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Conditional Error Functions · Tutorial 249 of 799

Unequal Information At Interim

Learn to derive and calculate conditional error from unequal stage information, and check the result against the normal-score model.

Advanced 12 min read

What You'll Learn

  • Distinguish interim information from the interim sample-size fraction.
  • Derive the final normal score’s conditional mean and variance given the interim score.
  • Calculate one-sided conditional error under an unequal-information design.
  • Check conditional-error calculations using limits, monotonicity, and variance diagnostics.
  • Recognize when the independent-increments normal model does not apply.

Why Unequal Information Matters

At an interim analysis, the evidence already observed contributes to the final test, while additional observations contribute new information. When the interim and remaining information differ, the final standardized score is not generally an equal-weight average of an interim score and a future score. Its weights—and its conditional variance—are determined by the information available at each stage.

The conditional error is the probability, under the null hypothesis and conditional on the interim data, that the originally specified final test will reject. For a normal-score test, this probability depends on the observed interim score and on the fraction of total information already accrued. Deriving that dependence correctly is important: using a sample-size fraction where information differs, or ignoring the conditional variance, changes the error function.

Definition: Suppose a one-sided test rejects for a final normal score \(Z_F\) greater than a fixed critical value \(c\). Given an interim score \(Z_1=z\), the conditional error is \(A(z)=P(Z_F>c\mid Z_1=z,H_0)\), evaluated under the model specified by the original design.

Model and Conditional Distribution

Assume information is additive across the two stages. Let \(I_1\) be the interim information and \(I_2\) the additional information planned after interim, so total information is \(I=I_1+I_2\). Define the information fraction \(t=I_1/I\), with \(0<t<1\). The fraction \(t\) need not equal the fraction of participants enrolled: that equality holds only in settings where information per participant is constant.

Use the canonical normal-score model under \(H_0\): the standardized interim score \(Z_1\) has a standard normal null distribution, and the standardized score increment from the remaining information, \(U_2\), is independent of \(Z_1\) and also standard normal under \(H_0\). The final score is the information-weighted combination

$$ Z_F=\sqrt{t}\,Z_1+\sqrt{1-t}\,U_2. $$

This model applies, for example, to a normal-score test with independent stagewise contributions and known variance, and to suitable large-sample canonical approximations. Its key requirements are that the information contributions add, the future score increment is independent of the interim score under the null, and the stated standard normal approximation is adequate. If these conditions fail, the conditional distribution must be derived for the model that actually applies.

Given \(Z_1=z\), the only random part of \(Z_F\) is \(U_2\). Since \(U_2\sim N(0,1)\), the conditional distribution follows directly:

$$ Z_F\mid Z_1=z,H_0 \sim N\!\left(\sqrt{t}\,z,\;1-t\right). $$

The second argument of \(N(\cdot,\cdot)\) is the variance. Thus the conditional mean is \(\sqrt{t}\,z\), and the conditional standard deviation is \(\sqrt{1-t}\). The variance is not \(1\): after conditioning, the interim contribution is fixed, and only the remaining information contributes randomness.

Deriving the Conditional Error Function

For the original upper-tail test, rejection occurs when \(Z_F>c\). Standardizing the conditional normal distribution gives the conditional error function:

$$ \begin{aligned} A(z) &=P(Z_F>c\mid Z_1=z,H_0)\\ &=1-\Phi\!\left( \frac{c-\sqrt{t}\,z}{\sqrt{1-t}} \right), \end{aligned} $$

where \(\Phi\) is the standard normal cumulative distribution function. The numerator is the distance from the conditional mean to the rejection threshold; the denominator is the conditional standard deviation. This expression is the null probability of crossing the original final boundary, given the observed interim score.

Result: Under the independent-increments normal-score model, with \(0<t<1\) and a fixed upper critical value \(c\), the conditional error after observing \(Z_1=z\) is \(A(z)=1-\Phi((c-\sqrt{t}\,z)/\sqrt{1-t})\).

A short derivation also makes the role of the weights clear. Under the null, \(Z_1\) and \(U_2\) are independent standard normal variables. The weighted sum has null mean zero and variance \(t+(1-t)=1\), so it is a standard normal final score. Conditional on \(Z_1=z\), the first term is the constant \(\sqrt{t}z\), while the remaining term has variance \(1-t\). Therefore \(Z_F\mid Z_1=z,H_0\) has exactly the mean and variance used above, and standardizing the event \(Z_F>c\) yields the formula.

Worked Example: One Quarter of the Information at Interim

Suppose the total information is 100 units, with 25 units accrued at interim and 75 units remaining. The original one-sided test rejects when \(Z_F>1.96\), and the observed interim score is \(z=0\). Calculate the conditional error.

The information fraction is \(t=25/100=0.25\), so the conditional mean is \(\sqrt{0.25}(0)=0\), and the conditional standard deviation is \(\sqrt{0.75}\). Substitution gives

$$ \begin{aligned} A(0) &=1-\Phi\!\left( \frac{1.96-\sqrt{0.25}(0)}{\sqrt{0.75}} \right)\\ &=1-\Phi(2.26321)\\ &\approx 0.0118. \end{aligned} $$

Thus, under the stated null model and conditional on an interim score of zero, the chance of crossing the original final boundary is about 1.18%. The calculation uses the remaining-information standard deviation, not the standard deviation of the unconditional final score.

Worked Example: A Favorable Interim Score With Unequal Information

Suppose 60 information units have accrued and 40 remain, so \(t=0.60\). The interim score is \(z=1\), and the original critical value is \(c=1.96\). Find the conditional error.

