Exact Intervals for a Single-Look Normal Design

Why the Stopping Rule Changes the Interval

A fixed-sample normal interval treats the reported estimate as though it could have been observed regardless of its value. That is not the sampling situation in a design that reports an estimate only when it crosses an efficacy threshold. Among reported results, estimates below the threshold are absent, so the relevant distribution is the normal distribution conditional on crossing. Inverting that conditional distribution gives an interval with a stated coverage guarantee for trials that stop at this single look.

The result is exact under the assumptions below. “Exact” refers to its conditional coverage under the specified normal model, not to a guarantee that the model or the design assumptions are realistic. The interval is generally different from an ordinary normal interval calculated after selection.

Model and Stopping Rule

Suppose \(X_1,\ldots,X_n\) are independent observations from \(N(\mu,\sigma^2)\), with known \(\sigma\), and the estimate is \(\bar X\). Then \(\bar X\sim N(\mu,\sigma^2/n)\), and its standard error is \(SE=\sigma/\sqrt{n}\). Assume \(n\) and \(SE\) are fixed and known. At the single planned look, the trial reports the estimate if and only if it crosses a fixed efficacy threshold \(a\):

S=\{\bar X\ge a\}.

For example, a one-sided test of a reference value \(\mu_0\) that stops for efficacy when \((\bar X-\mu_0)/SE\ge c\) has \(a=\mu_0+cSE\). For inference about \(\mu\), the threshold \(a\) is fixed by the design; it is not recalculated using the unknown value of \(\mu\). We consider a reported value \(y=\bar x\) with \(y\ge a\).

Definition: A single-look, stopping-adjusted conditional confidence interval is obtained by inverting tests whose sampling distribution is that of \(\bar X\) conditional on \(S\). Its coverage statement is conditional: under the model, the probability that the interval contains \(\mu\), given that the trial crosses the threshold, is the nominal confidence level.

This target is appropriate when the estimate is available only for trials that cross the threshold. It does not claim coverage for all planned trials, including trials that do not cross and therefore produce no reported estimate.

The Conditional Sampling Distribution

Write \(\Phi\) for the standard normal cumulative distribution function. Under a candidate value \(\mu\), the probability of crossing is \(P_\mu(S)=1-\Phi((a-\mu)/SE)\). For any \(y\ge a\), the conditional cumulative distribution function of the estimate is therefore

F_\mu(y\mid S) =P_\mu(\bar X\le y\mid \bar X\ge a) =\frac{\Phi((y-\mu)/SE)-\Phi((a-\mu)/SE)} {1-\Phi((a-\mu)/SE)}.

The numerator is the probability that the estimate falls between the threshold and \(y\); the denominator is the probability that it crosses the threshold. Dividing is essential: using the unconditioned normal probability in the numerator alone would ignore the selection rule.

For a central interval with confidence level \(1-\alpha\), invert the two conditional tail probabilities. For each fixed reported \(y>a\), solve the following equations for \(\mu\):

F_{\mu_L}(y\mid S)=1-\frac{\alpha}{2}, \qquad F_{\mu_U}(y\mid S)=\frac{\alpha}{2}.

These are usually numerical equations. Their order may seem reversed: increasing \(\mu\) shifts the conditional distribution to the right and lowers its cumulative probability at a fixed \(y\). Thus the larger conditional cumulative probability gives the lower endpoint. The resulting interval is \([\mu_L,\mu_U]\).

Why the Inversion Has Exact Conditional Coverage

The conditional density on \(x\ge a\) is the normal density divided by its probability of crossing. To see that the conditional distribution moves to the right as \(\mu\) increases, compare its densities at \(\mu_2>\mu_1\). Apart from a positive factor that does not depend on \(x\), their ratio is

\frac{f_{\mu_2}(x\mid S)}{f_{\mu_1}(x\mid S)} \ \propto\ \exp\left\{\frac{(\mu_2-\mu_1)x}{SE^2}\right\}, \qquad x\ge a.

