Conditional Maximum Likelihood at a Single Look

Why the Estimating Equation Changes After a Stop

At a planned analysis, a trial stops for efficacy only when its test statistic crosses a specified boundary. Once the analysis has stopped, the observed statistic is therefore not an unrestricted draw from its sampling distribution: it is known to have passed that boundary. A likelihood that conditions on this fact must account for the resulting truncation. In particular, the conditional maximum-likelihood estimate is generally not the same as the ordinary maximum-likelihood estimate.

This tutorial considers one normal-theory efficacy look. The focus is the estimating equation for the conditional maximum-likelihood estimate, not a general correction for every form of stopping-induced bias. The conditional estimate is defined relative to a particular event and model; changing what is conditioned on changes the likelihood and potentially the estimate.

Model and assumptions: At one planned look, let $Z$ be a standardized statistic with $Z\sim N(\delta,1)$, where $\delta=\theta\sqrt{I}$, $\theta$ is the treatment effect on its original scale, and $I$ is known information. The trial stops for efficacy if and only if $Z\ge c$, where $c$ is a fixed, known boundary. We observe $Z=z\ge c$, and the conditional likelihood conditions on the event $A=\{Z\ge c\}$. This setup assumes the normal model and known $I$ are appropriate, and that no additional data-dependent selection event is being conditioned on.

Write $\phi$ and $\Phi$ for the standard normal density and distribution functions, and define the upper-tail probability $Q(t)=1-\Phi(t)$. Under parameter $\delta$, the probability of crossing the boundary is

P_\delta(A)=P_\delta(Z\ge c)=Q(c-\delta).

The observed value has normal density $\phi(z-\delta)$. Given that the boundary was crossed, its conditional density is the original density divided by the probability of crossing:

f_\delta(z\mid Z\ge c) = \frac{\phi(z-\delta)}{Q(c-\delta)},\qquad z\ge c.

The denominator is the essential change. The ordinary likelihood uses the numerator alone; the conditional likelihood also accounts for the fact that only values at or above $c$ could have been observed at this stopping event.

Deriving the Conditional Score Equation

For a fixed observation $z$, terms that do not depend on $\delta$ can be dropped from the log likelihood. The conditional log likelihood is therefore

\ell_C(\delta;z) = -\frac{1}{2}(z-\delta)^2-\log Q(c-\delta)+\text{constant}.

Define the upper inverse Mills ratio by $\lambda(t)=\phi(t)/Q(t)$. Differentiating the conditional log likelihood gives

\frac{\partial \ell_C}{\partial \delta} = z-\delta-\lambda(c-\delta).

To check the sign of the last term, note that $dQ(c-\delta)/d\delta=\phi(c-\delta)$. Thus the derivative of $-\log Q(c-\delta)$ is $-\phi(c-\delta)/Q(c-\delta)$, giving the negative truncation correction in the score.

Definition: The single-look conditional maximum-likelihood estimate $\hat{\delta}_C$ is a value of $\delta$ maximizing the likelihood of the observed statistic conditional on crossing the efficacy boundary. For an observation $z>c$, it solves $$ z=\hat{\delta}_C+\lambda(c-\hat{\delta}_C). $$ If $I$ is known and $\delta=\theta\sqrt I$, the corresponding treatment-effect estimate is $\hat{\theta}_C=\hat{\delta}_C/\sqrt I$.

The correction has a direct interpretation. Given $Z\ge c$, the conditional mean of $Z$ is larger than its untruncated mean $\delta$. The conditional score sets the observed $z$ equal to this truncated mean, rather than to $\delta$ alone. The short derivation below establishes that interpretation and also helps justify numerical solution of the equation.

Why the Equation Has a Unique Root

Let $Y=Z-\delta$, so $Y\sim N(0,1)$. The event $Z\ge c$ is equivalent to $Y\ge t$, where $t=c-\delta$. Since $\phi'(y)=-y\phi(y)$,

E(Y\mid Y\ge t) = \frac{\int_t^\infty y\phi(y)\,dy}{Q(t)} = \frac{\phi(t)}{Q(t)} = \lambda(t).

It follows that $E_\delta(Z\mid Z\ge c)=\delta+\lambda(c-\delta)$, exactly the right side of the score equation. For the conditional variance, integration by parts gives $E(Y^2\mid Y\ge t)=1+t\lambda(t)$. Subtracting the squared conditional mean yields

\operatorname{Var}_\delta(Z\mid Z\ge c) = 1+t\lambda(t)-\lambda(t)^2 = 1-\lambda'(t) > 0.

The strict inequality holds because this is the variance of a non-degenerate truncated normal distribution. Consequently, the derivative with respect to $\delta$ of the conditional mean $\delta+\lambda(c-\delta)$ is the conditional variance and is positive. This mean increases continuously from $c$ as $\delta$ tends to negative infinity to infinity as $\delta$ tends to infinity. Hence, for any observed $z>c$, there is one finite solution to the score equation. The conditional log likelihood is strictly concave there, because its second derivative is the negative of that conditional variance, so this solution is the unique maximum.

Solving the Equation in Practice

Worked Example: A Statistic Just Above a Boundary

Suppose a hypothetical single-look design has boundary $c=2.00$ and observes $z=2.50$. Find the conditional estimate on the standardized scale. The model assumptions are those stated above: $Z\sim N(\delta,1)$, the boundary is fixed, and crossing occurs exactly when $Z\ge2.00$.

The conditional score equation is

2.50=\delta+\lambda(2.00-\delta).

For instance, at $\delta=0.879$, $t=2.00-0.879=1.121$, and $\lambda(1.121)\approx1.6229$. The right side is $0.879+1.6229=2.5019$, slightly above 2.50. Solving the equation numerically gives $\hat{\delta}_C\approx0.8688$. At this value, $t=1.1312$, $\lambda(t)\approx1.6312$, and $0.8688+1.6312\approx2.5000$, up to rounding.

