Hotelling's T² - Statistics Graduate Level

1. Which statement correctly describes the relationship between this search and Hotelling's statistic?
   Answer: The maximum is $T^2$ and is attained by a vector proportional to $S^{-1}d$. Whenever you see a question about maximizing a standardized linear combination across all directions, think of the Cauchy-Schwarz inequality applied in a Mahalanobis geometry — this is exactly where Hotelling's $T^2$ comes from. The standardized squared mean of a linear combination $a$ is the ratio $\frac{n(a^\top d)^2}{a^\top S a}$. To find the vector $a$ that maximizes this expression, apply the generalized Cauchy-Schwarz inequality: for any positive definite matrix $S$, we have $(a^\top d)^2 \leq (a^\top S a)(d^\top S^{-1} d)$, with equality when $a \propto S^{-1}d$. Substituting back, the maximum value of the ratio is exactly $n \cdot d^\top S^{-1} d = T^2$, which is precisely Hotelling's statistic. This confirms A is correct: the maximum equals $T^2$ and is achieved at $a \propto S^{-1}d$. B is wrong because it ignores the covariance structure. Simply projecting along $d$ does not account for correlations and scale differences among variables — you must "whiten" the direction using $S^{-1}$, not ignore it. C is wrong on two counts: the maximum is $T^2$, not its square root, and the optimal direction is $S^{-1}d$, not $Sd$ (which would point in the wrong Mahalanobis direction). D confuses this optimization with PCA. The eigenvector of $S$ with the largest eigenvalue maximizes variance, not the discrepancy from $\mu_0$ — these are entirely different objectives. As a study tip, remember: maximizing a quadratic ratio over all directions almost always invokes Cauchy-Schwarz, and the optimal direction involves the inverse of the relevant matrix, not the matrix itself.
2. How does the Hotelling statistic computed from the transformed observations compare with the original statistic?
   Answer: It is unchanged because the transformed mean difference and covariance cancel algebraically. Whenever you see a question about Hotelling's $T^2$ and linear transformations, your instinct should be to work through the algebra of how the statistic is constructed — because the answer almost always comes from careful cancellation. Recall that Hotelling's $T^2$ is defined as $T^2 = n(\bar{X} - \mu_0)^\top S^{-1} (\bar{X} - \mu_0)$, where $S$ is the sample covariance matrix. Under the transformation $Y = AX + b$, the sample mean transforms as $\bar{Y} = A\bar{X} + b$ and the null hypothesis mean becomes $A\mu_0 + b$. So the mean difference becomes $\bar{Y} - (A\mu_0 + b) = A(\bar{X} - \mu_0)$. The sample covariance of $Y$ transforms as $S_Y = AS_XA^\top$. Plugging into the statistic: $T_Y^2 = n \cdot (A(\bar{X}-\mu_0))^\top (AS_XA^\top)^{-1} (A(\bar{X}-\mu_0))$. Since $(AS_XA^\top)^{-1} = (A^\top)^{-1}S_X^{-1}A^{-1}$, the $A$ terms cancel completely, leaving exactly $T_X^2$. This confirms B — the statistic is invariant under any nonsingular affine transformation. Choice A is wrong because $|A|$ never appears in the algebraic cancellation; determinants govern volume scaling but not this quadratic form. Choice C is wrong because invariance holds for any nonsingular $A$, not just orthogonal matrices — the key is invertibility, not orthogonality. Choice D is wrong for the same reason as A: squaring the determinant has no basis in the cancellation algebra. A useful study tip: affine invariance of $T^2$ is a fundamental property that mirrors how Mahalanobis distance is scale-invariant. Memorizing why — the covariance inverse absorbs the transformation — will help you handle any variant of this question.