This ratio increases with \(x\), so the conditional distributions are ordered from left to right as \(\mu\) increases. Consequently \(F_\mu(y\mid S)\) decreases with \(\mu\) for fixed \(y>a\). The limits as \(\mu\) tends to negative and positive infinity are respectively 1 and 0, so each of the two equations has a unique solution.

Under the true value \(\mu\), the probability integral transform gives \(F_\mu(\bar X\mid S)\sim\text{Uniform}(0,1)\), conditional on \(S\), because the conditional distribution is continuous. The inverted interval includes \(\mu\) exactly when this quantity lies between the two tail cutoffs. Hence

P_\mu(\mu_L(\bar X)\le\mu\le\mu_U(\bar X)\mid S) =P\left(\frac{\alpha}{2}\le F_\mu(\bar X\mid S) \le 1-\frac{\alpha}{2}\,\middle|\,S\right) =1-\alpha.

This argument justifies the coverage conditional on crossing. Since the interval is defined only for reported estimates, it is not a statement about unconditional coverage across both crossing and noncrossing trials.

Worked Example: A 90% Interval After Crossing

Hypothetical data. Suppose \(SE=1\) outcome unit, the efficacy rule is \((\bar X-0)/SE\ge1.645\), and the reported estimate is \(y=2.645\) outcome units. Thus \(a=1.645\), and the trial has crossed. Find the central 90% conditional interval for \(\mu\), so \(\alpha=0.10\).

Substitute \(y=2.645\), \(a=1.645\), and \(SE=1\) into the conditional distribution. The endpoint equations are

\frac{\Phi(2.645-\mu)-\Phi(1.645-\mu)} {1-\Phi(1.645-\mu)} =0.95 \quad\text{and}\quad \frac{\Phi(2.645-\mu)-\Phi(1.645-\mu)} {1-\Phi(1.645-\mu)} =0.05.

Numerically solving the first equation gives \(\mu_L\approx-0.54\) outcome units; solving the second gives \(\mu_U\approx4.25\) outcome units. Therefore the 90% conditional interval is approximately \([-0.54,4.25]\) outcome units. Under the assumed model and stopping rule, 90% of intervals formed this way would contain \(\mu\) among trials that cross the threshold. This does not mean there is a 90% probability that this fixed parameter lies in this particular interval.

Worked Example: A Larger Distance Above the Threshold

Hypothetical data. Keep \(SE=1\) and \(a=1.645\), but suppose the reported estimate is \(y=3.645\), two standard errors above the threshold. Again seek a 90% conditional interval.

The equations now use \(y-a=2\). For example, on writing \(\eta=(\mu-a)/SE\), the conditional cumulative probability at the observed value becomes

F_\mu(y\mid S) =\frac{\Phi(2-\eta)-\Phi(-\eta)}{\Phi(\eta)}.

Set this expression equal to 0.95 and 0.05, respectively. Numerical solutions are \(\eta_L\approx0.06\) and \(\eta_U\approx3.65\). Converting back to the original units using \(\mu=a+\eta SE\) gives

\mu_L=1.645+(0.06)(1)\approx1.71, \qquad \mu_U=1.645+(3.65)(1)\approx5.30.

The conditional 90% interval is approximately \([1.71,5.30]\) outcome units. In this example the reported estimate is farther above the boundary, and the inverted endpoints are correspondingly different. The endpoints are not obtained by simply adding and subtracting a fixed normal margin from \(y\).

Worked Example: Converting the Interval to a Different Standard Error

Hypothetical data. Suppose the same efficacy rule has \(\mu_0=0\), \(c=1.645\), and \(SE=0.5\) outcome units. Then \(a=1.645(0.5)=0.8225\). A hypothetical reported estimate is \(y=1.3225\), which is one standard error above the boundary.

Because \((y-a)/SE=1\), the endpoint equations in standardized coordinates are the same as in the first example: their solutions are approximately \(\eta_L=-2.18\) and \(\eta_U=2.60\), where \(\eta=(\mu-a)/SE\). Convert each endpoint using the actual threshold and standard error:

\mu_L=0.8225+(-2.18)(0.5)\approx-0.27, \qquad \mu_U=0.8225+(2.60)(0.5)\approx2.12.

