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Simultaneous Multivariate Inference · Tutorial 230 of 975

Critical Values for Correlated Gaussian Maxima

Compute and interpret Gaussian maximum critical values for simultaneous inference, including how the full correlation structure affects interval calibration.

Intermediate 10 min read

What You'll Learn

  • Define a critical value using the distribution of the maximum absolute Gaussian component.
  • Relate the Gaussian maximum quantile to asymptotic simultaneous interval coverage.
  • Compute critical values for independent and perfectly correlated components.
  • Calibrate a two-component example with positive correlation using a rectangle probability.
  • Explain why componentwise normal quantiles and independence-based shortcuts can miscalibrate coverage.

Why the Gaussian Maximum Sets the Critical Value

When several estimates are reported together, simultaneous coverage concerns all of their errors at once. A critical value for one normal component does not generally provide the desired coverage for a vector of estimates: the chance that every component lies within its interval depends on their joint distribution. In the asymptotic setting, that dependence is summarized by the correlation matrix of the standardized errors.

The relevant random quantity is the largest absolute component of a correlated Gaussian vector. Its quantile calibrates a common multiplier for the componentwise standard errors, producing intervals whose joint coverage is approximately the target level. The calibration uses the entire correlation structure, not merely the number of components.

Definition: Let \(Z=(Z_1,\ldots,Z_m)^\mathsf{T}\sim N_m(0,R)\), where \(R\) is a positive semidefinite correlation matrix, so each \(Z_j\) has variance 1. Define \(M=\max_{1\leq j\leq m}|Z_j|\). For \(0<\alpha<1\), the Gaussian maximum critical value is \(c_{1-\alpha}(R)=\inf\{c\geq 0:P(M\leq c)\geq 1-\alpha\}\). Equivalently, it is the \(1-\alpha\) quantile of \(M\). For a nonsingular Gaussian vector, the distribution of \(M\) is continuous and strictly increasing on \(c>0\), so the critical value is the unique solution of \(P(-c\leq Z_j\leq c\text{ for all }j)=1-\alpha\).

The event defining the quantile is a rectangle probability: \(P(M\leq c)=P(-c\leq Z_j\leq c,\ j=1,\ldots,m)\). Thus, even though every marginal distribution is standard normal, the probability of this rectangle changes with \(R\). For general \(R\), the quantile is usually found by numerical multivariate normal probability evaluation followed by a one-dimensional root search.

From a Gaussian Quantile to Simultaneous Intervals

Suppose \(\hat\theta=(\hat\theta_1,\ldots,\hat\theta_m)^\mathsf{T}\) estimates \(\theta=(\theta_1,\ldots,\theta_m)^\mathsf{T}\). Let \(s_j\) estimate the asymptotic standard deviation of \(\hat\theta_j\), and suppose the standardized error vector satisfies \(( (\hat\theta_j-\theta_j)/s_j )_{j=1}^m\xrightarrow{d}N_m(0,R)\), with \(s_j\) consistent for the corresponding standard error scale and \(R\) a fixed correlation matrix. This is the joint-normal setting used in asymptotic simultaneous intervals.

Theorem: Under the stated joint convergence, if \(c=c_{1-\alpha}(R)\) and the limit distribution of \(M\) is continuous at \(c\), the intervals \(\hat\theta_j\pm c s_j\), \(j=1,\ldots,m\), have joint coverage tending to \(1-\alpha\).

Proof. Write \(T_n=\max_j|(\hat\theta_j-\theta_j)/s_j|\). The map \(x\mapsto\max_j|x_j|\) is continuous, since it is the maximum of finitely many continuous functions. The continuous mapping theorem therefore gives \(T_n\xrightarrow{d}M\). The event that all intervals cover their respective parameters is exactly \(\{|\hat\theta_j-\theta_j|\leq c s_j\text{ for every }j\}=\{T_n\leq c\}\). Convergence in distribution implies \(P(T_n\leq c)\to P(M\leq c)\) at every continuity point \(c\). By the definition of the critical value and continuity, \(P(M\leq c)=1-\alpha\). Hence the limiting joint coverage is \(1-\alpha\). \(\square\)

This is an asymptotic calibration, not an assertion of exact finite-sample coverage. Its validity depends on the joint normal approximation and on using standard errors and a correlation matrix that represent the standardized errors in that approximation. The earlier material on asymptotic simultaneous intervals gives the interval form; the Gaussian maximum quantile determines its multiplier.

