Conditional Power on the Information Scale

Why Information, Rather Than Enrollment, Sets the Scale

At an interim analysis, conditional power asks how likely the trial is to cross its final success boundary, given the data observed so far and a specified assumption about the effect that will govern future observations. The calculation depends on how much information has accrued and how much remains. The proportion of participants enrolled can be a useful proxy, but it is not the quantity that determines the conditional distribution of the future score.

The earlier material on Conditional Power at an Interim Analysis and on deriving the conditional distribution from independent increments motivates conditioning on the interim statistic. Here, the key step is to express that distribution in terms of interim information, final information, and their ratio. This makes clear why the information fraction controls both the contribution of the observed interim result and the uncertainty in the future standardized score.

Definition: Let $I_1$ be the information available at the interim analysis and $I_2$ the planned information at the final analysis, with $0<I_1<I_2$. The information fraction is $t=I_1/I_2$, so $0<t<1$. The remaining information is $I_2-I_1=(1-t)I_2$.

The information scale is chosen so that score increments have variances equal to their information increments. For example, information may be related to the inverse variance of an effect estimate. It need not increase in direct proportion to the number enrolled: unequal allocation, changing event rates, or differing amounts of outcome information per participant can make the two fractions differ.

Model and Conditional Distribution

For this calculation, assume a canonical normal score model with independent increments. Write $U(I)$ for the score at information level $I$. Under a fixed effect scenario $\theta$, suppose

U(I_1)\sim N(\theta I_1,I_1), \qquad U(I_2)-U(I_1)\sim N\!\left(\theta(I_2-I_1),I_2-I_1\right),

and that the future increment $U(I_2)-U(I_1)$ is independent of the interim score $U(I_1)$. Here, the second argument of $N(\mu,\sigma^2)$ is the variance. Define the standardized score at information $I$ as $Z(I)=U(I)/\sqrt{I}$. Let the observed interim value be $z_1=Z(I_1)$, and write $Z_2=Z(I_2)$ for the final statistic.

Result: Conditional on the observed interim score $z_1$, under the specified fixed effect $\theta$, $$ Z_2\mid Z(I_1)=z_1 \sim N\!\left(\sqrt{t}\,z_1+\theta\frac{I_2-I_1}{\sqrt{I_2}},\,1-t\right). $$ Equivalently, if $\Delta_2=\theta\sqrt{I_2}$ is the assumed mean of the final standardized score under that effect scenario, then $$ E(Z_2\mid z_1)=\sqrt{t}\,z_1+(1-t)\Delta_2, \qquad \operatorname{Var}(Z_2\mid z_1)=1-t. $$

The effect scenario must be specified to calculate the conditional mean. This tutorial treats $\theta$, or equivalently $\Delta_2$, as given; it does not estimate the effect from the interim data. The conditional variance, in contrast, follows from the remaining information under the model and does not depend on the specified effect.

Derivation on the Information Scale

Once $z_1$ is observed, the interim score is known: $U(I_1)=z_1\sqrt{I_1}$. The final score is the sum of this fixed interim score and the future increment. Dividing by the final standard deviation gives

Z_2 =\frac{U(I_1)+[U(I_2)-U(I_1)]}{\sqrt{I_2}}.

Conditionally, only the bracketed future increment is random. Its mean is $\theta(I_2-I_1)$ and its variance is $I_2-I_1$. Therefore, adding the known interim contribution and dividing by $\sqrt{I_2}$ gives the conditional mean and variance:

E(Z_2\mid z_1) =\frac{z_1\sqrt{I_1}+\theta(I_2-I_1)}{\sqrt{I_2}} =z_1\sqrt{\frac{I_1}{I_2}}+\theta\frac{I_2-I_1}{\sqrt{I_2}}, $$ $$ \operatorname{Var}(Z_2\mid z_1) =\frac{I_2-I_1}{I_2} =1-\frac{I_1}{I_2}.

