Why Critical Values and Tail Allocation Matter
An asymptotic confidence interval combines an estimate, an estimated standard error, and a critical value from a reference distribution. The critical value determines how far an endpoint lies from the estimate. For a two-sided interval, the confidence level alone does not determine both endpoints: one must also decide how to divide the total noncoverage probability between the two tails.
Equal allocation is the familiar default, but it is not the only choice. Unequal allocation can be useful when the two directions of error have different practical importance. It changes the critical value at each endpoint, and therefore can change both the interval’s location and its width. The calculations below assume a studentized estimator that is asymptotically standard normal, as in the earlier material on Wald intervals.
For an estimator \(\hat{\theta}\) with estimated standard error \(\widehat{\operatorname{se}}(\hat{\theta})\), the allocated two-sided interval is \[ \left[\hat{\theta}-c_L\widehat{\operatorname{se}}(\hat{\theta}),\ \hat{\theta}+c_U\widehat{\operatorname{se}}(\hat{\theta})\right]. \] Here \(\alpha_L\) names the probability of missing the parameter because the lower endpoint is too high; that event corresponds to the upper tail of the standardized estimator. Likewise, \(\alpha_U\) names the probability of missing because the upper endpoint is too low; that event corresponds to the lower tail of the standardized estimator. These endpoint-based labels help keep the signs straight.
Equal-Tail Two-Sided Intervals
For an equal-tail interval, each tail receives half of the total noncoverage probability: \(\alpha_L=\alpha_U=\alpha/2\). The two critical values are then equal, \(c_L=c_U=z_{1-\alpha/2}\), and the interval is symmetric around the estimate. For example, a nominal 95% interval has \(\alpha=0.05\), so its critical value is \(z_{0.975}\), approximately \(1.96\).
Worked Example: A 95% Equal-Tail Interval
Suppose an estimator is \(12.4\), and its estimated standard error is \(0.8\). Construct an equal-tail 95% asymptotic interval using \(z_{0.975}\approx 1.96\).
The total noncoverage probability is \(0.05\), so each tail receives \(0.025\). The margin of error is \[ 1.96(0.8)=1.568. \] Thus the endpoints are \(12.4-1.568=10.832\) and \(12.4+1.568=13.968\). The resulting interval is approximately \([10.832,13.968]\). Its equal distances from the estimate reflect the equal tail allocation.
A two-sided interval and a one-sided confidence limit answer different questions. A two-sided interval puts finite endpoints on both sides of the estimate. A one-sided limit leaves one side unbounded and spends the entire noncoverage probability on the direction that matters for that limit.
Worked Example: A One-Sided 90% Confidence Limit
Suppose an estimate is \(5.8\) with estimated standard error \(0.6\). For a one-sided 90% limit, use \(z_{0.90}\approx 1.28155\). The margin is \[ 1.28155(0.6)=0.76893. \]
The upper confidence bound is \(5.8+0.76893=6.56893\), giving the set \((-\infty,6.56893]\). The lower confidence bound is \(5.8-0.76893=5.03107\), giving \([5.03107,\infty)\). These are two different one-sided statements; neither is the two-sided interval \([5.03107,6.56893]\), which would use a different allocation and critical value.
Unequal Allocation and Endpoint Changes
With unequal allocation, the lower and upper endpoints use different quantiles. If \(\alpha_L\) is smaller, then \(1-\alpha_L\) is larger and \(c_L\) increases, moving the lower endpoint downward. If \(\alpha_U\) is smaller, \(c_U\) increases and the upper endpoint moves upward. A tail receiving more noncoverage probability gets a smaller critical value; while that critical value remains positive, its corresponding endpoint moves closer to the estimate.
The resulting interval need not be centered at \(\hat{\theta}\). Its width is \((c_L+c_U)\widehat{\operatorname{se}}(\hat{\theta})\). Thus changing the allocation can change the width as well as shift the endpoints. The following calculation uses a fixed estimate and standard error so that these effects can be seen directly.
Worked Example: Comparing Equal and Unequal Allocation
Let the estimate be \(0.42\), with estimated standard error \(0.022913\), and consider a 95% interval. The equal-tail critical value is approximately \(1.96\), giving a margin of \[ 1.96(0.022913)=0.04490948. \] The equal-tail interval is therefore approximately \([0.37509052,\,0.46490948]\).
Now allocate \(\alpha_L=0.01\) and \(\alpha_U=0.04\). The critical values are \(z_{0.99}\approx 2.32635\) and \(z_{0.96}\approx 1.75069\). Rounding these critical values first gives \(2.326\) and \(1.751\). Using those rounded operands, \[ 2.326(0.022913)=0.053295638,\qquad 1.751(0.022913)=0.040120663. \] The unequal-tail endpoints are \(0.42-0.053295638=0.366704362\) and \(0.42+0.040120663=0.460120663\), or approximately \([0.36670,\,0.46012]\) to five decimal places.
