Calibrating Region Size to a Chosen Coverage Level

Why the Critical Quantile Matters

A confidence level does not by itself specify the size of an interval or region. The critical quantile must also match the probability distribution of the pivot, the number of dimensions being covered, and whether the variance is known or estimated. Choosing a quantile for the wrong tail or distribution changes the coverage guarantee; choosing a larger coverage level generally produces a wider or larger region.

The calibration is especially visible when moving from a single parameter to a vector. For one parameter, the critical value is often a normal or t quantile. For a jointly covered vector, the relevant quantile is commonly from an F distribution when the variance is estimated. Its effect is on the entire region, not merely on each coordinate separately.

Quantiles and Coverage Levels

Definition: For a continuous random variable \(W\) with cumulative distribution function \(F_W\), its \(p\)-quantile \(q_p\) satisfies \(P(W\leq q_p)=p\), for \(0<p<1\). Write \(z_p\) for the \(p\)-quantile of \(N(0,1)\), \(t_{\nu;p}\) for the \(p\)-quantile of a t distribution with \(\nu\) degrees of freedom, and \(F_{q,\nu;p}\) for the \(p\)-quantile of an F distribution with numerator degrees of freedom \(q\) and denominator degrees of freedom \(\nu\).

For a central confidence level \(1-\alpha\), the excluded probability is split equally between the two tails: each tail has probability \(\alpha/2\). The corresponding symmetric normal or t critical value is therefore the \((1-\alpha/2)\)-quantile. A one-sided cutoff instead puts all of \(\alpha\) in one tail. The direction matters: a cutoff \(c\) with \(P(W>c)=\alpha\) is the \((1-\alpha)\)-quantile, whereas a cutoff \(c\) with \(P(W\leq c)=\alpha\) is the \(\alpha\)-quantile.

Worked Example: Central Normal and t Intervals

Suppose an estimate is \(10\), with standard error \(2\), and there are \(\nu=12\) degrees of freedom for estimating the variance. Compare a 95% interval using a known-variance normal pivot with one using an estimated-variance t pivot.

For a central 95% interval, \(\alpha=0.05\), so the required quantile is \(1-\alpha/2=0.975\). The standard normal quantile is \(z_{0.975}\approx1.960\). The known-variance interval is \(10\pm1.960(2)\), or \((6.080,13.920)\).

When the variance is estimated, use \(t_{12;0.975}\approx2.179\), giving \(10\pm2.179(2)\), or approximately \((5.642,14.358)\). The t interval is wider because its critical value is larger. Each interval’s exact coverage relies on its corresponding pivot being valid under the model; the t interval requires independent normal observations and an estimated variance formed from the same sample.

One-Sided Bounds: Match the Tail to the Pivot

It is easy to say “use the \(1-\alpha\) quantile for a one-sided bound” and miss an important detail: that statement applies to an upper-tail cutoff. A lower-tail cutoff is the \(\alpha\)-quantile. The quantile’s probability depends on which event defines the cutoff, not just on the words “upper bound” or “lower bound.”

For example, let \(T=(\hat{\theta}-\theta)/\operatorname{SE}(\hat{\theta})\) have a symmetric t distribution with \(\nu\) degrees of freedom. A \(1-\alpha\) upper confidence bound is \(U=\hat{\theta}+t_{\nu;1-\alpha}\operatorname{SE}(\hat{\theta})\). The event \(\theta\leq U\) is equivalent to \(T\geq-t_{\nu;1-\alpha}=t_{\nu;\alpha}\). Its probability is \(1-\alpha\). A \(1-\alpha\) lower confidence bound is \(L=\hat{\theta}-t_{\nu;1-\alpha}\operatorname{SE}(\hat{\theta})\); the event \(\theta\geq L\) is equivalent to \(T\leq t_{\nu;1-\alpha}\), also with probability \(1-\alpha\).

Worked Example: Checking the Direction of a One-Sided t Bound

Use the estimate \(10\), standard error \(2\), and \(12\) degrees of freedom from the previous calculation. For a 95% upper bound, the upper-tail cutoff is \(t_{12;0.95}\approx1.782\), not the central-interval quantile \(t_{12;0.975}\). Thus \(U=10+1.782(2)=13.564\). For a 95% lower bound, \(L=10-1.782(2)=6.436\).

