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Estimation and Confidence Intervals After Adaptation · Tutorial 296 of 799

Confidence Intervals With Estimated Variance

Learn how adaptive sample size changes the distribution of estimated variances, and construct a conservative interval whose unconditional coverage is protected after adaptation.

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What You'll Learn

  • State a two-stage normal model and distinguish stagewise means from variance estimates.
  • Explain why a stage 2 variance estimate can depend unconditionally on the stage 1-based adaptation branch.
  • Calculate a plug-in interval and identify why it may not retain its nominal coverage.
  • Construct branch-aware variance upper bounds and combine them into a conservative interval.
  • Distinguish unconditional coverage from coverage conditional on a particular adaptation branch.
  • Diagnose when normal and variance-estimation assumptions are doubtful.

Why Estimating Variances Matters After Adaptation

An interval based on a known variance can use a standard normal quantile when its estimator is normal. In practice, the variances are usually estimated. That changes the distribution of the interval statistic, and adaptation can change it again: if the stage 1 variance estimate determines the stage 2 sample size, the stage 2 variance estimate has a different sampling distribution on different branches of the design.

The issue is not resolved merely by substituting estimated variances into a standard error. A plug-in interval may be useful as an approximation, but its coverage needs justification under the actual adaptation rule. This tutorial develops a conservative variance-adjusted interval for a two-stage normal model and makes clear which coverage guarantee it provides.

Definition: Coverage is the long-run probability that an interval procedure contains the fixed parameter \(\theta\) under repeated trials generated according to the stated model and adaptation rule. Unconditional coverage averages over all possible adaptation branches. Conditional coverage on branch \(b\) is the probability of coverage given that branch \(b\) was selected. These are different guarantees.

A Two-Stage Normal Model

Suppose the target is a common treatment effect \(\theta\), estimated in two independent stages. For stage \(j\), observations are normally distributed with mean \(\theta\) and variance \(\sigma_j^2\). The variances may differ between stages, and both are unknown. Write \(\bar X_j\) for the stage mean and \(s_j^2\) for its sample variance. Within each stage, observations are independent and identically distributed, and the stage mean is independent of the sample variance.

Stage 1 has fixed size \(n_1\). An adaptation rule uses \(s_1^2\) to select the stage 2 size \(n_{2,b}\), where \(b\) indexes the branch. Conditional on the selected size, stage 2 observations are generated independently of stage 1. The stage 2 sample variance therefore has a chi-square distribution conditional on the selected branch, with degrees of freedom determined by that branch:

$$ \frac{(n_{2,b}-1)s_2^2}{\sigma_2^2}\,\bigg|\,B=b \sim \chi^2_{n_{2,b}-1}. $$

If \(n_{2,b}\) varies across branches, these conditional distributions differ. Consequently, stage 2 data being generated independently of stage 1 does not imply that \(s_2^2\) is unconditionally independent of stage 1: its distribution depends on the stage 1-based branch. The conditional statement is the appropriate one.

For clarity, consider a prespecified combined estimator \(\hat\theta=a_1\bar X_1+a_2\bar X_2\), where \(a_1+a_2=1\) and the coefficients are fixed, not chosen from the observed stage 1 mean. Conditional on branch \(b\), its variance is \(V_b=a_1^2\sigma_1^2/n_1+a_2^2\sigma_2^2/n_{2,b}\). Because stage 1 normal sample mean and sample variance are independent, selection based on \(s_1^2\) does not truncate \(\bar X_1\). Under the model, \(\hat\theta-\theta\), conditional on branch, is normal with mean zero and variance \(V_b\). This setup isolates how variance estimation and branch-dependent sample size affect interval coverage; it does not cover every possible adaptive estimator.

Model assumptions: The stagewise observations follow the stated normal models; stage 1 and newly generated stage 2 observations are independent; the adaptation rule and branch sizes are specified; and the combined estimator uses fixed coefficients summing to one. If the rule selects using a stage 1 treatment-effect estimate, or if the estimator’s weights depend on the observed data, the conditional distribution of the estimator may require a separate selection adjustment.

Why the Usual Plug-In Interval Is Not Exact

A common plug-in standard error is \(\widehat{SE}_b=\sqrt{a_1^2s_1^2/n_1+a_2^2s_2^2/n_{2,b}}\). One may place a normal critical value around \(\hat\theta\). But the estimated variance in this expression is random, and its distribution is not generally a single scaled chi-square variable. In addition, conditional on a branch selected using \(s_1^2\), the stage 1 variance estimate has a truncated distribution. Thus the ratio of estimation error to plug-in standard error is not generally standard normal or an ordinary \(t\) statistic.