The lower endpoint is farther from the estimate than in the equal-tail interval, while the upper endpoint is nearer. The unequal-tail interval’s width is \(0.053295638+0.040120663=0.093416301\), compared with \(0.08981896\) for the equal-tail interval. The difference follows from the chosen quantiles; it is not a change in the estimate or its standard error.
Asymptotic Coverage With Allocated Tails
The coverage justification uses a studentized pivot. Suppose \(\hat{\theta}\) estimates a scalar parameter \(\theta\), the estimated standard error is positive with probability tending to one, and \[ T_n=\frac{\hat{\theta}-\theta}{\widehat{\operatorname{se}}(\hat{\theta})} \xrightarrow{d}N(0,1). \] The critical values below are fixed as \(n\) increases. This assumption is the relevant large-sample condition; it does not assert that the interval has exact coverage at a particular finite sample size.
Proof. Write \(c_L=z_{1-\alpha_L}\) and \(c_U=z_{1-\alpha_U}\). Let \(A_n=\{\widehat{\operatorname{se}}(\hat{\theta})>0\}\). On \(A_n\), the event that the interval covers \(\theta\) is equivalent to \[ \hat{\theta}-c_L\widehat{\operatorname{se}}(\hat{\theta}) \leq \theta \leq \hat{\theta}+c_U\widehat{\operatorname{se}}(\hat{\theta}). \] Subtracting \(\theta\), dividing by the positive standard error, and rearranging gives \[ -c_U\leq T_n\leq c_L. \] The coverage event and this pivot event agree on \(A_n\), so their probabilities differ by at most \(P(A_n^c)\to0\). Since \(T_n\xrightarrow{d}Z\), where \(Z\sim N(0,1)\), and a normal random variable assigns probability zero to either fixed endpoint, the probabilities of these intervals converge: \[ P(-c_U\leq T_n\leq c_L) \longrightarrow P(-c_U\leq Z\leq c_L). \] By the quantile definitions, \(\Phi(c_L)=1-\alpha_L\) and \(\Phi(-c_U)=\alpha_U\). Therefore, \[ P(-c_U\leq Z\leq c_L) =\Phi(c_L)-\Phi(-c_U) =1-\alpha_L-\alpha_U. \] This proves the stated limiting coverage. \(\square\)
The same pivot also explains one-sided limits. For the upper bound \(\hat{\theta}+z_{1-\alpha}\widehat{\operatorname{se}}(\hat{\theta})\), coverage of \(\theta\) is the event \(T_n\geq -z_{1-\alpha}\). Its limiting probability is \(\Phi(z_{1-\alpha})=1-\alpha\). For the lower bound \(\hat{\theta}-z_{1-\alpha}\widehat{\operatorname{se}}(\hat{\theta})\), coverage is \(T_n\leq z_{1-\alpha}\), with the same limiting probability. These claims follow from convergence in distribution and continuity of the normal distribution at the cutoff.
Choosing an Allocation and Avoiding Common Errors
Equal-tail allocation is a transparent default when neither direction of error has special priority. An unequal allocation can be justified when the consequences of the two errors differ, but it should be selected for a reason connected to the scientific question, not simply because it produces a preferred endpoint after looking at the data. The nominal confidence level alone does not specify the allocation.
- Using the wrong two-sided quantile. A 95% equal-tail interval uses \(z_{0.975}\), not \(z_{0.95}\). The latter leaves 5% in one tail and is the critical value for a 95% one-sided bound.
- Putting the critical values on the wrong endpoints. In the allocated interval, \(c_L\) multiplies the subtraction for the lower endpoint, and \(c_U\) multiplies the addition for the upper endpoint.
- Assuming unequal allocation preserves symmetry. Different tail probabilities generally give different quantiles, so the endpoints are not equally distant from the estimate.
- Calling an asymptotic interval an exact finite-sample interval. The coverage result is a limit as sample size grows under the stated pivot condition. It need not equal the nominal level in a small sample.
A practical check is to verify that \(\alpha_L+\alpha_U=\alpha\), then confirm the endpoint directions: a smaller allocation at an endpoint gives it a larger critical value; when that critical value remains positive throughout the change, the endpoint moves farther from the estimate. Finally, distinguish the critical-value calculation from the validity of the normal approximation itself. Correct arithmetic cannot compensate for a poor approximation or an unsuitable standard error.
Check Your Understanding
Answer each question by identifying the relevant tail probability and critical value.
- For an equal-tail 90% interval, what probability is assigned to each tail, and which standard normal quantile supplies the critical value?
- A 95% interval allocates \(0.02\) to one tail and \(0.03\) to the other. Write both critical values in quantile notation and state which one multiplies the subtraction from the estimate.
- For an upper one-sided 99% confidence bound, which normal quantile is used? Why is it not the equal-tail 99% interval quantile?
- State the studentized convergence condition that supports the limiting coverage result for an allocated interval.
- If the noncoverage allocation for the lower endpoint is reduced while the estimate and standard error stay fixed, what happens to that endpoint? Explain using the quantile.