Confidence region for a normal mean vector

Section of Estimation and inference for the multivariate normal, part of Statistical learning

Type	Section
Taught in	Estimation and inference for the multivariate normal
Part	Statistical learning

A confidence region can now be built without inventing another statistic. Run the level-\alpha test for every proposed mean and retain the values it does not reject. Under the Gaussian model, the resulting random region contains the fixed true mean in a proportion 1-\alpha of repeated samples. Once the data are observed, this statement is not a probability that the parameter lies in this particular region.

n(\bar{\mathbf{x}}-\boldsymbol{\mu})^\top S^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu}) \le\frac{(n-1)p}{n-p}\,F_{\alpha;\,p,\,n-p}.

A level set of (\mathbf{x}-\boldsymbol{\mu})^\top\Sigma^{-1}(\mathbf{x}-\boldsymbol{\mu}) is an ellipsoid whose axes are the eigenvectors of \Sigma. The same algebra, with S in place of \Sigma and with centre \bar{\mathbf{x}}, shows that the inequality is an ellipsoid centred at the sample mean, with axes along the eigenvectors of S. The quadratic form carries a factor n, so the ellipsoid describes uncertainty about the mean and shrinks as n grows. The scatter of the individual observations is what S itself measures. The numbers in the next example are a case in which this ellipsoid excludes the origin while each separate t interval still contains it. The two measurements move together, and the sample mean leaves the origin along a direction in which the sample hardly varies.

Example 1 (Two paired measurements on twelve subjects) Twelve subjects each provide the change in two correlated measurements, (x_1,x_2):

Subject	1	2	3	4	5	6	7	8	9	10	11	12
x_1	-1.8	6.2	-1.5	-0.9	0.3	1.4	2.6	1.6	-2.3	0.5	3.4	0.9
x_2	-2.1	3.8	-2.2	-1.4	-1.3	-1.9	0.8	0.8	-2.5	-0.2	0.3	-1.0

Question. Test H_0:\boldsymbol{\mu}=\mathbf{0} at the 5\% level, and compare the result with two separate t tests.

Solution. The sums are \sum x_1=10.4 and \sum x_2=-6.9, so

\bar{\mathbf{x}}=(10.4/12,\ -6.9/12)^\top=(0.8667,\ -0.5750)^\top.

The corrected cross-product matrix and the sample covariance S=A/11 are

A=\begin{pmatrix}65.0067&43.9600\\43.9600&35.6425\end{pmatrix}, \qquad S=\begin{pmatrix}5.9097&3.9964\\3.9964&3.2402\end{pmatrix}.

The sample correlation is

r=\frac{3.9964}{\sqrt{5.9097\times 3.2402}}=0.913.

Then

|S|=5.9097\times 3.2402-3.9964^2=3.1778

and

S^{-1}=\frac{1}{3.1778}\begin{pmatrix}3.2402&-3.9964\\-3.9964&5.9097\end{pmatrix} =\begin{pmatrix}1.0196&-1.2576\\-1.2576&1.8597\end{pmatrix}.

Hence

\begin{aligned} T^2 &=12\,\bar{\mathbf{x}}^\top S^{-1}\bar{\mathbf{x}}\\ &=12\Bigl[1.0196(0.8667)^2-2(1.2576)(0.8667)(-0.5750)+1.8597(0.5750)^2\Bigr]\\ &=12\times 2.6342=31.61. \end{aligned}

The three products round to 0.7659, 1.2535 and 0.6149. Their unrounded sum is 2.6342, and 12\times 2.6342=31.61.

The scaled statistic is

F=\frac{12-2}{2\times 11}\times 31.61=\frac{10}{22}\times 31.61=14.37

on (2,10) degrees of freedom. The critical value is F_{0.05;\,2,\,10}=4.103 and the p-value is 0.0011, so H_0 is rejected.

The separate t tests give

t_1=\frac{0.8667}{\sqrt{5.9097/12}}=1.235 \quad(p=0.243), \qquad t_2=\frac{-0.5750}{\sqrt{3.2402/12}}=-1.107 \quad(p=0.292).

Neither coordinate is individually significant. The two coordinates are strongly positively correlated (r=0.913) while the sample mean departs from the origin in opposite directions, (+,-). Such a departure lies along the shorter principal axis of the data cloud, where there is little variability. The T^2 test takes the correlation into account. The marginal tests cannot. The 95\% Hotelling ellipsoid for the mean excludes the origin. The rectangle of Bonferroni-adjusted t intervals includes it.

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Taught in Estimation and inference for the multivariate normal
Also used in Covariance and the multivariate normal
All 9 questions on Estimation and inference for the multivariate normal

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