Substituting $t=I_1/I_2$ and $\Delta_2=\theta\sqrt{I_2}$ yields the stated information-fraction form. The observed score is weighted by $\sqrt{t}$, not by $t$, because $z_1$ is standardized by $\sqrt{I_1}$, whereas the final statistic is standardized by $\sqrt{I_2}$. The effect contribution is $(1-t)\Delta_2$, and the future variance is $1-t$.

From the Conditional Distribution to Conditional Power

Consider a prespecified one-sided final success rule: declare success when $Z_2\ge c$, for a fixed critical value $c$. Conditional power under the specified effect scenario is the conditional probability of this event. If $m=\sqrt{t}\,z_1+(1-t)\Delta_2$, standardizing the normally distributed future score gives

\operatorname{CP}(z_1;\theta) =P(Z_2\ge c\mid z_1;\theta) =1-\Phi\!\left(\frac{c-m}{\sqrt{1-t}}\right) =\Phi\!\left(\frac{m-c}{\sqrt{1-t}}\right),

where $\Phi$ is the standard normal cumulative distribution function. This is a conditional probability of crossing a specified boundary under the assumed effect. It is not the probability that the assumed effect is true, nor is it an unconditional probability of trial success.

Worked Example: Low Conditional Power at an Interim

Suppose the information fraction is $t=0.4000$, the observed interim score is $z_1=0$, and the assumed mean final standardized score is $\Delta_2=0.9000$. The one-sided final boundary is $c=1.960$. Calculate conditional power.

The conditional mean and variance are

m=\sqrt{0.4000}(0)+ (1-0.4000)(0.9000)=0.5400, \qquad \operatorname{Var}(Z_2\mid z_1)=1-0.4000=0.6000.

Thus the conditional standard deviation is $\sqrt{0.6000}$, and

\operatorname{CP} =\Phi\!\left(\frac{0.5400-1.960}{\sqrt{0.6000}}\right) =\Phi\!\left(\frac{-1.4200}{0.7746}\right) \approx \Phi(-1.8332) \approx 0.0334.

Under this assumed effect, the probability of crossing the final boundary, conditional on the interim score of zero, is about 3.34%. This result follows from the assumed effect and the observed interim information; it does not establish that the effect is absent.

Worked Example: A Different Information Fraction and Boundary

Suppose $t=0.2500$, $z_1=0.8000$, and the assumed final mean is $\Delta_2=0.6667$. Let the one-sided final boundary be $c=1.645$. Find the conditional power, retaining sufficient precision in the intermediate calculation.

The conditional mean is

m=\sqrt{0.2500}(0.8000)+(1-0.2500)(0.6667) =0.4000+0.5000 =0.9000

to the displayed precision. The conditional variance is $1-0.2500=0.7500$. Therefore,

\operatorname{CP} =\Phi\!\left(\frac{0.9000-1.645}{\sqrt{0.7500}}\right) =\Phi\!\left(\frac{-0.745}{\sqrt{0.7500}}\right) \approx\Phi(-0.8600) \approx 0.1949.

The conditional probability of crossing the boundary is approximately 19.49% under the specified effect scenario. A common arithmetic source of error is rounding the standardized value too early; here the quotient is about $-0.8600$, which corresponds to a lower-tail probability of about 0.1949.

Interpreting Information Fractions and Unequal Accrual

For a concrete information-scale calculation, suppose interim information is $I_1=40$ units and planned final information is $I_2=100$ units. Then $t=40/100=0.4000$, regardless of the number enrolled at either analysis. If participants contribute differing amounts of information, an enrollment fraction of, say, 0.5000 does not justify substituting $t=0.5000$. The information fraction must be calculated from the information relevant to the score model.

Worked Example: Using Information Values Directly

Suppose $I_1=40$, $I_2=100$, $z_1=1.200$, and the specified effect is $\theta=0.1500$ per square root of an information unit. The final boundary is $c=1.960$. Calculate conditional power from the information values.