The ordinary maximum-likelihood estimate under the untruncated normal likelihood is $\hat{\delta}=z=2.50$, whereas the conditional estimate is about $0.869$. This difference is not an arithmetic adjustment applied after estimation: it follows from maximizing a different likelihood, one that conditions on crossing $c=2.00$.

Worked Example: A Larger Observed Statistic

In another hypothetical single-look analysis, let $c=2.00$ and $z=3.00$. Again assume a known-information normal model and that the sole stopping event being conditioned on is $Z\ge2.00$. The equation is

3.00=\delta+\lambda(2.00-\delta).

A numerical root is approximately $\hat{\delta}_C=2.481$. To check it, $t=2.00-2.481=-0.481$, and $\lambda(-0.481)=\phi(0.481)/\Phi(0.481)\approx0.519$. Thus the right side is approximately $2.481+0.519=3.000$. The ordinary estimate is $z=3.00$, so the conditional estimate is again lower, but the difference is smaller than in the preceding example.

This check also illustrates a useful computational practice: after solving numerically, substitute the candidate into the original score equation and report the residual at the stated precision. A root solver's output alone does not verify that the intended boundary or tail probability was used.

From the Standardized Scale to the Treatment-Effect Scale

Worked Example: Converting the Conditional Estimate

Return to the first hypothetical analysis, where $c=2.00$, $z=2.50$, and $\hat{\delta}_C\approx0.8688$. Suppose the known information is $I=100$, so $\sqrt I=10$. The model links the standardized parameter to the treatment effect by $\delta=\theta\sqrt I$. Therefore,

\hat{\theta}_C = \frac{\hat{\delta}_C}{\sqrt I} = \frac{0.8688}{10} = 0.08688 \approx0.0869.

The conditional estimate is approximately $0.0869$ treatment-effect units. If $\theta$ is a mean difference measured in, for example, points on a specified outcome scale, the estimate is $0.0869$ points on that scale. The units come from the definition of $\theta$; $\delta$ itself is standardized and dimensionless.

Diagnostics, Comparisons, and Limitations

Several checks help identify a mistaken or poorly specified calculation. For a finite observation $z>c$, the correction $\lambda(c-\hat{\delta}_C)$ is positive, so the equation implies $\hat{\delta}_C<z$. The candidate should also reproduce $z$ when substituted into $\delta+\lambda(c-\delta)$. A result above $z$, a substantial score residual, or use of $Q(c)$ in place of $Q(c-\delta)$ signals a calculation or parameterization error. When $z$ is only slightly above $c$, the solution can be far below the observed statistic; numerical work may then be sensitive to tail-probability accuracy. Use a stable upper-tail calculation rather than computing a very small $Q(t)$ as the difference of two nearly equal numbers.

The main alternative in this setting is the ordinary, unconditional likelihood for the observed statistic, which gives $\hat{\delta}=z$. It answers a different likelihood question because it does not divide by the probability of crossing. The conditional likelihood instead describes the distribution of $Z$ among outcomes that cross the specified boundary. Neither estimate should be described as a universal correction to the other; the choice depends on the inferential target and the conditioning event.

The conditional maximum-likelihood estimate is not generally conditionally unbiased, and its construction alone does not guarantee a confidence interval with a desired coverage probability. It also depends on the assumed normal model, known information, and the exact fixed boundary. If the standard error is estimated, the boundary is random, or stopping depends on additional looks or criteria, this single-look equation does not automatically apply. Those settings require a likelihood that represents their actual sampling and stopping structure.

Common mistake: Using the ordinary score $z-\delta=0$ after an efficacy stop gives $\hat{\delta}=z$. The reasoning is incomplete because it uses the untruncated density while ignoring that the observed value is known to satisfy $Z\ge c$. For the conditional likelihood, include the normalizing probability $Q(c-\delta)$; its derivative contributes $-\lambda(c-\delta)$ to the score, and the estimate solves $z=\delta+\lambda(c-\delta)$.

Key takeaway: At a single normal-theory efficacy look, conditioning on crossing changes the score by the inverse Mills ratio of the distance from the parameter-specific mean to the boundary. The conditional maximum-likelihood estimate is the unique solution of $z=\delta+\lambda(c-\delta)$ for $z>c$, under the stated model.

Check Your Understanding

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

For a fixed boundary $c$, write the conditional density of an observed $z\ge c$ under $Z\sim N(\delta,1)$, and identify its normalizing factor.
Why is the conditional score equation not $z-\delta=0$? State the equation that must instead be solved.
For $c=2.00$, $z=2.50$, and $I=64$, use $\hat{\delta}_C\approx0.8688$ to calculate the conditional treatment-effect estimate.
Suppose a numerical candidate satisfies $\hat{\delta}_C>z$. Is that compatible with the single-look conditional score equation? Explain.

Solutions. (1) The conditional density is $f_\delta(z\mid Z\ge c)=\phi(z-\delta)/Q(c-\delta)$; the normalizing factor is the crossing probability $Q(c-\delta)$. (2) The ordinary score ignores the conditioning event. Differentiating the conditional log likelihood gives $z-\delta-\lambda(c-\delta)=0$, or $z=\delta+\lambda(c-\delta)$. (3) Since $\sqrt{64}=8$, $\hat{\theta}_C=0.8688/8=0.1086$ treatment-effect units. (4) No. Since $\lambda(c-\hat{\delta}_C)>0$, the score equation implies $z=\hat{\delta}_C+\lambda(c-\hat{\delta}_C)>\hat{\delta}_C$, so a candidate above $z$ cannot solve it.