The One-Sample Gaussian Mean Hypothesis

The One-Sample Mean as a Vector Parameter

In a multivariate study, each observational unit contributes several measurements. The target of a one-sample mean hypothesis is therefore not usually one scalar average, but a vector of population means. For example, a clinical study might record systolic blood pressure, diastolic blood pressure, and heart rate for every participant. The inferential question can ask whether the population mean of all three outcomes equals a specified target vector.

A precise hypothesis requires a model for the repeated observations, a definition of the parameter being tested, and a clear account of what is assumed about their covariance. In particular, saying that the mean vector is specified does not by itself say whether the covariance matrix is known.

Definition: Let \(X_1,\ldots,X_n\) be independent and identically distributed random vectors in \(\mathbb{R}^p\), with \[ X_i\sim N_p(\boldsymbol\mu,\boldsymbol\Sigma),\qquad i=1,\ldots,n. \] Here \(\boldsymbol\mu\in\mathbb{R}^p\) is the population mean vector and \(\boldsymbol\Sigma\) is a symmetric positive-definite \(p\times p\) covariance matrix. The one-sample Gaussian mean null hypothesis at target \(\boldsymbol\mu_0\) is \(H_0:\boldsymbol\mu=\boldsymbol\mu_0\). The usual two-sided alternative is \(H_1:\boldsymbol\mu\ne\boldsymbol\mu_0\).

The notation \(N_p(\boldsymbol\mu,\boldsymbol\Sigma)\) means that each observation is a \(p\)-dimensional multivariate normal vector with mean \(\boldsymbol\mu\) and covariance \(\boldsymbol\Sigma\). The diagonal entries of \(\boldsymbol\Sigma\) are the coordinate variances; its off-diagonal entries are covariances between coordinates. Positive definiteness ensures a nonsingular covariance and an ordinary density with respect to \(p\)-dimensional volume.

f(\mathbf{x};\boldsymbol\mu,\boldsymbol\Sigma) = (2\pi)^{-p/2}|\boldsymbol\Sigma|^{-1/2} \exp\!\left\{ -\frac12(\mathbf{x}-\boldsymbol\mu)^{T} \boldsymbol\Sigma^{-1} (\mathbf{x}-\boldsymbol\mu) \right\}.

The sample consists of \(n\) vectors, not \(np\) unrelated observations: coordinates within a vector may be correlated, while the iid assumption states that the vectors from different units are independent and share the same distribution. The mean vector is a population parameter. An observed average, even if it happens to equal the target, is not the same thing as the assertion that the population parameter equals that target.

Known and Unknown Covariance Models

The hypothesis \(H_0:\boldsymbol\mu=\boldsymbol\mu_0\) specifies a restriction on the mean parameter. Whether the covariance is known is a separate modeling choice. If the covariance is known, it is fixed at a specified positive-definite matrix \(\boldsymbol\Sigma_0\). If it is unknown, it remains an unrestricted nuisance parameter in the set of positive-definite matrices; it is not silently fixed by stating the mean null.

Model specification	Parameters under the null	Parameters under the alternative
Known covariance \(\boldsymbol\Sigma_0\)	\(\boldsymbol\mu=\boldsymbol\mu_0,\ \boldsymbol\Sigma=\boldsymbol\Sigma_0\)	\(\boldsymbol\mu\ne\boldsymbol\mu_0,\ \boldsymbol\Sigma=\boldsymbol\Sigma_0\)
Unknown covariance	\(\boldsymbol\mu=\boldsymbol\mu_0,\ \boldsymbol\Sigma\) any positive-definite matrix	\(\boldsymbol\mu\ne\boldsymbol\mu_0,\ \boldsymbol\Sigma\) any positive-definite matrix

In the unknown-covariance model, the same covariance parameter space is allowed under the null and alternative. Thus, the null is not the joint assertion that both the mean and covariance equal specified values. It restricts the mean and leaves covariance unspecified. A test procedure may use the covariance information differently in these two model specifications, but the mean restriction itself is unchanged.

