Conditional Means Under Truncated-Normal Selection

Why a Fixed Retention Rule Changes the Mean

An interim treatment-effect estimate can be unbiased before any decision is made and still be systematically high among the estimates that survive a retention rule. The reason is direct: when an arm is retained only if its estimate exceeds a threshold, estimates on the lower side of the sampling distribution are excluded. The retained estimates therefore have a different conditional distribution from the original estimates.

The introductory material on winner’s-curse bias motivates this phenomenon. Here the focus is the exact conditional mean for a single normal estimate selected by a fixed upper threshold. This calculation describes the selection effect itself; it does not require comparing several arms or modeling their dependence.

Definition: Let $\hat{\theta}$ be an interim estimate of a treatment effect $\theta$, with $\hat{\theta}\sim N(\theta,\sigma^2)$, where $\sigma$ is the known standard error for this calculation. Suppose the arm is retained when $A=\{\hat{\theta}>c\}$, for a fixed threshold $c$. The conditional mean is $E(\hat{\theta}\mid A)$, and the conditional selection bias is $E(\hat{\theta}\mid A)-\theta$. The threshold $c$, $\theta$, and $\sigma$ are expressed in the same treatment-effect units.

The normal model is exact if the estimate is normally distributed with fixed standard error. It can also be an approximation when the estimate is approximately normal and its standard error is treated as fixed. The threshold must be fixed for the calculation, and the selection event must be determined by whether this same estimate exceeds that threshold. If the standard error is random, the threshold changes with the data, or retention depends on additional statistics, this one-variable formula need not apply as written.

Deriving the Truncated-Normal Conditional Mean

Standardize the estimate by writing $Z=(\hat{\theta}-\theta)/\sigma$. Under the stated model, $Z\sim N(0,1)$. Retention occurs precisely when $Z>a$, where

a=\frac{c-\theta}{\sigma}.

Thus $a$ is the threshold’s distance above the true effect, measured in standard errors. Let $\phi$ and $\Phi$ denote the standard normal density and cumulative distribution functions. The retention probability is $P(A)=P(Z>a)=1-\Phi(a)$. The conditional expectation is obtained by integrating the standard normal density over the retained region and dividing by the probability of that region:

E(Z\mid Z>a) =\frac{\int_a^\infty z\phi(z)\,dz}{1-\Phi(a)} =\frac{\phi(a)}{1-\Phi(a)} =\lambda(a).

For the middle equality, use $\phi'(z)=-z\phi(z)$, so that $\int_a^\infty z\phi(z)\,dz=[-\phi(z)]_a^\infty=\phi(a)$. The function $\lambda(a)=\phi(a)/[1-\Phi(a)]$ is the inverse Mills ratio for upper truncation. Returning to the original scale gives the key result:

Theorem: If $\hat{\theta}\sim N(\theta,\sigma^2)$ and an estimate is retained exactly when $\hat{\theta}>c$, then $$ E(\hat{\theta}\mid\hat{\theta}>c) =\theta+\sigma\lambda(a), \qquad \operatorname{Bias}(\hat{\theta}\mid\hat{\theta}>c) =\sigma\lambda(a), \qquad a=\frac{c-\theta}{\sigma}. $$ For every finite $a$, $\lambda(a)>0$; hence upper-threshold selection gives a positive conditional bias under this model.

The result separates the effect into two parts. The standardized threshold $a$ determines the size of the truncation on the normal scale, while $\sigma$ converts the resulting shift back into treatment-effect units. The conditional bias is not generally equal to the distance between the threshold and the true effect: it depends on the entire retained tail through $\lambda(a)$.

Worked Example: A Threshold One Standard Error Above the True Effect

Suppose a hypothetical treatment-effect estimate has true value $\theta=1.00$ units and standard error $\sigma=0.50$ units. The arm is retained if the interim estimate exceeds $c=1.50$ units. Assume the normal model with fixed standard error is appropriate.

Standardize the threshold. $a=(1.50-1.00)/0.50=1$. Thus the estimate is retained only when its standardized error exceeds 1.

Find the retention probability and inverse Mills ratio. $P(A)=1-\Phi(1)\approx0.158655$, and $\lambda(1)=\phi(1)/[1-\Phi(1)]\approx0.2419707/0.1586553\approx1.525135$.

Calculate the bias and conditional mean. The bias is $\sigma\lambda(a)=(0.50)(1.525135)=0.762568$ units, or approximately $0.7626$ units. Therefore, $E(\hat{\theta}\mid A)=1.00+0.762568=1.762568$, approximately $1.7626$ units.

The selected estimates average above the true effect by about $0.7626$ units, even though the estimate has mean $1.00$ units before selection. This is a conditional statement about the subset of estimates that pass the rule, not a claim that the true treatment effect has changed.

How the Threshold and Standard Error Shape the Bias

The standardized threshold $a$ determines how selective the rule is relative to the estimate’s sampling variability. If $a=0$, the rule retains estimates above their sampling mean. If $a$ is positive, the cutoff lies above that mean and selection is more stringent. If $a$ is negative, the cutoff lies below the mean and most estimates are retained, but the conditional bias remains positive.