Two Useful Exact Cases

When the coordinates are independent, the rectangle probability factors into \(m\) identical marginal probabilities. When they are perfectly correlated in the same direction, all coordinates are copies of one standard normal variable, and the maximum absolute value is just the absolute value of that variable. These cases provide useful checks on a numerical calculation.

Result: If \(Z_1,\ldots,Z_m\) are independent standard normal variables, then \(P(M\leq c)=[2\Phi(c)-1]^m\), and \(c_{1-\alpha}(I_m)=\Phi^{-1}\!\left((1+(1-\alpha)^{1/m})/2\right)\). If instead \(Z_1=\cdots=Z_m\) almost surely, then \(P(M\leq c)=2\Phi(c)-1\), and \(c_{1-\alpha}=\Phi^{-1}(1-\alpha/2)\).

Derivation. For each standard normal component, symmetry gives \(P(|Z_j|\leq c)=\Phi(c)-\Phi(-c)=2\Phi(c)-1\). Independence makes the probability of the intersection the product of these probabilities, yielding the first formula. Solving \([2\Phi(c)-1]^m=1-\alpha\) gives the stated quantile. In the perfectly correlated case, \(M=|Z_1|\); solving \(2\Phi(c)-1=1-\alpha\) gives the second formula. \(\square\)

Worked Example: Two Independent Components

Suppose \(Z_1,Z_2\) are independent standard normal variables and the desired simultaneous coverage is 0.95. The maximum is below \(c\) exactly when both absolute components are below \(c\), so

$$ P(M\leq c)=[2\Phi(c)-1]^2. $$

Set this probability equal to 0.95 and solve:

$$ 2\Phi(c)-1=\sqrt{0.95},\qquad \Phi(c)=\frac{1+\sqrt{0.95}}{2} =\frac{1+0.9746794}{2}\approx0.9873397. $$

Therefore \(c=\Phi^{-1}((1+\sqrt{0.95})/2)\approx2.237\). This value is larger than the single-component 95% two-sided critical value, approximately 1.960, because both components must be covered. Substituting the unrounded quantile into \([2\Phi(c)-1]^2\) gives 0.95 by construction.

Worked Example: Perfectly Correlated Components

Let \(Z_1=Z_2\) almost surely, so the correlation matrix is \(R=\begin{pmatrix}1&1\\1&1\end{pmatrix}\). Although there are two reported coordinates, there is only one distinct standardized error: \(M=\max(|Z_1|,|Z_2|)=|Z_1|\). For 95% coverage,

$$ P(M\leq c)=2\Phi(c)-1=0.95,\qquad c=\Phi^{-1}(0.975)\approx1.960. $$

Using the two-independent-component value 2.237 here would be unnecessarily conservative: the resulting joint coverage is \(P(|Z_1|\leq2.237)\), about 0.975, because the two intervals succeed or fail together. This example also shows why the definition permits a positive semidefinite, possibly singular, correlation matrix. In practice, singularity can indicate redundant standardized coordinates; their joint maximum remains well-defined.

Computing a Correlated Gaussian Maximum

For a general correlation matrix, evaluate the probability of the centered rectangle \([-c,c]^m\) under \(N_m(0,R)\). A practical procedure is to choose a candidate \(c\), calculate that probability, and adjust \(c\) until it equals \(1-\alpha\). The probability calculation may use a multivariate normal CDF routine, deterministic numerical integration, or a carefully controlled simulation method. Root-finding then locates the quantile. Accuracy should be checked by evaluating the rectangle probability again at the reported critical value.