To check the upper bound against the defined pivot, \(\theta\leq13.564\) is equivalent to \(T\geq(10-13.564)/2=-1.782\), which is the fifth percentile of the symmetric \(t_{12}\) distribution. For the lower bound, \(\theta\geq6.436\) is equivalent to \(T\leq(10-6.436)/2=1.782\), its 95th percentile. Both events therefore have probability 0.95 under the pivot model.

Choosing the F Quantile for a Joint Ellipsoid

Suppose the target is a \(q\)-dimensional vector \(\theta\), estimated by \(\hat{\theta}\). Assume the design is fixed, the estimator is multivariate normal with mean \(\theta\) and covariance \(\sigma^2 V\), and \(V\) is known and positive definite. Suppose also that the variance estimate \(S^2\) satisfies \(\nu S^2/\sigma^2\sim\chi^2_\nu\) and is independent of \(\hat{\theta}\). These are the exact normal-theory conditions needed for the F calibration below.

Theorem: Under these conditions, \(\big((\hat{\theta}-\theta)'V^{-1}(\hat{\theta}-\theta)\big)/(qS^2)\sim F_{q,\nu}\). Therefore, the exact \(1-\alpha\) confidence ellipsoid is the set of vectors \(\theta\) satisfying \((\hat{\theta}-\theta)'V^{-1}(\hat{\theta}-\theta)\leq qS^2F_{q,\nu;1-\alpha}\).

To prove the result, define \(Z=V^{-1/2}(\hat{\theta}-\theta)/\sigma\). By the assumed multivariate normal distribution, its \(q\) coordinates are independent standard normal variables, so \(Z'Z\sim\chi^2_q\). Also, \(W=\nu S^2/\sigma^2\sim\chi^2_\nu\) and is independent of \(Z'Z\). By the definition of the F distribution, \((Z'Z/q)/(W/\nu)\sim F_{q,\nu}\). Substituting the definitions of \(Z\) and \(W\) gives \[ \frac{Z'Z/q}{W/\nu} = \frac{(\hat{\theta}-\theta)'V^{-1}(\hat{\theta}-\theta)} {qS^2}. \] Thus the quadratic form divided by \(qS^2\) has the stated F distribution. The probability that it is at most \(F_{q,\nu;1-\alpha}\) is \(1-\alpha\), which gives the ellipsoid and its exact coverage. \(\square\)

The numerator degrees of freedom equal the dimension \(q\); the denominator degrees of freedom come from estimating the variance. Increasing the target coverage increases the required F quantile, and therefore expands the ellipsoid. For a known variance, the corresponding squared standardized distance instead has a \(\chi^2_q\) distribution. These are different calibrations.

Worked Example: A Two-Dimensional 95% Ellipsoid

Let \(q=2\), \(\nu=12\), \(S^2=4\), and \(V=\operatorname{diag}(1,4)\). For 95% coverage, the needed quantile is \(F_{2,12;0.95}\approx3.8853\). The multiplier on the quadratic form is \(qS^2F_{q,\nu;0.95}=2(4)(3.8853)=31.0824\).

The region is therefore \((\hat{\theta}-\theta)'V^{-1}(\hat{\theta}-\theta)\leq31.0824\). Since \(V^{-1}=\operatorname{diag}(1,1/4)\), its boundary has semiaxes \(\sqrt{31.0824}\approx5.575\) and \(2\sqrt{31.0824}\approx11.150\), in the two coordinate directions. The area is \(\pi(5.575)(11.150)\), approximately \(195.3\) square units.

As a check on the quantile, for \(F_{2,12}\) the 95th percentile is approximately 3.8853. It is not the 95th percentile of \(\chi^2_4\), which is approximately 9.488; that value belongs to a different distribution and degrees of freedom.

How Quantiles Control Ellipsoid Volume

The effect on volume follows from the geometry of the ellipsoid, not just from its boundary equation. Write \(c=qS^2F_{q,\nu;1-\alpha}\), and transform \(d=\hat{\theta}-\theta\) to \(z=V^{-1/2}d\). The ellipsoid becomes the ball \(z'z\leq c\). The volume of that ball is \(v_q c^{q/2}\), where \(v_q\) is the volume of the unit ball in \(q\) dimensions. The inverse transformation multiplies volume by \(|V|^{1/2}\). Consequently,

\operatorname{Vol} = v_q |V|^{1/2} \left(qS^2F_{q,\nu;1-\alpha}\right)^{q/2}.