Proposition: In the unknown-covariance model, the null parameter set is \(\{(\boldsymbol\mu_0,\boldsymbol\Sigma):\boldsymbol\Sigma\text{ is positive definite}\}\), while the alternative parameter set is \(\{(\boldsymbol\mu,\boldsymbol\Sigma):\boldsymbol\mu\ne\boldsymbol\mu_0,\ \boldsymbol\Sigma\text{ is positive definite}\}\). Consequently, changing the covariance from known to unknown changes the model parameter space, not the meaning of equality in the mean null.

Proof. The full unknown-covariance model has parameter pairs \((\boldsymbol\mu,\boldsymbol\Sigma)\), where \(\boldsymbol\mu\in\mathbb{R}^p\) and \(\boldsymbol\Sigma\) is positive definite. The null retains exactly those pairs whose first component equals \(\boldsymbol\mu_0\), with no additional restriction on the second component. The alternative retains exactly those pairs whose first component differs from \(\boldsymbol\mu_0\), again with no additional restriction on the second component. These sets differ only in the restriction on the mean coordinate; the covariance remains positive definite in each. Fixing \(\boldsymbol\Sigma\) at \(\boldsymbol\Sigma_0\) instead intersects each set with the constraint \(\boldsymbol\Sigma=\boldsymbol\Sigma_0\). This changes the parameter space while leaving the mean equality condition intact. \(\square\)

What the iid Assumption Says About the Sample

The model gives a compact description of the joint density of all observations. This follows directly from independence and is useful for keeping track of which parameters describe the sample.

Result: Under the iid model, the joint density of \(X_1,\ldots,X_n\) at \(\mathbf{x}_1,\ldots,\mathbf{x}_n\) is \[ \prod_{i=1}^{n} f(\mathbf{x}_i;\boldsymbol\mu,\boldsymbol\Sigma) = (2\pi)^{-np/2}|\boldsymbol\Sigma|^{-n/2} \exp\!\left\{ -\frac12\sum_{i=1}^{n} (\mathbf{x}_i-\boldsymbol\mu)^T \boldsymbol\Sigma^{-1} (\mathbf{x}_i-\boldsymbol\mu) \right\}. \]

Derivation. Independence means that the joint density is the product of the \(n\) individual densities. Substituting the multivariate normal density into that product multiplies the constants \((2\pi)^{-p/2}\) \(n\) times, giving \((2\pi)^{-np/2}\), and multiplies \(|\boldsymbol\Sigma|^{-1/2}\) \(n\) times, giving \(|\boldsymbol\Sigma|^{-n/2}\). The product of the exponential terms is the exponential of the sum of their exponents, which gives the displayed expression. No distributional claim about a sample average or sample covariance is needed for this joint-density formula. \(\square\)

This result also makes the roles of the assumptions explicit: every observation has the same \(\boldsymbol\mu\) and \(\boldsymbol\Sigma\), and the observations contribute through a product because they are independent. If measurements from different units are dependent, or their distributions differ, this joint density does not describe the stated iid model.

Worked Specifications

Worked Example: A Two-Outcome Study With Known Covariance

Suppose each participant contributes a vector \(\mathbf{X}_i=(X_{i1},X_{i2})^T\) of two clinical measurements. The proposed target is \(\boldsymbol\mu_0=(120,80)^T\), and the covariance is assumed known to be \[ \boldsymbol\Sigma_0= \begin{pmatrix} 4&1\\ 1&3 \end{pmatrix}. \] State the model and mean hypotheses, and check that the given covariance is admissible.

The model is \(X_i\stackrel{\mathrm{iid}}{\sim}N_2(\boldsymbol\mu,\boldsymbol\Sigma_0)\). The covariance matrix is symmetric. Its first diagonal entry is \(4>0\), and its determinant is \(4(3)-1(1)=11>0\). For a symmetric \(2\times2\) matrix, these leading principal minors being positive imply positive definiteness, so this is a valid covariance matrix.

The mean hypotheses are \(H_0:\boldsymbol\mu=(120,80)^T\) and \(H_1:\boldsymbol\mu\ne(120,80)^T\). The alternative includes any mean vector other than the target, for example \((121,80)^T\) or \((120,79)^T\). The covariance is fixed at the displayed matrix under both hypotheses.

