Deriving Sample Size From a Conditional-Power Target

From a Conditional-Power Target to an Information Requirement

A conditional-power rule can specify a desired probability of success given the data observed at an interim analysis. Turning that target into a sample size requires solving the conditional-power expression for the final information level. The calculation is not generally a direct algebraic rearrangement: the final critical value is multiplied by the square root of final information, so the resulting equation is usually solved numerically.

The derivation below uses a one-sided normal test for a positive treatment effect, with known variance and a fixed final critical value. This makes the inversion explicit while keeping the interim estimate fixed. The resulting sample size is conditional on the assumed effect used to calculate future information; it is not a guarantee that the target power will be achieved under the unknown true effect.

Model and assumptions: Let $\theta$ be the treatment effect, with positive values favoring the treatment. At interim information $I_1$, the observed estimate is $\hat{\theta}_1$, and $U_1=I_1\hat{\theta}_1$ is its observed score. Information is positive and accumulates through independent normal increments. The variance is known, and the final test rejects $H_0:\theta\leq 0$ when the final standardized estimate exceeds the fixed cutoff $c$, commonly $c=z_{1-\alpha}$. For the conditional-power calculation, future increments are modeled under a specified effect $\theta^*>0$. The interim data are held fixed.

Write $I>I_1$ for the proposed final information and $\Delta I=I-I_1$ for the additional information. Under the assumptions above, the final estimate is the information-weighted combination of the interim estimate and the future increment. Conditional on the interim data, its numerator has mean $U_1+\theta^*\Delta I$ and variance $\Delta I$. Thus the final standardized estimate has conditional mean $(U_1+\theta^*\Delta I)/\sqrt{I}$ and conditional variance $\Delta I/I$.

The final test succeeds when that standardized estimate exceeds $c$. Standardizing this conditional normal distribution gives the conditional power:

\operatorname{CP}(I\mid U_1,\theta^*)= \Phi\!\left( \frac{U_1+\theta^*(I-I_1)-c\sqrt{I}} {\sqrt{I-I_1}} \right).

Let $q$ be the desired conditional power, with $0<q<1$, and let $z_q=\Phi^{-1}(q)$. Since $\Phi$ is increasing, setting conditional power equal to $q$ is equivalent to setting its argument equal to $z_q$. Define $x=\sqrt{I-I_1}$, the square root of additional information. The equation to solve becomes:

\theta^*x^2+U_1-c\sqrt{I_1+x^2}=z_qx,\qquad x>0.

After a positive solution $x$ is found, the required final information is $I=I_1+x^2$. This transformation is useful both for numerical computation and for interpretation: the root measures the additional information, not the total information.

Why the Equation Usually Needs Numerical Solution

The unknown $x$ appears linearly, quadratically, and inside $\sqrt{I_1+x^2}$. Squaring the equation to remove the square root can produce a higher-degree equation and may introduce extraneous roots, so a numerical root finder is usually clearer and safer. A bracketed method such as bisection is robust; a Newton method can be faster if its starting value is sensible and the result is checked.

The equation need not behave monotonically for every combination of interim data and planning effect. A numerical procedure should therefore seek a positive root that meets the target in the original conditional-power expression, and, when the design requires the least information that meets the target, identify the first feasible crossing. A root obtained after squaring should never be accepted without substituting it into the unsquared equation.

Derivation check: The conditional mean of the final standardized estimate is $(U_1+\theta^*\Delta I)/\sqrt I$, and its conditional standard deviation is $\sqrt{\Delta I/I}$. Therefore its standardized distance above the cutoff is $$ \frac{(U_1+\theta^*\Delta I)/\sqrt I-c}{\sqrt{\Delta I/I}} = \frac{U_1+\theta^*\Delta I-c\sqrt I}{\sqrt{\Delta I}}. $$ This is exactly the argument of $\Phi$ in the conditional-power formula.

Worked Example: Solving for the Information Level

Suppose a hypothetical one-sided normal trial has interim information $I_1=1.00$ inverse squared outcome units and observed score $U_1=1.167796$ inverse outcome units. The assumed effect is $\theta^*=2.00$ outcome units, the final cutoff is $c=1.644854$, and the conditional-power target is $q=0.8000$. Thus $z_q=0.841621$. Find the required final information.

Set $x=\sqrt{I-I_1}$. Substitution into the target equation gives

2.00x^2+1.167796-1.644854\sqrt{1.00+x^2}=0.841621x.

A numerical root search gives $x\approx1.0000$. Substituting $x=1.0000$ into the two sides yields

\text{left side}=2.0000+1.167796-1.644854\sqrt{2} \approx0.841621, \qquad \text{right side}=0.841621.

The required final information is therefore

I=I_1+x^2=1.00+1.0000^2=2.00 \quad\text{inverse squared outcome units}.

Checking the original conditional-power expression gives $\Phi(0.841621)\approx0.8000$, as targeted. The check confirms the root and the interpretation: the required additional information is $1.00$ inverse squared outcome units, and the required total information is $2.00$.

Converting Information to Total Sample Size

Information is not itself an enrollment count. The conversion depends on the design and the relationship between sample size and information. If information is proportional to total sample size $N$, write $I(N)=\kappa N$, where $\kappa$ is the information contributed per participant under the specified allocation and variance assumptions. Then the continuous total sample-size requirement is $N=I/\kappa$. In a group-sequential design, this is the total sample size by the final analysis, not the number to enroll after the interim look.

For example, in a hypothetical two-arm normal trial with equal allocation and known common variance $\sigma^2$, the variance of the difference in group means is $4\sigma^2/N$. Its information is therefore $I=N/(4\sigma^2)$, or $N=4\sigma^2 I$. The design must use the correct information relationship for its endpoint, allocation, and variance assumptions; a conversion based on the wrong variance or allocation will give the wrong enrollment.

