Schoenfeld Approximation For Required Events

Why Estimate Events Rather Than Participants?

In a time-to-event trial, the amount of information for comparing groups is driven chiefly by the number of observed events, not simply by the number enrolled. A trial can recruit many participants yet have limited information if few have experienced the event by the analysis cutoff. Schoenfeld’s approximation provides a convenient first estimate of how many events are needed to detect a specified hazard ratio with a chosen significance level and power.

The result is an approximate event count for a conventional fixed-design logrank test. It does not, by itself, give the number of participants to enroll, and it does not account for the extra design requirements of a group sequential test. Those distinctions matter when using the calculation as a planning benchmark.

Definition: Let $D$ be the total number of observed events, $q$ the proportion assigned to the treatment group, $\mathrm{HR}$ the treatment-to-control hazard ratio, $\alpha$ the type I error level, and $1-\beta$ the desired power. For a specified proportional-hazards effect, Schoenfeld’s approximation is $$ D \approx \frac{\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{q(1-q)\left[\log(\mathrm{HR})\right]^2} $$ for a two-sided test. Here $z_u$ is the $u$ quantile of the standard normal distribution. For a prespecified one-sided test, replace $z_{1-\alpha/2}$ with $z_{1-\alpha}$, using the tail in the direction of the alternative.

If the planned treatment-to-control allocation ratio is $r:1$, then $q=r/(r+1)$, and $1-q=1/(r+1)$. Thus $q(1-q)=r/(r+1)^2$. With equal allocation, $q=0.5$ and $q(1-q)=0.25$, the largest possible value of this allocation term. Departing from equal allocation therefore generally increases the required event count for the same effect, alpha, and power.

Where the Approximation Comes From

Write $\theta=\log(\mathrm{HR})$. Under the alternative, the logrank or Cox score statistic has, approximately, a normal distribution with mean $\theta\sqrt{I}$ and variance 1, where $I$ is the information for the treatment effect. Under the planning approximation for the risk sets, $I\approx Dq(1-q)$. The magnitude of the standardized signal is therefore approximately $|\theta|\sqrt{Dq(1-q)}$.

For a two-sided test, the upper critical value in the direction of the planned effect is $z_{1-\alpha/2}$. The one-sided power in that direction is approximately the probability that a standard normal variable with mean $|\theta|\sqrt{Dq(1-q)}$ exceeds this critical value. To make that probability $1-\beta$, the mean signal must be approximately $z_{1-\alpha/2}+z_{1-\beta}$. Solving for $D$ gives the formula above.

|\theta|\sqrt{Dq(1-q)} \approx z_{1-\alpha/2}+z_{1-\beta} \quad\Longrightarrow\quad D \approx \frac{\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{q(1-q)\theta^2}.

This derivation uses the normal approximation and focuses on the rejection tail in the direction of the planned effect. For typical design alternatives, the probability of crossing the opposite two-sided boundary is small; the formula is an approximation, not an exact power equation. A one-sided test uses its one-sided critical value instead. The sign of $\log(\mathrm{HR})$ indicates the direction of the effect, while its square in the event-count formula means the required count depends on the magnitude of the log hazard ratio.

Assumptions and interpretation: The calculation assumes a clinically specified, approximately constant hazard ratio over follow-up; an event-driven logrank-type comparison; an allocation fraction in the relevant risk sets that is reasonably represented by planned $q$; and censoring that is non-informative for the event process, conditional on the design factors. It also uses large-sample normal approximations. The output estimates events needed for the stated fixed-design test under those assumptions; it is not a guarantee of achieved power in a particular trial.

Worked Event-Count Calculations

Worked Example: Equal Allocation, Two-Sided Test

Suppose a hypothetical trial plans equal allocation, a two-sided $\alpha=0.05$ test, 90% power, and a target hazard ratio of 0.70. Assume the proportional-hazards and censoring conditions above are reasonable for planning, and that the risk-set allocation is expected to remain near 1:1. Then $q=0.5$, $z_{0.975}=1.960$, and $z_{0.90}=1.2816$.

Substitute these values, with $\log(0.70)=-0.356675$:

D \approx \frac{(1.960+1.2816)^2}{(0.5)(0.5)(-0.356675)^2} = \frac{10.5080}{0.0318043} \approx 330.4.

Because an event count must be a whole number and the target must not fall below the approximation, round up: the estimated requirement is 331 events. This is the total event target, not 331 events in each group.

Worked Example: Unequal Allocation and a One-Sided Test

Consider a hypothetical trial with two treatment participants for every control participant, a prespecified one-sided $\alpha=0.025$ test, 80% power, and target $\mathrm{HR}=0.75$. Suppose the scientific alternative is that treatment lowers the hazard, and the assumed proportional-hazards and censoring conditions are acceptable. The allocation fraction is $q=2/3$, so $q(1-q)=2/9$. The one-sided critical value is $z_{0.975}=1.960$, and $z_{0.80}=0.8416$.

D \approx \frac{(1.960+0.8416)^2}{(2/9)[\log(0.75)]^2} = \frac{7.84896}{(2/9)(0.287682)^2} \approx \frac{7.84896}{0.0183913} \approx 426.8.

Rounding up gives an estimated target of 427 events. A one-sided test is appropriate only if its direction was specified in advance and an effect in the opposite direction will not be treated as evidence for the intended claim.

