose an orthogonal matrix \(A\) with \(Ae_1=u\). Since \(A'WA\) has the same Wishart law as \(W\), \(u'W^{-1}u\) has the same law as \(e_1'W^{-1}e_1\). This proves the result. \(\square\)
The degrees of freedom \(\nu-p+1\) arise from the dimension left after projecting away the other \(p-1\) columns. This is why the inverse quadratic form does not have a chi-square denominator with \(\nu\) degrees of freedom.
The Normal-Wishart Ratio Law
The Gaussian vector contributes its squared length, while the inverse-Wishart matrix contributes a random factor in the vector’s direction. Isotropy makes that directional factor have the same distribution in every direction.
Proof. Write \(R^2=Z'Z\) and, on the probability-one event \(Z\ne 0\), \(D=Z/R\). For a standard Gaussian vector, \(R^2\sim\chi^2_p\), and its length \(R\) is independent of its direction \(D\). The pair \((D,W)\) is independent of \(R\), since \(W\) is independent of all of \(Z\). Consequently, \(R^2\) is independent of \(D'W^{-1}D\). Conditional on any value of the unit direction \(D\), the directional result just proved gives \(D'W^{-1}D\sim 1/\chi^2_{\nu-p+1}\); this conditional law is constant in \(D\), so the unconditional law is the same. Therefore $$ Q=Z'W^{-1}Z=R^2D'W^{-1}D \sim \frac{\chi^2_p}{\chi^2_{\nu-p+1}}, $$ where the two chi-square variables in the ratio are independent. By definition, if \(U\sim\chi^2_p\) and \(V\sim\chi^2_{\nu-p+1}\) independently, then \((U/p)/(V/(\nu-p+1))\sim F_{p,\nu-p+1}\). Rearranging gives the stated law. \(\square\)
Worked Example: A Two-Dimensional Ratio
Suppose \(p=2\) and \(\nu=5\). Then the denominator degrees of freedom in the directional inverse-Wishart factor are \(5-2+1=4\), and the ratio law is $$ Q\sim \frac{2}{4}F_{2,4}=\frac12 F_{2,4}. $$
For example, the event \(Q>1\) is equivalent to \(F_{2,4}>2\). The survival function for \(F_{2,4}\) is \(P(F_{2,4}>x)=(1+x/2)^{-2}\) for \(x\ge 0\). Hence $$ P(Q>1)=P(F_{2,4}>2)=(1+2/2)^{-2}=\frac14. $$ The factor \(1/2\) is essential: using \(F_{2,4}\) directly for \(Q\) would give a different tail probability.
Worked Example: Checking the Scalar Case
Set \(p=1\) and \(\nu=4\). Then \(W\) is a \(\chi^2_4\) variable, \(Z^2\sim\chi^2_1\), and \(Q=Z^2/W\), with independent numerator and denominator. The general formula gives $$ Q\sim \frac{1}{4}F_{1,4}. $$
Indeed, by the definition of the \(F\) distribution, \(F_{1,4}=(Z^2/1)/(W/4)=4Z^2/W\). Dividing by \(4\) recovers \(Q=Z^2/W\). This check confirms that the general scaling agrees with the familiar ratio of two independent chi-square variables in one dimension.
Hotelling’s One-Sample \(T^2\) Law
Let \(X_1,\ldots,X_n\) be independent observations from \(N_p(\mu,\Sigma)\), where \(\Sigma\) is positive definite, and assume \(n>p\). Define the sample mean and unbiased sample covariance by $$ \bar X=\frac1n\sum_{i=1}^n X_i,\qquad S=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)'. $$ For testing a specified mean \(\mu\), the one-sample Hotelling statistic is \(T^2=n(\bar X-\mu)'S^{-1}(\bar X-\mu)\).
Derivation. Set \(Z=\sqrt n\,\Sigma^{-1/2}(\bar X-\mu)\) and \(W=\Sigma^{-1/2}\left[\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)'\right]\Sigma^{-1/2}\). For a multivariate normal sample, the sample mean is independent of the centered scatter, \(Z\sim N_p(0,I_p)\), and \(W\sim\operatorname{Wishart}_p(I_p,n-1)\). These are the Gaussian sample-mean and centered-scatter results. Thus the normal-Wishart ratio theorem applies with \(\nu=n-1\).
