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Simultaneous Multivariate Inference · Tutorial 233 of 975

Worked Comparison of Bonferroni and Scheffé Bounds

Learn to calculate both procedures for the same Gaussian model, compare their interval widths, and identify exactly which contrast families their guarantees cover.

Intermediate 10 min read

What You'll Learn

  • Calculate Bonferroni intervals for a finite family of prespecified Gaussian contrasts.
  • Derive Scheffé bounds from a simultaneous confidence ellipsoid for the mean vector.
  • Compare the two procedures through their critical multipliers and standard errors.
  • Explain the different scopes of the Bonferroni and Scheffé guarantees.
  • Handle zero contrasts and check the sample-size conditions behind the formulas.

Two Ways to Protect a Family of Contrasts

Suppose a study measures a \(p\)-dimensional Gaussian response and asks for several scientifically chosen linear combinations of its mean. A simultaneous confidence procedure aims to cover all the requested contrasts at once, rather than giving each interval a separate nominal confidence level and treating those levels as a joint guarantee. Bonferroni and Scheffé bounds both provide familywise coverage, but they protect different scopes of contrasts.

Bonferroni calibrates intervals for a finite, prespecified list. Scheffé starts from a confidence region for the entire mean vector and thereby protects every linear contrast, including ones not listed in advance. That broader guarantee can cost width. The comparison is clearest when both methods are applied to the same data, confidence level, and contrast family.

Model and notation: Let \(X_1,\ldots,X_n\) be independent and identically distributed \(p\)-dimensional random vectors with \(X_i\sim N_p(\mu,\Sigma)\), where \(\Sigma\) is positive definite. Write \(\bar X=n^{-1}\sum_{i=1}^n X_i\) and \(S=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)(X_i-\bar X)'\). For a contrast vector \(c\in\mathbb R^p\), the target is \(c'\mu\), its estimate is \(c'\bar X\), and its estimated standard error is \(\sqrt{c'Sc/n}\).

For a nonzero fixed vector \(c\) and \(n\ge2\), the usual Gaussian one-sample result gives \((c'\bar X-c'\mu)/\sqrt{c'Sc/n}\sim t_{n-1}\). The positive-definite covariance assumption ensures the denominator is positive with probability one. A zero contrast is different: its target and estimate are both zero, so its interval is simply \([0,0]\); the ratio above is undefined and no \(t\) distribution is asserted for it.

Bonferroni Bounds for a Prespecified List

Let \(c_1,\ldots,c_m\) be a fixed finite list of nonzero contrasts, specified independently of the observed data. For a desired familywise error level \(\alpha\in(0,1)\), define \(t_{\nu,q}\) as the \(q\)-quantile of a \(t\) distribution with \(\nu\) degrees of freedom. Allocate error \(\alpha/m\) to each two-sided interval. The resulting intervals are

$$ c_j'\bar X\ \pm\ t_{n-1,\,1-\alpha/(2m)} \sqrt{\frac{c_j'Sc_j}{n}}, \qquad j=1,\ldots,m. $$

If the list also includes zero contrasts, their intervals are \([0,0]\). They are covered with probability one. They may be omitted before defining \(m\), which avoids spending any of the error allocation on intervals that cannot fail.

Theorem (Bonferroni simultaneous coverage): Under the Gaussian model, if \(n\ge2\) and \(c_1,\ldots,c_m\) are fixed nonzero contrasts, the intervals above satisfy \(P(c_j'\mu\text{ is in its interval for every }j)\ge1-\alpha\). No independence among the contrast estimates is required.

Proof. For each \(j\), the marginal \(t_{n-1}\) result implies that the probability its two-sided interval misses \(c_j'\mu\) is \(\alpha/m\). Let \(A_j\) denote that miss event. The event of at least one miss is \(\bigcup_{j=1}^m A_j\). The union bound gives

$$ P\left(\bigcup_{j=1}^m A_j\right) \le \sum_{j=1}^m P(A_j) =m\frac{\alpha}{m} =\alpha. $$

Taking complements proves the stated simultaneous coverage. This argument uses the individual error probabilities, not independence; correlation among the contrasts can affect the actual joint coverage but does not invalidate the bound. \(\square\)