A two-dimensional case makes the dependence effect especially visible. If \(Z_1,Z_2\) are standard normal with correlation \(\rho\), write \(\Phi_2(a,b;\rho)=P(Z_1\leq a,Z_2\leq b)\). Symmetry and inclusion-exclusion give

$$ P(|Z_1|\leq c,\ |Z_2|\leq c) =\Phi_2(c,c;\rho)-\Phi_2(-c,c;\rho) -\Phi_2(c,-c;\rho)+\Phi_2(-c,-c;\rho). $$

This is the rectangle probability to solve for the desired coverage. It is not the product of marginal probabilities unless \(\rho=0\). For this bivariate symmetric rectangle, an alternative useful identity is

$$ P(|Z_1|,|Z_2|\leq c) =[2\Phi(c)-1]^2+ \int_0^\rho \frac{\exp\{-c^2/(1+r)\}-\exp\{-c^2/(1-r)\}} {\pi\sqrt{1-r^2}}\,dr. $$

The identity follows by differentiating the bivariate normal rectangle probability with respect to its correlation parameter \(r\), which gives the integrand, and integrating from \(r=0\), where independence gives the product term. It is useful for checking a two-component numerical calculation.

Worked Example: Two Components With Correlation 0.5

Let \(R=\begin{pmatrix}1&0.5\\0.5&1\end{pmatrix}\) and seek 95% simultaneous coverage. The independent critical value 2.237 is not the answer, since the rectangle probability does not factor. Use the bivariate identity above and find \(c\) satisfying

$$ [2\Phi(c)-1]^2+ \int_0^{0.5} \frac{\exp\{-c^2/(1+r)\}-\exp\{-c^2/(1-r)\}} {\pi\sqrt{1-r^2}}\,dr=0.95. $$

Numerical quadrature and one-dimensional root-finding give \(c\approx2.208\). For a direct check, at \(c=2.210\), \(\Phi(c)\approx0.98645\), so the independence term is \([2(0.98645)-1]^2\approx0.9465\). The integral from 0 to 0.5 is approximately 0.00321, giving a total of about 0.94973. This is slightly below 0.95 because \(c=2.210\) is slightly below the critical value. The positive correlation makes the two components more likely to be simultaneously near the same side of the rectangle than under independence, so in this example the required critical value is smaller than 2.237. The contrast with the perfectly correlated case, whose value is 1.960, shows that partial correlation reduces but does not eliminate the multiplicity adjustment.

How Correlation Matters—and a Common Pitfall

The correlation structure determines how often large componentwise errors occur together. With independent components, extreme values can arise in different coordinates, so the maximum tends to require a larger cutoff as the number of coordinates grows. With perfectly correlated copies, all coordinates move together and adding copies does not change the maximum. Intermediate dependence lies between these familiar examples in particular models, but there is no universal rule that a certain sign or average size of correlations determines the critical value for every correlation matrix. The full matrix matters.

A common mistake is to use a single-component normal critical value for every coordinate and then call the resulting intervals jointly 95% intervals. For two independent coordinates, using 1.960 gives joint coverage \([2\Phi(1.960)-1]^2\approx0.95^2=0.9025\), not 0.95. Another mistake is to use the independent-components formula when the coordinates are correlated. That formula is correct only when the standardized Gaussian components are independent; it is not justified merely because every marginal is standard normal.

Key takeaway: Calibrate simultaneous asymptotic intervals with the quantile of \(M=\max_j|Z_j|\) for the relevant joint Gaussian limit. Independence and perfect correlation give useful exact checks, while other correlation structures require their own rectangle-probability calculation.

Check Your Understanding

Use the definitions and calculations above to check your understanding of Gaussian maximum calibration.

  1. Write the rectangle event whose probability defines \(c_{1-\alpha}(R)\) for a vector with four components.
  2. Why does the independence formula for the maximum fail when two standard normal components have correlation 0.5?
  3. For three independent standard normal components and 95% coverage, write the exact expression for the critical value in terms of \(\Phi^{-1}\).
  4. If two components are identical almost surely, explain why adding the second component does not increase the maximum’s critical value.
  5. State the convergence argument that connects a Gaussian maximum quantile to limiting joint coverage of asymptotic intervals.
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