Since \(S\) is the centered scatter divided by \(n-1\), substituting the definitions yields $$ T^2=(n-1)Z'W^{-1}Z. $$ In detail, \(\bar X-\mu=\Sigma^{1/2}Z/\sqrt n\), and the centered scatter equals \(\Sigma^{1/2}W\Sigma^{1/2}\). Its inverse is \(\Sigma^{-1/2}W^{-1}\Sigma^{-1/2}\), so the factors of \(n\) cancel in \(n(\bar X-\mu)'S^{-1}(\bar X-\mu)\), leaving \((n-1)Z'W^{-1}Z\). The ratio theorem now gives $$ \frac{n-p}{p(n-1)}T^2 =\frac{n-p}{p}Z'W^{-1}Z \sim F_{p,\,n-p}, $$ as claimed.
Worked Example: Converting a Hotelling Statistic
Suppose \(n=12\), \(p=3\), and the observed one-sample statistic is \(T^2=11\). The exact \(F\)-scale factor is $$ \frac{n-p}{p(n-1)}=\frac{12-3}{3(12-1)}=\frac{9}{33}=\frac{3}{11}. $$
Therefore the transformed observed value is \((3/11)(11)=3\), and under the null mean the transformed statistic has an \(F_{3,9}\) distribution. An upper-tail test uses \(P(F_{3,9}\ge 3)\). The degrees of freedom are \(3\) and \(9\), not \(3\) and \(11\): the denominator degrees of freedom are the Wishart degrees of freedom \(n-1\), reduced by \(p-1\).
Worked Example: Computing \(T^2\) From Sample Summaries
Suppose \(n=5\), \(p=2\), the tested mean is \(\mu\), and the observed summaries are \(\bar X-\mu=(1,2)'\) and \(S=\begin{pmatrix}2&0\\0&1\end{pmatrix}\). Then \(S^{-1}=\begin{pmatrix}1/2&0\\0&1\end{pmatrix}\), so $$ (\bar X-\mu)'S^{-1}(\bar X-\mu) =(1,2)\begin{pmatrix}1/2&0\\0&1\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix} =(1,2)\begin{pmatrix}1/2\\2\end{pmatrix} =\frac12+4=\frac92. $$
Thus \(T^2=n(9/2)=45/2\). Its \(F\)-scaled value is $$ \frac{n-p}{p(n-1)}T^2 =\frac{5-2}{2(5-1)}\cdot\frac{45}{2} =\frac38\cdot\frac{45}{2} =\frac{135}{16}. $$ Under the multivariate normal model and the tested mean, this scaled statistic has an \(F_{2,3}\) distribution.
Interpretation and a Common Pitfall
The exact Hotelling law makes a multivariate mean test possible without estimating a separate reference distribution for every coordinate. The statistic accounts for the estimated covariance, including correlations among coordinates, and its \(F\) scaling incorporates the finite-sample uncertainty from estimating that covariance. The law is exact under independent multivariate normal sampling with positive-definite \(\Sigma\); it is not a distribution-free result.
- Wrong reasoning: “\(W\) has \(\nu\) degrees of freedom, so \(Z'W^{-1}Z\) should be a \(\chi^2_p/\chi^2_\nu\) ratio.” This ignores the effect of taking a directional quadratic form in the inverse of a \(p\)-dimensional Wishart matrix.
- Correction: The directional inverse-Wishart factor has \(\nu-p+1\) degrees of freedom, so \(Q\) is \(p/(\nu-p+1)\) times an \(F_{p,\nu-p+1}\) variable.
- Another scaling error: Using \(S\) as if it were the centered scatter changes the multiplier. Here \(S\) is the centered scatter divided by \(n-1\), which is why \(T^2=(n-1)Q\).
Check Your Understanding
Use the degrees of freedom and scale factors explicitly in each answer.
- For \(p=4\) and \(\nu=9\), state the distribution of \(Z'W^{-1}Z\), including its multiplier.
- Why does the directional inverse-Wishart factor have \(\nu-p+1\) degrees of freedom rather than \(\nu\)?
- For a one-sample Gaussian model with \(n=20\) and \(p=5\), give the exact \(F\) distribution of the appropriately scaled \(T^2\).
- Explain why the Gaussian norm \(Z'Z\) is independent of the directional factor \(D'W^{-1}D\).
- If a sample covariance is defined using the centered scatter divided by \(n-1\), what is the relation between \(T^2\) and the normal-Wishart ratio \(Q\)?