Scheffé Bounds for All Linear Contrasts

Scheffé's method uses the full \(p\)-dimensional mean vector, not just the \(m\) listed contrasts. Assume \(n>p\), so the usual Hotelling statistic is defined with an invertible sample covariance matrix with probability one. Let \(F_{p,n-p;q}\) denote the \(q\)-quantile of the \(F\) distribution with \(p\) and \(n-p\) degrees of freedom. Define

$$ K_{\mathrm{Sch}} = \sqrt{\frac{p(n-1)}{n-p}\,F_{p,n-p;1-\alpha}}. $$

For any contrast vector \(c\), Scheffé's interval is

$$ c'\bar X\ \pm\ K_{\mathrm{Sch}}\sqrt{\frac{c'Sc}{n}}. $$
Theorem (Scheffé simultaneous coverage): Under the stated Gaussian model with positive-definite \(\Sigma\) and \(n>p\), with probability at least \(1-\alpha\), every \(c\in\mathbb R^p\) satisfies \(\lvert c'\bar X-c'\mu\rvert\le K_{\mathrm{Sch}}\sqrt{c'Sc/n}\). Consequently, all contrasts in any finite or infinite collection are covered simultaneously.

Derivation. The exact Hotelling \(T^2\) result for this model states that

$$ T^2=n(\bar X-\mu)'S^{-1}(\bar X-\mu), \qquad \frac{n-p}{p(n-1)}T^2\sim F_{p,n-p}. $$

Therefore, the event \(T^2\le \frac{p(n-1)}{n-p}F_{p,n-p;1-\alpha}=K_{\mathrm{Sch}}^2\) has probability \(1-\alpha\). On this event, put \(d=\bar X-\mu\). For every \(c\), Cauchy–Schwarz applied to \(S^{1/2}c\) and \(S^{-1/2}d\) gives

$$ |c'd|^2 =\left|(S^{1/2}c)'(S^{-1/2}d)\right|^2 \le (c'Sc)(d'S^{-1}d) \le (c'Sc)\frac{K_{\mathrm{Sch}}^2}{n}. $$

The last inequality follows from \(n d'S^{-1}d=T^2\le K_{\mathrm{Sch}}^2\). Taking square roots yields the claimed bound for every \(c\) on the same event. This is why the guarantee covers contrasts chosen after the data are observed as well as contrasts specified beforehand: the event is a single simultaneous statement over all \(c\), not a collection of separately calibrated intervals. \(\square\)

Worked Comparison on One Data Set

For a given contrast, both methods use the same estimated standard error. Their interval widths differ only through their critical multipliers: \(t_{n-1,1-\alpha/(2m)}\) for Bonferroni and \(K_{\mathrm{Sch}}\) for Scheffé. The following example uses the same two contrasts for both procedures.

Worked Example: Two Contrasts in a Bivariate Gaussian Model

Suppose \(p=2\), \(n=20\), \(\alpha=0.05\), and the observed summaries are \(\bar X=(10,12)'\) and \(S=\begin{pmatrix}4&1\\1&9\end{pmatrix}\). Use the fixed family \(c_1=(1,-1)'\) and \(c_2=(1,1)'\).

Step 1: Calculate the estimates and standard errors. The estimated contrasts are \(c_1'\bar X=10-12=-2\) and \(c_2'\bar X=10+12=22\). Their estimated variances are \(c_1'Sc_1=4+9-2(1)=11\) and \(c_2'Sc_2=4+9+2(1)=15\). Thus the standard errors are \(\sqrt{11/20}\approx0.7416\) and \(\sqrt{15/20}=\sqrt{0.75}\approx0.8660\).

Step 2: Calculate the Bonferroni multiplier. Here \(m=2\), so the per-interval two-sided tail probability is \(\alpha/(2m)=0.0125\). The multiplier is \(t_{19,0.9875}\approx2.43\). The margins of error are approximately \(2.43(0.7416)=1.80\) and \(2.43(0.8660)=2.10\). The intervals are therefore approximately \([-3.80,-0.20]\) and \([19.90,24.10]\).

Step 3: Calculate the Scheffé multiplier. Here \(n>p\), and \(F_{2,18;0.95}\approx3.555\). Hence \(K_{\mathrm{Sch}}=\sqrt{\frac{2(19)}{18}(3.555)} =\sqrt{7.505}\approx2.740\). The margins are approximately \(2.739(0.7416)=2.03\) and \(2.739(0.8660)=2.37\). The Scheffé intervals are approximately \([-4.03,0.03]\) and \([19.63,24.37]\).