First, $t=40/100=0.4000$, and the assumed final mean is $\Delta_2=\theta\sqrt{I_2}=0.1500(10)=1.500$. The conditional mean and variance are

m=\sqrt{0.4000}(1.200)+(1-0.4000)(1.500) =0.7589+0.9000 =1.6589, \qquad \operatorname{Var}(Z_2\mid z_1)=0.6000.

The conditional power is

\operatorname{CP} =\Phi\!\left(\frac{1.6589-1.960}{\sqrt{0.6000}}\right) =\Phi(-0.3887) \approx 0.3488.

Under the stated effect, the probability of meeting the final one-sided boundary, conditional on the observed score and information, is about 34.88%. The information units need not be participant counts; what matters is that $I_1$, $I_2$, and $\theta$ use a consistent score-information scale.

Assumptions, Diagnostics, and Alternatives

The formula is exact for the stated normal independent-increments model. In many applications, a canonical joint normal model is an approximation supported by large-sample theory. Its use requires that the chosen information adequately describes score variance, the future score increment is independent of the interim score under the model, and the planned final information and success boundary are defined consistently with the analysis design. If these assumptions are doubtful, the calculation may not describe the actual conditional distribution.

Before interpreting a result, check the following:

Information consistency. Confirm that interim and final information are on the same scale and that the information fraction lies between zero and one.
Variance check. The conditional variance must equal the remaining information fraction $1-t$. It should be positive when a future analysis remains, and smaller when more information has accrued.
Mean check. The interim contribution is $\sqrt{t}\,z_1$, while the assumed future-effect contribution is $(1-t)\Delta_2$. Confirm that the assumed effect and its units are explicitly stated.
Design check. Verify that $c$ is the applicable final boundary and that the rule really is the stated one-sided criterion. A different success rule requires a corresponding probability calculation.

Conditional power is often compared with unconditional power. Unconditional power averages over the possible interim results under a specified effect, whereas conditional power fixes the observed interim score and evaluates the future probability under an effect scenario. Neither quantity automatically incorporates uncertainty about the effect scenario. Another tempting alternative is to use the enrollment fraction in place of $t$; this is appropriate only when that fraction correctly represents the information fraction under the design.

The information-scale expression is useful because it separates the influence of the observed interim evidence, the assumed future effect, and the amount of information remaining. Its limitations are equally important: misspecified information, dependence not captured by the independent-increments model, or a mismatched final boundary can make the numerical probability misleading. Conditional power is a model-based calculation, not a guarantee of eventual success or a substitute for the prespecified analysis plan.

Common Mistake

Common mistake: “Forty percent of participants have been enrolled, so the information fraction is $t=0.4000$, and the future standardized score has variance $0.6000$.” This reasoning is incorrect unless enrollment fraction is known to equal information fraction. Correction: calculate $t=I_1/I_2$ from the score information, then use $1-t$ for the conditional variance. Also, standardize the difference $m-c$ by $\sqrt{1-t}$, not by $1-t$.

Check Your Understanding

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

If $t=0.6400$, what is the conditional variance of the final standardized score?
For $t=0.2500$, $z_1=1.000$, and $\Delta_2=1.200$, calculate the conditional mean.
Why can the proportion enrolled differ from the information fraction?
For a one-sided boundary $c$, write the conditional-power expression using the conditional mean $m$ and information fraction $t$.

Solutions. (1) The variance is $1-t=1-0.6400=0.3600$. (2) The mean is $\sqrt{0.2500}(1.000)+(1-0.2500)(1.200)=0.5000+0.9000=1.400$. (3) Participants may contribute unequal amounts of score information, so enrollment counts need not be proportional to information. (4) The expression is $\operatorname{CP}=\Phi((m-c)/\sqrt{1-t})$, because the future score has conditional variance $1-t$.