3. Which value is the appropriate paired-sample Hotelling statistic?
   Answer: $T^2=32/3$, obtained from the covariance of the differences Whenever you see a paired multivariate design, the key insight is that pairing eliminates between-subject variability by working directly with the covariance matrix of the differences, not the individual occasion covariances alone. For paired Hotelling's $T^2$, you first compute the covariance matrix of $D = X - Y$. Using the standard formula for the variance of a difference: $S_D = S_X + S_Y - C - C' = \begin{pmatrix}3&1\\1&3\end{pmatrix} + \begin{pmatrix}3&1\\1&3\end{pmatrix} - \begin{pmatrix}1&0.5\\0.5&1\end{pmatrix} - \begin{pmatrix}1&0.5\\0.5&1\end{pmatrix} = \begin{pmatrix}4&1\\1&4\end{pmatrix}$ Then the test statistic is $T^2 = n\,\bar{d}'\,S_D^{-1}\,\bar{d}$. The inverse of $\begin{pmatrix}4&1\\1&4\end{pmatrix}$ is $\frac{1}{15}\begin{pmatrix}4&-1\\-1&4\end{pmatrix}$, so: $\bar{d}'\,S_D^{-1}\,\bar{d} = \frac{1}{15}(1,-1)\begin{pmatrix}4&-1\\-1&4\end{pmatrix}\begin{pmatrix}1\\-1\end{pmatrix} = \frac{1}{15}(5,-5)\begin{pmatrix}1\\-1\end{pmatrix} = \frac{10}{15} = \frac{2}{3}$ Multiplying by $n = 16$ gives $T^2 = \frac{32}{3}$, confirming D. Choice A is wrong because it ignores cross-occasion covariance entirely — using $S_X + S_Y$ only — which overcounts variance and distorts the statistic. Choice B uses only $C$, discarding the within-occasion variances altogether, producing a nonsensical and inflated result. Choice C correctly computes $\bar{d}'\,S_D^{-1}\,\bar{d} = 2/3$ but forgets to multiply by $n$, omitting the sample-size scaling that makes the statistic comparable to its reference distribution. Your study tip: always derive $S_D = S_X + S_Y - C - C'$ explicitly in paired multivariate problems — the cross-covariance term is the most commonly dropped component, and dropping it changes both the direction and magnitude of the result.
4. What is the primary obstacle to using the standard Hotelling statistic and its usual exact $F$ calibration?
   Answer: The sample covariance is necessarily singular, so its ordinary inverse and the standard exact test are unavailable. Whenever you encounter a question about Hotelling's $T^2$ statistic, your first instinct should be to check the relationship between $n$ (sample size) and $p$ (dimension), because this determines whether the sample covariance matrix is even invertible. The standard one-sample Hotelling statistic is $T^2 = n(\bar{\mathbf{x}} - \boldsymbol{\mu}_0)^\top \mathbf{S}^{-1} (\bar{\mathbf{x}} - \boldsymbol{\mu}_0)$, which requires computing $\mathbf{S}^{-1}$. The sample covariance matrix $\mathbf{S}$ is constructed from $n-1 = 19$ deviation vectors, each living in $\mathbb{R}^{25}$. Because you only have 19 linearly independent vectors spanning at most a 19-dimensional subspace of $\mathbb{R}^{25}$, the matrix $\mathbf{S}$ has rank at most 19 and is therefore necessarily singular — regardless of the population. You cannot invert a singular matrix, so $T^2$ is undefined and the exact $F$ calibration never even gets off the ground. This confirms B as the correct answer. A is wrong because once $\mathbf{S}$ is singular, the statistic doesn't exist at all — there is no reference distribution to adjust. C fabricates a nonexistent correction formula; dividing by $p - n$ has no basis in standard Hotelling theory. D is the subtlest trap: positive definiteness of the population covariance $\boldsymbol{\Sigma}$ does not save you — singularity here is a property of the sample covariance determined purely by the geometry of having fewer observations than dimensions. As a study rule: whenever $p \geq n$, the sample covariance is singular — memorize this rank condition, as high-dimensional testing questions will repeatedly exploit it.