The conditional mean is \(\sqrt{0.60}(1)=0.774597\), and the conditional standard deviation is \(\sqrt{0.40}=0.632456\). Therefore,

$$ \begin{aligned} A(1) &=1-\Phi\!\left( \frac{1.96-\sqrt{0.60}(1)}{\sqrt{0.40}} \right)\\ &=1-\Phi(1.87429)\\ &\approx 0.0305. \end{aligned} $$

Conditional on this interim result, the original test has about a 3.05% probability of rejection under the null model. This exceeds the conditional error at \(z=0\), because a positive interim score shifts the conditional mean of the final score toward the upper rejection boundary.

Worked Example: Information Fraction Is Not Necessarily Sample Fraction

Consider a hypothetical trial with 30 information units accrued and 70 additional information units planned. The interim score is \(z=1.5\), and the final test uses \(c=1.96\). Even if the number of participants accrued were not 30% of the planned total, the specified information gives \(t=30/(30+70)=0.30\). Calculate the conditional error.

The conditional mean is \(\sqrt{0.30}(1.5)=0.821584\), and the conditional standard deviation is \(\sqrt{0.70}=0.836660\). Hence,

$$ \begin{aligned} A(1.5) &=1-\Phi\!\left( \frac{1.96-\sqrt{0.30}(1.5)}{\sqrt{0.70}} \right)\\ &=1-\Phi(1.36067)\\ &\approx 0.0868. \end{aligned} $$

The conditional error is about 8.68%. The calculation is based on the information ratio, not a head count. If information per participant differs between stages, replacing \(t=0.30\) with a sample-size fraction would generally give the wrong conditional distribution.

Checks, Diagnostics, and Interpretation

Several checks help catch mistakes in an unequal-information calculation:

  • Check the information fraction. Confirm that \(t=I_1/(I_1+I_2)\), using information on the same scale, and that \(0<t<1\) when both stages contribute.
  • Check the conditional variance. It must be \(1-t\) under this canonical model. It decreases as more information has accrued because less random information remains.
  • Check the direction in \(z\). For \(0<t<1\), \(A(z)\) increases with \(z\): a larger interim score raises the conditional mean while leaving the conditional standard deviation unchanged. Differentiating gives \(A'(z)=\phi(q)\sqrt{t/(1-t)}>0\), where \(q=(c-\sqrt{t}z)/\sqrt{1-t}\) and \(\phi\) is the standard normal density.
  • Check limiting behavior. As \(t\) approaches zero, the interim score contributes almost nothing and the conditional error approaches the unconditional tail probability \(1-\Phi(c)\). As \(t\) approaches one, little future randomness remains; for a fixed \(z<c\), the conditional rejection probability approaches zero, and for \(z>c\) it approaches one.

The conditional error is not the probability that \(H_0\) is true. It is a probability computed under \(H_0\), conditional on the interim score. It also differs from an unconditional type I error rate: the latter averages rejection probability over all possible interim scores under the null.

Comparison and Limits

With equal information in the two stages, \(t=0.5\), and the formula uses the corresponding weights \(\sqrt{0.5}\) for both the interim score and the independent future increment. Unequal information changes both weights: the interim contribution is \(\sqrt{t}\,z\), and the future standard deviation is \(\sqrt{1-t}\). Simply averaging two stage scores with equal weights when their information differs does not generally produce the correctly standardized final score.

The formula is exact under the stated normal independent-increments model and an approximate result when that model is only an asymptotic approximation. It does not automatically cover discrete statistics, dependent stage increments, a changed test statistic, or a change in the correlation structure. In those settings the conditional distribution must be re-derived rather than assumed to retain this mean and variance.

Common Mistakes

  • Using the wrong information fraction. The reasoning “30% of participants are enrolled, so \(t=0.30\)” is valid only if information is proportional to enrollment. Use the actual information ratio.
  • Keeping the unconditional variance after conditioning. The final score has null variance 1 before observing interim data, but its conditional variance is \(1-t\), because the interim component is fixed.
  • Using an unweighted score average. Equal weights are appropriate only for equal information under the stated construction. For unequal information, use the square-root information weights.
  • Calling conditional error a posterior probability. The calculation conditions on observed data but evaluates a rejection probability under the null; it does not assign a probability to the null hypothesis.
Key takeaway: Under the canonical normal model, unequal information determines both the conditional mean \(\sqrt{t}\,z\) and the remaining conditional variance \(1-t\). Standardizing the original final boundary with these conditional parameters gives the conditional error.

Check Your Understanding

Work each problem using the stated normal-score model. Solutions follow the questions.

  1. If \(t=0.40\), \(z=0\), and \(c=1.96\), write the conditional distribution and calculate the conditional error.
  2. For fixed \(t=0.25\) and \(c=1.96\), should conditional error be larger at \(z=1\) or at \(z=0\)? Explain using the formula.
  3. A trial has interim information \(I_1=45\) and remaining information \(I_2=55\). What is \(t\), and what is the conditional variance of the final score?
  4. Why is the conditional variance not 1 after observing \(Z_1=z\)?

Solutions. (1) The conditional distribution is \(Z_F\mid Z_1=0,H_0\sim N(0,0.60)\). Thus \(A(0)=1-\Phi(1.96/\sqrt{0.60})=1-\Phi(2.52982)\approx0.0057\). (2) It is larger at \(z=1\): increasing \(z\) increases the conditional mean \(\sqrt{0.25}z\), so the probability of exceeding the fixed boundary increases. (3) \(t=45/(45+55)=0.45\), and the conditional variance is \(1-t=0.55\). (4) Once \(Z_1=z\) is observed, its contribution to the final score is fixed. Only the independent future increment remains random, and its weighted variance is \(1-t\), not 1.

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