Worked Example: A Plug-In Interval on One Branch

Suppose, hypothetically, \(n_1=10\), \(n_2=10\), and the adaptation rule selected the larger stage 2 size. The observed means are \(\bar X_1=5.2\) and \(\bar X_2=6.0\), with \(s_1^2=4.00\) and \(s_2^2=2.25\). Use equal fixed weights \(a_1=a_2=0.5\). The combined estimate is \(\hat\theta=0.5(5.2)+0.5(6.0)=5.60\).

The plug-in variance estimate and standard error are \(\widehat V=0.25(4.00/10)+0.25(2.25/10)=0.15625\) and \(\widehat{SE}=\sqrt{0.15625}=0.3953\). A nominal 90% normal interval uses \(z_{0.95}=1.645\), giving margin \(1.645(0.3953)=0.6503\) and interval \((4.950,6.250)\) in the units of \(\theta\).

The arithmetic is valid, but the nominal label does not establish 90% coverage after adaptation. It ignores the sampling uncertainty in both variance estimates and the branch-dependent distribution of \(s_2^2\). Coverage of this plug-in procedure would need to be evaluated or justified for the particular design.

A Conservative Variance-Adjusted Interval

One way to protect unconditional coverage is to form upper confidence bounds for each variance, using the correct stagewise degrees of freedom, and then use those bounds to construct a conservative standard error. For any chi-square degrees of freedom \(d\), let \(q_{\gamma,d}\) be the lower \(\gamma\) quantile, defined by \(P(\chi^2_d\le q_{\gamma,d})=\gamma\). An upper confidence bound for a variance based on sample variance \(s^2\) and \(d\) degrees of freedom is \(d s^2/q_{\gamma,d}\), since the event that this bound is at least the true variance is equivalent to \(\chi^2_d\ge q_{\gamma,d}\), with probability \(1-\gamma\).

For stage 1, \(d_1=n_1-1\). Its usual chi-square bound has unconditional coverage \(1-\gamma_1\). It need not have that coverage conditional on the selected branch, because the branch was chosen using \(s_1^2\). For stage 2, use the branch-specific degrees of freedom \(d_{2,b}=n_{2,b}-1\) and a lower quantile for those degrees of freedom. Conditional on branch \(b\), this stage 2 bound has coverage \(1-\gamma_2\). Thus its failure probability is at most \(\gamma_2\) unconditionally as well, by averaging over branches.

$$ U_1=\frac{d_1s_1^2}{q_{\gamma_1,d_1}}, \qquad U_{2,b}=\frac{d_{2,b}s_2^2}{q_{\gamma_2,d_{2,b}}}, \qquad SE_{U,b}= \sqrt{\frac{a_1^2U_1}{n_1}+\frac{a_2^2U_{2,b}}{n_{2,b}}}. $$

By the union bound, the probability that either variance bound fails is no greater than \(\gamma_1+\gamma_2\). When both bounds hold, the true \(V_b\) is no greater than \(SE_{U,b}^2\). Conditional on the branch and variance estimates, the standardized estimation error \((\hat\theta-\theta)/\sqrt{V_b}\) is standard normal and independent of the variance estimates under this model. Therefore a central normal event with probability \(1-\alpha_*\), together with both variance bounds holding, guarantees that \(\theta\) is in the interval. The resulting unconditional coverage is at least \((1-\alpha_*)(1-\gamma_1-\gamma_2)\).

$$ \hat\theta\ \pm\ z_{1-\alpha_*/2}\,SE_{U,b}. $$

For a target coverage of at least \(1-\alpha\), one convenient conservative allocation is \(\alpha_*+\gamma_1+\gamma_2\le\alpha\), since \((1-\alpha_*)(1-\gamma_1-\gamma_2)\ge1-\alpha_*-\gamma_1-\gamma_2\). The price of this protection can be a wide interval, particularly with small samples or very conservative variance-bound quantiles.

Worked Example: Constructing the Variance-Adjusted Interval

Continue with the hypothetical data above, and suppose the adaptation rule chose \(n_2=10\), so both stages have \(d=9\) degrees of freedom. Construct an interval with at least 90% unconditional coverage by choosing \(\gamma_1=\gamma_2=0.025\) and \(\alpha_*=0.05\). The allocation satisfies \(\alpha_*+\gamma_1+\gamma_2=0.10\). For \(d=9\), use the lower chi-square quantile \(q_{0.025,9}\approx2.700\).