Comparison. In this family the Scheffé intervals are wider: their common multiplier is about \(2.739\), compared with about \(2.43\) for Bonferroni. Both procedures guarantee at least 95% simultaneous coverage of the two stated contrasts under the model. Scheffé additionally protects every other linear contrast in the two-dimensional mean vector.

Worked Example: What the Broader Scheffé Guarantee Adds

Keep the summaries and confidence level above, but consider the contrast \(c_3=(1,0)'\), which was not in the two-contrast Bonferroni family. Its estimate is \(c_3'\bar X=10\), and its standard error is \(\sqrt{c_3'Sc_3/20}=\sqrt{4/20}\approx0.4472\).

The Scheffé interval is \(10\pm2.739(0.4472)=10\pm1.225\), or approximately \([8.78,11.22]\). The simultaneous Scheffé theorem covers this contrast along with every other linear contrast on its single probability-\(0.95\) event. The Bonferroni guarantee calculated for \(c_1,c_2\) alone makes no corresponding promise for \(c_3\). To guarantee the three listed contrasts by Bonferroni, include \(c_3\) in the prespecified family and use \(m=3\) in its critical value.

Worked Example: Equality in One Dimension

Let \(p=1\), \(n=12\), \(\alpha=0.05\), and consider the single contrast \(c=1\). The Bonferroni multiplier is \(t_{11,0.975}\approx2.201\). Scheffé's multiplier becomes \(\sqrt{\frac{1(11)}{11}F_{1,11;0.95}}=\sqrt{F_{1,11;0.95}}\). Since the square of a \(t_{11}\) random variable has an \(F_{1,11}\) distribution, \(F_{1,11;0.95}=t_{11,0.975}^2\), so the Scheffé multiplier is also approximately \(2.201\).

Thus the methods coincide for this one-dimensional, single-contrast case. This check also confirms that the Scheffé formula reduces to the ordinary two-sided \(t\) interval when there is only one coordinate and one contrast.

Interpreting Width and Scope

For a fixed nonzero contrast \(c\), both interval widths equal twice the method's multiplier times \(\sqrt{c'Sc/n}\). Therefore, comparing widths for the same data and contrast is a direct comparison of critical values. Bonferroni is designed for the stated finite list; Scheffé pays for the stronger protection of the whole \(p\)-dimensional contrast space. Neither multiplier is guaranteed to be smaller for every possible choice of \(p,n,m,\alpha\), so the critical values should be compared rather than relying on a slogan.

Bonferroni can be attractive when the scientific questions define a modest prespecified family. Scheffé is useful when many contrasts may be examined or when the family is effectively all linear combinations. Its guarantee does not mean every interval will be equally informative, and it does not remove the need to define meaningful contrasts. Both methods are exact finite-sample procedures under the stated Gaussian model; their coverage statements are lower bounds, so actual simultaneous coverage can be larger than \(1-\alpha\).

Common Pitfall

Wrong reasoning: “Every contrast has a \(t_{n-1}\) statistic, so I can apply that statement even to a zero contrast and treat all interval errors as independent.” The first clause fails for a zero vector: its numerator and standard error are both zero, so the ratio is undefined. The independence claim is also unnecessary and generally false for correlated contrasts.

Correction: State the marginal \(t\) result only for nonzero contrasts and \(n\ge2\); cover a zero contrast trivially by \([0,0]\). Then use the union bound for Bonferroni, which needs no independence. For Scheffé, use the simultaneous Hotelling ellipsoid argument, which supplies a guarantee over all contrasts at once.

Key takeaway: Bonferroni protects a prespecified finite family by allocating error across its members. Scheffé protects every linear contrast through a joint confidence region for the mean vector. For the same contrast, their widths differ through the critical multiplier, while their scopes of guarantee differ fundamentally.

Check Your Understanding

Use the assumptions and formulas in this tutorial to answer the following questions.

  1. Why does the Bonferroni coverage proof not require the contrast estimates to be independent?
  2. For \(p=3\) and a family of \(m=4\) nonzero contrasts, what degrees of freedom and quantile determine the Bonferroni multiplier at level \(1-\alpha\)?
  3. Why is the Scheffé condition \(n>p\) relevant to the stated formula?
  4. What interval should be reported for the zero contrast, and why is its studentized statistic undefined?
  5. If a contrast is selected after examining the data, which of the two stated guarantees directly covers it, and what feature of the proof provides that protection?
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