Thus the interval is approximately \([-0.27,2.12]\) outcome units. The standardized limits can be reused here because the standardized distance between the observation and the threshold is unchanged; the endpoints in outcome units must still be rescaled.

Practical Computation and Checks

A numerical solver can find the roots of the conditional-probability equations. A reliable calculation keeps the threshold, estimate, and standard error in consistent units and evaluates the normal tail probabilities accurately, especially when crossing is rare. Check that the lower endpoint solves the equation with \(1-\alpha/2\), that the upper endpoint solves the equation with \(\alpha/2\), and that \(\mu_L<\mu_U\). Substitution of the computed endpoints back into the conditional CDF is a direct numerical check.

A useful computational simplification is to work with \(z=(y-a)/SE\) and \(\eta=(\mu-a)/SE\). Then

F_\mu(y\mid S)= \frac{\Phi(z-\eta)-\Phi(-\eta)}{\Phi(\eta)}.

Solve in \(\eta\), then return to the original scale with \(\mu=a+\eta SE\). This form also makes it easier to identify a unit mismatch: \(z\) and \(\eta\) must be dimensionless.

Common Mistakes, Alternatives, and Limitations

Ignoring selection. The ordinary interval \(\bar x\pm z_{1-\alpha/2}SE\) uses the unconditioned normal law. It does not generally have the stated conditional coverage after crossing. The correction is to divide by the crossing probability in \(F_\mu(y\mid S)\) before inverting.
Using the wrong endpoint equation. Since the conditional CDF decreases with \(\mu\), the \(1-\alpha/2\) equation produces the lower endpoint, not the upper endpoint. Check the ordering by substitution.
Confusing conditional and unconditional coverage. Exact coverage here is given crossing the threshold. It does not provide an interval for an unreported noncrossing trial or automatically establish a coverage target that averages over all planned trials.
Applying the calculation to a different design. The formula assumes one fixed look and one fixed threshold. Multiple looks, futility stopping, a threshold that changes with the data, or reporting rules beyond crossing require the corresponding stopped sampling law; this one-look calculation is not a substitute.
Treating an estimated standard error as known. The derivation assumes a normal estimate with known, fixed \(SE\). If the variance is estimated, the relevant sampling distribution and conditioning calculation change.

The main alternative is to report an ordinary fixed-sample interval, which is simpler but does not adjust for selecting estimates by crossing. A conditional interval instead targets coverage among the selected, reported results and may be wide, particularly when crossing is rare. Its guarantee depends on the normal model, known standard error, fixed boundary, and accurate record of the stopping rule. If any of these fail, the numerical interval can still be calculated, but its exact conditional coverage claim no longer follows from this derivation.

Key takeaway: For a single-look normal estimate reported only when \(\bar X\ge a\), use the normal distribution truncated below at \(a\), then invert its conditional cumulative probabilities. The resulting interval is exact conditional on crossing under the stated known-variance normal model.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

In the model above, write the conditional cumulative distribution function of \(\bar X\) at a reported value \(y\ge a\).
For a 95% central conditional interval, which conditional CDF value gives the lower endpoint, and which gives the upper endpoint?
Suppose the reported estimate is below the fixed threshold. Can the single-look conditional interval described here be applied to it? Explain.
What does the exact 95% coverage statement mean, and what does it not mean?

Solutions. (1) \(F_\mu(y\mid S)=[\Phi((y-\mu)/SE)-\Phi((a-\mu)/SE)]/[1-\Phi((a-\mu)/SE)]\). (2) Solve \(F_{\mu_L}(y\mid S)=0.975\) for the lower endpoint and \(F_{\mu_U}(y\mid S)=0.025\) for the upper endpoint, because the CDF decreases as \(\mu\) increases. (3) No. Under the stated rule, such an estimate is not reported after crossing, and the conditional distribution and inversion are defined for \(y\ge a\). (4) In repeated trials under the assumed model, 95% of intervals constructed this way contain \(\mu\) among trials that cross. It is not a 95% probability statement about the fixed \(\mu\) after observing one particular interval, nor does it assert 95% coverage over all planned trials without conditioning on crossing.