Worked Example: A Moderate Effect With Unequal Allocation

For a hypothetical two-sided design, suppose $\alpha=0.05$, power is 80%, the target $\mathrm{HR}=0.80$, and there is one treatment participant for every two controls. Assume the intended hazard ratio is approximately constant and the planned allocation is a reasonable representation of the event risk sets. Here $q=1/3$, $q(1-q)=2/9$, $z_{0.975}=1.960$, and $z_{0.80}=0.8416$.

D \approx \frac{(1.960+0.8416)^2}{(2/9)[\log(0.80)]^2} = \frac{7.84896}{(2/9)(0.223144)^2} \approx \frac{7.84896}{0.0110651} \approx 709.5.

The target is therefore 710 events after rounding up. This relatively large target reflects both the moderate hazard reduction and the unequal allocation. The calculation does not imply that 710 participants are sufficient: some participants may be censored before an event, so enrollment must be planned using assumptions about follow-up and event occurrence.

Using the Estimate in a Trial Design

The Schoenfeld calculation is useful for comparing candidate designs and for translating a clinically meaningful hazard ratio into a first event target. The following sequence helps prevent common input and interpretation errors.

Specify the testing plan.
State whether the test is one-sided or two-sided, the significance level, and the desired power. Use a one-sided formula only when the direction and testing plan are prespecified.

Set the effect and allocation.
Choose a clinically justified hazard ratio and convert the planned treatment-to-control ratio to $q$. Do not substitute the allocation ratio itself for the fraction $q$.

Calculate and round the event target.
Evaluate the log hazard ratio and allocation term carefully, retain sufficient precision during calculation, and round the final event count up to the next whole event.

Translate events into enrollment separately.
Use assumptions about accrual, follow-up, event incidence, and censoring to estimate how many participants and how much calendar time may be required. The event formula alone does not perform this conversion.

For a group sequential trial, this calculation is a fixed-design reference point, not automatically the final group sequential event target. Sequential boundaries and the desired overall operating characteristics can change the amount of information required. The event target should therefore be coordinated with the sequential design rather than treated as an exact sequential sample-size calculation.

Diagnostics, Alternatives, and Limitations

A useful diagnostic is to examine whether the inputs plausibly describe the planned trial. The approximation can be misleading if proportional hazards is doubtful, if treatment and control risk sets become substantially imbalanced, or if censoring depends on prognosis in a way not represented in the planning assumptions. In a stratified design, varying allocation or event patterns across strata may also make a single overall $q$ a poor summary. These concerns call for a design-specific information calculation or simulation rather than false precision from one closed-form number.

Compared with planning directly from an information target, Schoenfeld’s formula makes the link between event count, allocation, and a specified hazard ratio explicit through $q(1-q)[\log(\mathrm{HR})]^2$. An information-based approach can be more flexible when the information per event is not well represented by this approximation. More detailed power calculations or simulations can account for features such as changing hazards, complex censoring, stratification, and sequential monitoring, but require more assumptions and inputs. None can rescue an implausible target effect or unexamined censoring assumptions.

Do not use participants in place of events. The formula returns an event count. Enrollment requires a separate model of event accrual and follow-up.
Do not confuse a ratio with a fraction. For a 2:1 treatment-to-control ratio, $q=2/3$, not 2.
Do not omit the log. The effect term is $[\log(\mathrm{HR})]^2$, not $\mathrm{HR}^2$. A hazard ratio closer to 1 has a smaller log-effect magnitude and therefore requires more events.
Do not round intermediate values too early. Small errors in the denominator can change the rounded event target. Keep precision through the calculation, then round up.
Do not call the result an exact guarantee. It is a planning approximation conditional on the effect, assumptions, and test specification.

Key takeaway: For a fixed-design logrank comparison, the approximate event requirement grows with the squared normal-quantile sum and shrinks with $q(1-q)[\log(\mathrm{HR})]^2$. Treat the rounded result as a planning benchmark, then check risk-set, censoring, and sequential-design assumptions before using it operationally.

Check Your Understanding

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

With equal allocation, a two-sided $\alpha=0.05$ test, 80% power, and target $\mathrm{HR}=0.75$, approximately how many events are required?
For a one-sided $\alpha=0.05$ test, 90% power, equal allocation, and target $\mathrm{HR}=0.70$, calculate the rounded-up event target.
For the same effect, alpha, and power, how does changing from equal allocation to a 3:1 treatment-to-control ratio affect the approximation? Give the multiplier on the event count.
Why can the Schoenfeld event target not be read as the required number of participants, and why is it not automatically a final sequential event target?

Solutions. (1) With $q(1-q)=0.25$, the numerator is $(1.960+0.8416)^2=7.84896$. Since $\log(0.75)=-0.287682$, the denominator is $0.25(0.287682)^2=0.0206902$. Thus $D\approx 7.84896/0.0206902\approx379.3$, so round up to 380 events. (2) A one-sided test uses $z_{0.95}=1.6449$, and 90% power uses $z_{0.90}=1.2816$. The numerator is $(1.6449+1.2816)^2\approx8.5644$; the denominator is $0.25[\log(0.70)]^2\approx0.0318043$. Hence $D\approx269.3$, giving 270 events. (3) Under 3:1 allocation, $q=3/4$ and $q(1-q)=3/16=0.1875$, compared with 0.25 for equal allocation. The event-count multiplier is $0.25/0.1875=4/3$, so the approximate requirement increases by one third. (4) Enrollment depends on how many participants experience an event during follow-up, which the formula does not model. A sequential design’s required information can also differ from the fixed-design benchmark because its monitoring plan must be incorporated into the design.