5. Which comparison of the two alternatives is correct?
   Answer: The low-variance direction has nine times the noncentrality and therefore greater power. When you see a question involving Hotelling's $T^2$ and directional mean shifts, the key concept to invoke is the noncentrality parameter, which governs power. For a one-sample Hotelling test, the noncentrality is $\delta^2 = n \cdot \boldsymbol{\mu}^T \boldsymbol{\Sigma}^{-1} \boldsymbol{\mu}$. Notice that the covariance matrix appears inverted — this is the critical insight. If the mean shift $\boldsymbol{\mu}$ lies entirely along an eigenvector of $\boldsymbol{\Sigma}$ with eigenvalue $\lambda$, then $\boldsymbol{\Sigma}^{-1}$ acts on that direction with eigenvalue $1/\lambda$. So the noncentrality becomes $\delta^2 = n \cdot h^2 / \lambda$. A smaller eigenvalue (less variance) means a larger noncentrality — the test finds it easier to detect signal in directions where the data naturally vary little. When $\lambda = 1$, the noncentrality is $nh^2$; when $\lambda = 9$, it drops to $nh^2/9$. The low-variance direction yields nine times the noncentrality, giving it substantially greater power. This confirms C is correct. A is wrong because Hotelling's test does not depend only on Euclidean length — it weights directions by the inverse covariance, so geometry relative to $\boldsymbol{\Sigma}$ matters enormously. B inverts the logic entirely: high variance means the signal is harder to distinguish from noise, reducing noncentrality, not increasing it. D gets the ratio right (nine, not three) but incorrectly attributes it to standard deviations entering linearly — the noncentrality involves $\boldsymbol{\Sigma}^{-1}$, so eigenvalues (not their square roots) determine the ratio. Remember: $\boldsymbol{\Sigma}^{-1}$ in the noncentrality means low variance = high power. The test is most sensitive in the directions the data vary least.
6. For testing equality of the two population mean vectors, which statement gives the correct two-sample Hotelling statistic and its exact $F$ transformation?
   Answer: $T^2=40/3$ and $F=6.4$ with degrees of freedom $2$ and $24$ When you encounter a two-sample Hotelling $T^2$ problem, your first move is computing the scalar coefficient, then the quadratic form, then converting to an $F$ statistic with the right degrees of freedom. The pooled scalar coefficient combines both sample sizes: $\frac{n_1 n_2}{n_1+n_2} = \frac{12 \cdot 15}{27} = \frac{180}{27} = \frac{20}{3}$. Next, invert $S_p$. For a $2\times2$ matrix with determinant $4-1=3$, you get $S_p^{-1} = \frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$. The quadratic form is $(1,-1)S_p^{-1}(1,-1)' = \frac{1}{3}[(2)(1)+(-1)(-1)+(-1)(1)+(2)(1)] = \frac{1}{3}(2+1+1+2) = 2$. So $T^2 = \frac{20}{3}\cdot 2 = \frac{40}{3}$. The exact $F$ transformation is $F = \frac{n_1+n_2-p-1}{(n_1+n_2-2)p}\,T^2 = \frac{24}{2\cdot 2}\cdot\frac{40}{3} = 6\cdot\frac{40}{3}\cdot\frac{1}{4} = 6.4\overline{6}$... more precisely, $F = \frac{24}{4}\cdot\frac{40}{3} = 6\cdot\frac{10}{3}\cdot\frac{1}{3}$—let's be clean: $F = \frac{24}{52}\cdot\frac{40}{3}$... The formula gives $F = \frac{n_1+n_2-p-1}{(n_1+n_2-2)\,p}\,T^2 = \frac{24}{4}\cdot\frac{40}{3} = 6\cdot\frac{40}{12}$... simplifying directly: $\frac{24}{4}\cdot\frac{40}{3} = 6\cdot\frac{40}{12} = \frac{240}{12}$... $= \frac{24\cdot 40}{4\cdot 3\cdot 12}$. Cleanly: $6 \times \frac{10}{3} = \frac{60}{3}\cdot\frac{1}{...}$. Simply: $\frac{24}{4}\cdot\frac{40}{3} = \frac{960}{12} = 80/...$. Direct: numerator $24\times40=960$, denominator $4\times3=12$... wait, $F=\frac{(n_1+n_2-p-1)}{(n_1+n_2-2)p}T^2=\frac{24}{(25)(2)}\cdot\frac{40}{3}=\frac{960}{150}=6.4$. The $F$ distribution has $p=2$ and $n_1+n_2-p-1=24$ degrees of freedom, confirming answer A. Choice B uses $df_2=25$, which incorrectly applies $n_1+n_2-p$ rather than $n_1+n_2-p-1$. Choice C uses a wrong $T^2$ value (likely from failing to invert $S_p$ correctly). Choice D computes $F$ without dividing by $(n_1+n_2-2)$ in the denominator, inflating the result. Remember: the $F$ degrees of freedom are always $p$ and $n_1+n_2-p-1$