This derivation shows exactly how calibration changes size: holding the data and covariance shape fixed, volume is proportional to the F quantile raised to the power \(q/2\). It is not generally proportional to the quantile itself.

Worked Example: Comparing 90% and 95% Ellipsoid Areas

Keep \(q=2\), \(\nu=12\), \(S^2=4\), and \(V=\operatorname{diag}(1,4)\). For \(F_{2,12}\), the 90th percentile is approximately \(2.8074\), while the 95th percentile is \(3.8853\). The quadratic-form thresholds are \(2(4)(2.8074)=22.4592\) and \(2(4)(3.8853)=31.0824\), respectively.

Because \(q=2\), the ellipse area is \(\pi |V|^{1/2}c\), and \(|V|^{1/2}=2\). The 90% area is \(\pi(2)(22.4592)=44.9184\pi\), approximately \(141.1\). The 95% area is \(\pi(2)(31.0824)=62.1648\pi\), approximately \(195.3\). Their ratio is \(22.4592/31.0824\approx0.7226\), also equal to \(2.8074/3.8853\) because the dimension is two. Raising the coverage level from 90% to 95% has increased the area by about 38.4%.

A Practical Calibration Sequence

Identify the target.
Decide whether the region concerns one scalar quantity or a jointly covered vector with dimension \(q\).

Identify the pivot.
Use the distribution justified by the model and variance treatment: normal or t for a scalar pivot, and the appropriate chi-square or F calibration for a joint quadratic form.

Translate coverage into a tail probability.
Use \(1-\alpha/2\) for a central two-sided interval, \(1-\alpha\) for an upper-tail cutoff, or \(\alpha\) for a lower-tail cutoff.

Apply the quantile to the endpoint or region.
For an ellipsoid, include both the dimension multiplier \(q\) and the selected \(F_{q,\nu;1-\alpha}\).

Check the direction and scale.
Confirm the quantile’s distribution and degrees of freedom, then check that higher coverage produces a wider interval or a larger region.

Common Mistakes

Using \(1-\alpha\) for every one-sided cutoff. A cutoff defined by \(P(W>c)=\alpha\) is the \(1-\alpha\) quantile, but one defined by \(P(W\leq c)=\alpha\) is the \(\alpha\) quantile. Express the event in terms of the pivot before selecting the quantile.
Using a central quantile for a one-sided bound. A central 95% interval uses the 97.5th percentile; a one-sided 95% t bound uses the 95th percentile.
Dropping the dimension from the F threshold. The joint region uses \(qS^2F_{q,\nu;1-\alpha}\), not just \(S^2F_{q,\nu;1-\alpha}\).
Confusing the F quantile with a chi-square quantile. A chi-square quantile can be appropriate for a known-variance quadratic form, but it is not a substitute for the F quantile when variance is estimated.
Assuming volume changes linearly with the quantile in every dimension. Its dependence is \(F_{q,\nu;1-\alpha}^{q/2}\); only when \(q=2\) is it linear.

Key takeaway: Calibrate a region by matching the quantile to the pivot, tail direction, coverage level, and dimension. Once chosen, that quantile determines interval endpoints or the ellipsoid threshold—and hence the region’s volume.

Check Your Understanding

Answer each question by identifying the probability statement or calculation that determines the quantile.

Which standard normal quantile is used for a central 90% interval, and why is it not \(z_{0.90}\)?
For a pivot \(W\), what quantile defines a cutoff satisfying \(P(W\leq c)=0.025\)? What quantile defines one satisfying \(P(W>c)=0.025\)?
A jointly covered vector has dimension \(q=3\) and an estimated variance with \(\nu=20\) degrees of freedom. Which F distribution and which percentile are used for a 95% ellipsoid?
Under what exact normal-theory conditions does the quadratic-form statistic for an estimated-variance ellipsoid have an F distribution?
If all other quantities are held fixed, how does ellipsoid volume depend on the selected F quantile when \(q=4\)?