Worked Example: The Same Mean Null With Unknown Covariance

Now retain the target \((120,80)^T\), but assume the covariance is unknown. Two candidate matrices are \[ \boldsymbol\Sigma_A= \begin{pmatrix} 4&1\\ 1&3 \end{pmatrix}, \qquad \boldsymbol\Sigma_B= \begin{pmatrix} 9&-2\\ -2&5 \end{pmatrix}. \] Check that both can belong to the unknown-covariance model and state what the null permits.

For \(\boldsymbol\Sigma_A\), the leading principal minors are \(4\) and \(4(3)-1=11\), both positive. For \(\boldsymbol\Sigma_B\), they are \(9\) and \(9(5)-(-2)^2=45-4=41\), both positive. Both matrices are symmetric positive definite and hence admissible covariance values.

Under \(H_0\), the mean is fixed at \((120,80)^T\), but either matrix—and any other positive-definite covariance matrix—is allowed. For instance, \(((120,80)^T,\boldsymbol\Sigma_A)\) and \(((120,80)^T,\boldsymbol\Sigma_B)\) are both null parameter values. The null does not select between these covariance values. Under \(H_1\), the mean differs from the target, while covariance is still unrestricted within the positive-definite matrices.

Worked Example: Reading a Three-Coordinate Null

A study records three outcomes per unit and uses target vector \(\boldsymbol\mu_0=(0,1,10)^T\). Suppose the covariance is unknown and is modeled by \[ \boldsymbol\Sigma= \begin{pmatrix} 1&0.4&0\\ 0.4&2&0.3\\ 0&0.3&1.5 \end{pmatrix}. \] Write the null coordinate by coordinate, check covariance admissibility, and classify \(\boldsymbol\mu=(0,1,10.2)^T\).

Vector equality means equality in every coordinate, so the null is \(\mu_1=0,\ \mu_2=1,\ \mu_3=10\). For the covariance matrix, the first two leading principal minors are \(1\) and \(1(2)-0.4^2=1.84\). Its determinant is \[ 1(2\cdot1.5-0.3^2)-0.4(0.4\cdot1.5-0\cdot0.3) =2.91-0.24=2.67>0. \] Thus all leading principal minors are positive, so the symmetric matrix is positive definite and admissible.

The proposed mean \((0,1,10.2)^T\) is in the alternative because its third coordinate, \(10.2\), is not \(10\). The first two coordinates matching the target do not make the full vectors equal. Correlations among the measured outcomes affect the covariance model, but they do not change the coordinate-by-coordinate meaning of the mean null.

Common Mistakes and Why the Setup Matters

Treating the null as separate unrelated claims. The equality \(H_0:\boldsymbol\mu=\boldsymbol\mu_0\) is one vector restriction, equivalent to requiring every coordinate to match. The alternative holds if at least one coordinate differs.
Assuming the null fixes covariance. It does so only if the model explicitly states that covariance is known. In the unknown-covariance model, the null restricts the mean and leaves covariance free within the positive-definite matrices.
Confusing an observed average with the population mean. A sample average is calculated from the observed vectors; \(\boldsymbol\mu\) is a parameter describing their distribution. A realized average equal to \(\boldsymbol\mu_0\) does not establish \(H_0\).
Ignoring the unit of independence. The iid observations are the full vectors \(X_i\), one per unit. Correlated coordinates within a vector are allowed and represented by off-diagonal covariance entries.

This model specification is the starting point for mean inference: it identifies the target parameter, defines exactly what the null asserts, and records whether covariance is fixed or unknown. By itself, it does not give a rejection rule, a test statistic, or a p-value; those require additional inferential steps.

Check Your Understanding

Answer each question by referring to the model and parameter restrictions.

For \(p=4\), what does \(H_0:\boldsymbol\mu=\boldsymbol\mu_0\) require of the four population means?
In an unknown-covariance model, does the mean null require \(\boldsymbol\Sigma\) to equal a specified matrix? Explain.
What does the iid assumption say about different observation vectors, and what does it allow about coordinates within one vector?
A proposed mean equals the target in all but one coordinate. Is it in the null or the alternative?
Why does observing a sample average equal to the target not establish that the population mean equals the target?