7. What is the value of the one-sample Hotelling statistic $T^2=n(\bar{x}-\mu_0)'S^{-1}(\bar{x}-\mu_0)$?
   Answer: $20$ Hotelling's $T^2$ statistic generalizes the univariate $t$-statistic to multivariate settings, and the core computational challenge is always inverting the sample covariance matrix $S$. When you see this type of question, focus first on getting $S^{-1}$ correct — that's where most errors occur. For a 2×2 matrix $S = \begin{pmatrix}4&1\\1&2\end{pmatrix}$, the inverse is $S^{-1} = \frac{1}{\det(S)}\begin{pmatrix}2&-1\\-1&4\end{pmatrix}$. The determinant is $\det(S) = (4)(2)-(1)(1) = 7$, so $S^{-1} = \frac{1}{7}\begin{pmatrix}2&-1\\-1&4\end{pmatrix}$. Now compute the quadratic form with $d = (1,2)'$: first find $S^{-1}d = \frac{1}{7}\begin{pmatrix}2(1)+(-1)(2)\\(-1)(1)+4(2)\end{pmatrix} = \frac{1}{7}\begin{pmatrix}0\\7\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}$. Then $d'S^{-1}d = (1,2)\begin{pmatrix}0\\1\end{pmatrix} = 2$. Finally, $T^2 = n \cdot 2 = 10 \times 2 = \mathbf{20}$, confirming answer B. Choice A (2) forgets to multiply by $n=10$ — it's just the quadratic form $d'S^{-1}d$ without the sample size scaling. Choice C (22.5) likely results from using the raw matrix entries without properly inverting $S$, perhaps computing $d'Sd = (1,2)\begin{pmatrix}4&1\\1&2\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix} = (6,5)\begin{pmatrix}1\\2\end{pmatrix}=16$... or mishandling the determinant. Choice D (200) inflates the result by skipping the $1/\det(S)$ scaling entirely, treating $S^{-1}$ as $\begin{pmatrix}2&-1\\-1&4\end{pmatrix}$ without dividing by 7. Your study tip: always write out the two-step matrix inversion explicitly — compute the determinant first, then the adjugate — before touching the quadratic form. Rushing through $S^{-1}$ is the single most common source of error on multivariate computation problems.
8. Which transformed statistic and null reference distribution should be used for the exact test?
   Answer: $F=3.58$ with degrees of freedom $3$ and $17$ Whenever you see a one-sample Hotelling's $T^2$ problem, your first instinct should be to recall the exact $F$ transformation. The key formula is: $F = \frac{n - p}{p(n-1)} \cdot T^2$ with degrees of freedom $p$ and $n - p$. This transformation is exact under multivariate normality — not approximate — which is why this question emphasizes "exact test." Plugging in $T^2 = 12$, $n = 20$, and $p = 3$: $F = \frac{20 - 3}{3(20 - 1)} \cdot 12 = \frac{17}{57} \cdot 12 = \frac{204}{57} \approx 3.58$ The degrees of freedom are $p = 3$ (numerator) and $n - p = 20 - 3 = 17$ (denominator). This confirms C is correct. Now trace where each wrong answer goes astray. A uses $F = T^2/p = 12/3 = 4.00$ with df $(3, 19)$ — this ignores the scaling factor entirely and uses the wrong denominator df. B gets $n - p$ wrong, using $16$ instead of $17$ (an off-by-one error, perhaps confusing $n - p - 1$), and the $F$ value $3.79$ doesn't match the correct formula either. D gets the $F$ value right at $3.58$ but uses $19$ as the denominator df — a tempting trap since $n - 1 = 19$ appears naturally in the denominator of the formula, but the denominator degrees of freedom for the $F$ distribution is $n - p$, not $n - 1$. As a memory anchor: the denominator df equals $n - p$, the same quantity that appears in the numerator of the scaling factor. Recognizing this symmetry helps you avoid the $n-1$ trap that snares many students.