Worked Example: Converting Required Information to Enrollment

Use the required final information $I=2.00$ from the preceding calculation in a hypothetical two-arm, equally allocated normal trial. Suppose the known standard deviation in each arm is $8.00$ outcome units. At the interim, $I_1=1.00$, corresponding to 256 participants in total. Find the final total sample size.

For equal allocation, $I=N/(4\sigma^2)$. Therefore

N=4\sigma^2 I =4(8.00)^2(2.00) =4(64.00)(2.00) =512\text{ participants}.

The interim enrollment is $N_1=4(8.00)^2(1.00)=256$ participants, so the information-based calculation calls for 256 additional participants and 512 in total. Because 512 is even, it permits equal allocation without further rounding. If the calculation instead produced an odd total, the operational rule would need to round to a feasible balanced enrollment and then recompute the information and conditional power at that rounded total.

Discrete enrollment can make the implemented sample size differ slightly from the continuous requirement. The appropriate procedure is to convert the continuous information requirement, apply the design’s allocation and integer constraints, and then evaluate conditional power at the resulting feasible sample size. Rounding down simply because it is convenient can leave the implemented design below its target.

Worked Example: Checking a Sample-Size Cap

Continue with the hypothetical setting of the first worked example, but suppose operational constraints cap total enrollment at 480 participants. With $\sigma=8.00$ outcome units, the cap corresponds to $I=480/[4(8.00)^2]=1.875$. At this information, $x=\sqrt{1.875-1.00}=\sqrt{0.875}\approx0.9354$. Evaluate the conditional power under the same interim data and assumed effect.

Substitution into the conditional-power argument gives

\frac{1.167796+2.00(0.875)-1.644854\sqrt{1.875}}{\sqrt{0.875}} \approx \frac{0.6654869}{0.9354143} \approx0.7114.

Thus the conditional power at the cap is approximately $\Phi(0.7114)=0.7616$, below the target of 0.8000. The cap does not attain the target at its maximum permitted enrollment under these assumptions. This check is more informative than reporting the unconstrained requirement alone: it identifies that the target and the operational limit are incompatible in this scenario.

Assumptions, Diagnostics, and Alternatives

The inversion is only as sound as the model used to calculate conditional power. The normal independent-increments model, known variance, fixed final cutoff, and assumed effect $\theta^*$ must be suitable for the planned analysis. The information-to-sample-size relationship must also match the actual design. In practice, variance may be estimated, outcomes may be non-normal, or information may not increase linearly with enrollment. Those issues can change the mapping from information to sample size or the conditional distribution itself.

Useful checks include verifying that $I>I_1$, confirming that the computed root is positive, substituting the result into the original conditional-power formula, and calculating the conditional power again after integer rounding. Also check that the assumed effect has the same direction and units as $\hat{\theta}_1$, that $c$ corresponds to the intended final test, and that the proposed enrollment is feasible. A numerical warning, a root that fails the original equation, or a large change in conditional power after rounding signals that the calculation needs review.

The principal alternative is to choose a final sample size directly from a conventional power calculation under a prespecified effect, without conditioning on the interim result. That produces a fixed design target rather than a conditional-power-driven update. A conditional-power calculation instead incorporates the observed interim score and a chosen $\theta^*$, so two trials with different interim results can yield different information requirements. Neither calculation makes the assumed effect known. The introductory material on target conditional power and target final power distinguishes these targets; here the task is specifically to invert the conditional-power target.

A target-driven calculation also has limits. It can imply a very large increase when the interim data are unfavorable, or a very small increase when they are favorable. A maximum sample size, a minimum worthwhile effect, clinical feasibility, and the stability of the interim estimate may therefore matter when turning the mathematical requirement into a rule. Conditional power is a probability under the specified conditional model; it is not the unconditional probability of success for the whole adaptive design, and meeting a conditional-power target by itself does not establish that all design operating characteristics are acceptable.

Common Mistake

Common mistake: Treating the target as though it were a required final $Z$-statistic, then setting the final information equal to $z_q^2$, or inserting the target effect into a fixed-sample formula while ignoring the observed interim score. The correction is to set the conditional normal argument equal to $z_q$, retaining $U_1$, $I_1$, and the final cutoff $c$. Solve that equation for additional information, convert to total information, and only then translate to an enrollment count.

Check Your Understanding

Answer each question, then compare with the solutions below.

In the normal conditional-power model, which quantities are held fixed when solving for a required final information level, and which quantity is varied?
For a target conditional power $q$, why can the conditional-power equation be converted into an equation involving $z_q$?
A numerical solution gives $x=1.5$ and interim information $I_1=2.0$. What is the final information, and what is the additional information?
In an equally allocated two-arm normal trial with known standard deviation $\sigma=5$, what total sample size corresponds to final information $I=1.2$?
Why should a numerical root be substituted into the original conditional-power expression after it has been found?

Solutions. (1) Hold the interim information $I_1$, observed score $U_1$, assumed effect $\theta^*$, target $q$, and final cutoff $c$ fixed. Vary $I$, or equivalently $x=\sqrt{I-I_1}$. (2) Since $\Phi$ is strictly increasing, $\Phi(a)=q$ is equivalent to $a=\Phi^{-1}(q)=z_q$. (3) The additional information is $x^2=1.5^2=2.25$, so final information is $I=2.0+2.25=4.25$. (4) Use $N=4\sigma^2I=4(25)(1.2)=120$ participants. (5) Numerical methods can return an inaccurate or extraneous root, particularly if algebraic transformations were used. Substitution verifies that the root is positive, satisfies the unsquared equation, and yields the stated conditional power.