The stage 1 variance upper bound is \(U_1=9(4.00)/2.700=13.33\). The stage 2 bound, using the branch-specific \(d_{2,b}=9\), is \(U_{2,b}=9(2.25)/2.700=7.500\). With equal weights and both sample sizes equal to 10, \(SE_{U,b}=\sqrt{0.25(13.33/10)+0.25(7.500/10)} =\sqrt{0.5208}=0.7217\).

Since \(z_{0.975}=1.960\), the margin is \(1.960(0.7217)=1.415\). The variance-adjusted interval is therefore \(5.60\pm1.415=(4.185,7.015)\), in the units of the treatment-effect estimate. Its guaranteed coverage is at least \(0.95(0.95)=0.9025\), or 90.25%, unconditionally under the model. This is a lower guarantee, not a claim that its actual coverage equals 90.25%.

Branch Dependence, Diagnostics, and Alternatives

Consider a rule that chooses \(n_2=6\) on one branch and \(n_2=10\) on another. Then \(s_2^2\), conditional on the first branch, has a scaled \(\chi^2_5\) distribution; conditional on the second, it has a scaled \(\chi^2_9\) distribution. For example, if \(\sigma_2^2=4\), the conditional variances of \(s_2^2\) are \(2(4^2)/5=6.400\) and \(2(4^2)/9=3.556\), respectively. The stage 2 observations are generated independently of stage 1, but the distribution of their variance estimate changes with the branch. This is why using a single, unqualified stage 2 variance distribution can give the wrong uncertainty calculation.

The conservative interval above targets unconditional coverage. It does not promise the same coverage after conditioning on each branch. In particular, the stage 1 variance bound is not calibrated conditionally on a branch defined using that same variance estimate. A conditional guarantee requires accounting for the truncated distribution induced by the rule; its calibration generally depends on the candidate variance and may require numerical inversion. Do not describe an unconditional guarantee as conditional coverage.

Other approaches have different trade-offs. A plug-in normal interval is simple but generally approximate. A Satterthwaite approximation replaces a weighted sum of estimated variance components by a scaled chi-square distribution and uses an effective degrees-of-freedom calculation; this can be useful, but branch selection and truncation must be incorporated if conditional accuracy is sought. Directly calibrating a branch-aware statistic under the full model can target conditional or unconditional coverage more closely, at the cost of design-specific calculations. The conservative upper-bound method is transparent and has a clear coverage argument, but often sacrifices interval length.

Useful diagnostics include checking the prespecified adaptation rule, the actual branch-specific sample sizes, residual patterns within each stage, and whether the normal and constant-within-stage variance assumptions are plausible. A variance estimate near zero, severe outliers, or a rule that also selects on the stage 1 mean are warnings that the stated model or coverage argument may not apply. Small-sample exact procedures and simulation-based assessment can be useful in such cases, but they must represent the adaptation mechanism rather than treating the realized sample sizes as if they had been fixed in advance.

Common pitfall: “Stage 2 observations are independent of stage 1, so \(s_2^2\) is unconditionally independent of stage 1.” This overlooks that the stage 2 sample size is selected from stage 1. The correct statement is that stage 2 observations are generated independently of stage 1 and that the stage 2 variance estimate has its stated chi-square distribution conditional on the selected sample size or branch. Its unconditional distribution is generally a mixture over branches.

Check Your Understanding

Use the model and results above. Solutions are included after the questions.

  1. If stage 2 size is selected using \(s_1^2\), what is the appropriate conditioning for stating the chi-square distribution of \(s_2^2\)?
  2. Why does the ordinary stage 1 upper variance bound not automatically have its nominal coverage conditional on a branch selected using \(s_1^2\)?
  3. With \(\alpha_*=0.04\), \(\gamma_1=0.02\), and \(\gamma_2=0.03\), what lower bound does the product argument give for unconditional coverage?
  4. For fixed weights summing to one, write the conditional variance of \(\hat\theta\) on branch \(b\).
  5. Does an unconditional 90% coverage guarantee imply 90% coverage on every adaptation branch?

Solutions. (1) State the distribution conditional on the selected branch or sample size; the degrees of freedom then equal \(n_{2,b}-1\). (2) Selection truncates the distribution of \(s_1^2\) within a branch, so its conditional distribution differs from its unselected chi-square distribution. (3) The lower bound is \((1-0.04)(1-0.02-0.03)=0.96(0.95)=0.912\), or 91.2%. (4) \(V_b=a_1^2\sigma_1^2/n_1+a_2^2\sigma_2^2/n_{2,b}\). (5) No. Unconditional coverage averages across branches and does not guarantee the same conditional coverage on each one.

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