9. For a fixed nonzero vector $a$, which interval for $a'\mu$ is implied by the simultaneous Hotelling confidence region with confidence level $1-\alpha$?
   Answer: $a'\bar{x}\mathbin{\pm}\sqrt{(c/n)\,a'Sa}$ Whenever you see a question about Hotelling's $T^2$ simultaneous confidence regions, your job is to translate the ellipsoidal region in $\mathbb{R}^p$ into a marginal interval for a linear combination $a'\mu$. The Hotelling region contains all $\mu$ satisfying $n(\bar{x}-\mu)'S^{-1}(\bar{x}-\mu)\leq c$. To find the implied interval for $a'\mu$, you project this ellipsoid onto the direction $a$. The key identity is that the extremes of $a'\mu$ over the ellipsoid are achieved when you maximize/minimize subject to the constraint. Applying the Cauchy-Schwarz inequality to the quadratic form yields: $|a'\bar{x}-a'\mu|\leq\sqrt{\frac{c}{n}\,a'Sa}$ which gives the interval $a'\bar{x}\pm\sqrt{(c/n)\,a'Sa}$, confirming C is correct. A is wrong because it omits the crucial $1/n$ factor. The sample mean has variance $\Sigma/n$, estimated by $S/n$, so dropping $n$ inflates the interval incorrectly — it treats the variance as if you had a single observation rather than a mean. B is a trap: it looks like a standard univariate $t$-interval for $a'\bar{x}$. While $t_{n-1}$ intervals are valid for a single pre-specified direction, they do not provide simultaneous coverage over all directions $a$. Hotelling uses the $F$-based critical value $c$ to account for searching over all linear combinations simultaneously. D replaces $a'Sa$ with $a'S^{-1}a$, which appears in the Hotelling statistic itself but not in the projected variance of $a'\bar{x}$. Confusing the role of $S^{-1}$ in the test statistic with the variance of the projection is a common error. Study tip: Always track whether you're working with the quadratic form for the test statistic (uses $S^{-1}$) versus the variance of a linear combination (uses $S$), and never forget the $1/n$ that comes from working with the sample mean.
10. Which claim about the null distribution is most accurate?
    Answer: The exact $F$ distribution need not hold for nonnormal populations, but $T^2$ converges in distribution to $\chi_p^2$ under the null as $n\to\infty$. When you see a question about Hotelling's $T^2$ test, the key distinction to keep in mind is between exact finite-sample results and asymptotic results, and precisely which distributional assumptions drive each. The exact $F$ distribution for Hotelling's $T^2$ relies critically on multivariate normality. Under $H_0$, you need the sample mean vector to be normal and the sample covariance matrix to follow a Wishart distribution — and the Wishart distribution is derived specifically under multivariate normality, not under broader assumptions. With a fixed $p$ and $n \to \infty$, the multivariate central limit theorem guarantees that $\sqrt{n}(\bar{\mathbf{x}} - \boldsymbol{\mu})$ converges to a multivariate normal, and standard arguments then show $T^2 = n(\bar{\mathbf{x}} - \boldsymbol{\mu}_0)^\top S^{-1}(\bar{\mathbf{x}} - \boldsymbol{\mu}_0) \xrightarrow{d} \chi^2_p$. This makes D correct: the exact $F$ need not hold for nonnormal populations, but the asymptotic $\chi^2_p$ null distribution does. A is wrong because marginal normality of each coordinate does not imply joint multivariate normality, which is what the exact result requires. B is wrong because elliptical symmetry alone does not guarantee the Wishart distribution for $S$ — that result is specific to the multivariate normal family. C is wrong on two counts: mutual independence of coordinates is neither required nor sufficient for the exact $F$ result under multivariate normality, and it misidentifies the actual assumption structure entirely. Your study tip: always ask whether a distributional result is exact (requiring full multivariate normality) or asymptotic (requiring only finite second moments plus the CLT). The exam frequently tests whether you conflate these two regimes.

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