For a fixed value of $a$, doubling $\sigma$ doubles the bias in treatment-effect units. But changing $\sigma$ while holding $c-\theta$ fixed also changes $a$, so the overall change need not be a simple doubling. This distinction matters when comparing designs or estimates with different standard errors: both the scale factor and the standardized distance to the threshold are relevant.

Worked Example: A Threshold Below the True Effect

Suppose a hypothetical estimate has $\theta=2.00$ treatment-effect units and $\sigma=0.40$ units. An arm is retained if $\hat{\theta}>1.60$ units. The estimate is assumed normal with this fixed standard error.

Here $a=(1.60-2.00)/0.40=-1$, so $P(A)=1-\Phi(-1)=\Phi(1)\approx0.841345$. The upper-tail inverse Mills ratio is $\lambda(-1)=\phi(-1)/[1-\Phi(-1)]\approx0.2419707/0.8413447\approx0.287600$.

The conditional bias is $(0.40)(0.287600)=0.115040$ units, and the conditional mean is $2.00+0.115040=2.115040$, or approximately $2.115$ units. Most estimates pass this relatively low threshold, so the upward shift is smaller than in the preceding example. It is still positive because the rule removes some estimates from the lower tail.

The bias increases as the standardized threshold increases. One way to check this is to differentiate the inverse Mills ratio. From $\lambda(a)=\phi(a)/[1-\Phi(a)]$, the quotient rule and $\phi'(a)=-a\phi(a)$ give $\lambda'(a)=\lambda(a)\{\lambda(a)-a\}$. For the standard normal upper tail, $\lambda(a)>a$, so the derivative is positive. This inequality is immediate for $a\leq0$ because $\lambda(a)>0$; for $a>0$, it also follows from $E(Z\mid Z>a)>a$, since a continuous variable conditioned to exceed $a$ has conditional mean strictly above $a$. Consequently, increasing $a$ increases $\sigma\lambda(a)$ when $\sigma$ is fixed.

Conditional Variability and a Numerical Check

The truncated distribution is not only shifted; its variance also changes. A useful check on the conditional mean calculation is to derive the second moment. Integration by parts gives $\int_a^\infty z^2\phi(z)\,dz=a\phi(a)+1-\Phi(a)$. Dividing by $1-\Phi(a)$ and subtracting the squared conditional mean yields

\operatorname{Var}(\hat{\theta}\mid\hat{\theta}>c) =\sigma^2\{1+a\lambda(a)-\lambda(a)^2\}.

This variance is conditional on passing the threshold; it is not the original variance $\sigma^2$. It can help check a simulation or a numerical implementation, but the conditional mean and bias remain the main quantities of interest here.

Worked Example: A Stringent Threshold and the Conditional Spread

Consider a hypothetical estimate with $\theta=0.80$ units and $\sigma=0.30$ units, retained if it exceeds $c=1.40$ units. The normal fixed-standard-error model is assumed. The standardized threshold is $a=(1.40-0.80)/0.30=2$, and the retention probability is $1-\Phi(2)\approx0.022750$.

Using $\lambda(2)\approx2.373216$, the conditional bias is $(0.30)(2.373216)=0.711965$ units, and the conditional mean is $0.80+0.711965=1.511965$ units. The conditional mean is above the cutoff, as it must be for a distribution restricted to values above that cutoff.

For the conditional variance, use the unrounded inverse Mills ratio in the formula; its value is approximately $2.3732155$. Thus $1+2\lambda(2)-\lambda(2)^2\approx0.114279$, and the conditional variance is $(0.30)^2(0.114279)\approx0.0102851$ square units. The corresponding conditional standard deviation is approximately $0.1014$ units, smaller than the original $0.30$-unit standard deviation. Rounding the inverse Mills ratio before squaring can produce a small arithmetic discrepancy, so retain extra precision for intermediate calculations.

Diagnostics, Alternatives, and Limitations

Before using the formula, check the model and the selection rule rather than treating truncation as an automatic explanation for every difference between selected and unselected estimates. In particular, verify that the threshold is fixed, the estimate has the stated normal distribution, and $\sigma$ is appropriate for the estimate’s sampling variability. The probability of retention should also agree with the standardized threshold: a large positive $a$ implies a small upper-tail probability.

Check the scale. $c-\theta$ and $\sigma$ must use the same units, making $a$ dimensionless. The final bias $\sigma\lambda(a)$ has the units of the treatment effect.
Check the event. This result is for $\hat{\theta}>c$. A rule based on a different statistic or on several data-dependent conditions defines a different conditioning event.
Check numerical stability. For a very high threshold, $1-\Phi(a)$ is tiny. Directly subtracting $\Phi(a)$ from 1 can lose numerical precision; use a stable upper-tail probability or a reliable inverse-Mills-ratio calculation.
Check the conditional mean’s location. It must exceed $c$. A computed conditional mean at or below the cutoff signals an error in the tail probability, standardization, or arithmetic.

The truncated-normal calculation is an analytic alternative to simulation when the single-estimate normal model and fixed threshold are adequate. Simulation can instead approximate the conditional mean by generating estimates from the assumed distribution, retaining those above $c$, and averaging the retained values. Agreement between the simulated mean and $\theta+\sigma\lambda(a)$, within Monte Carlo error, is a useful implementation check. Simulation does not remove a model mismatch: if the assumed distribution or selection rule is wrong, both approaches can describe the wrong conditional behavior.

A tempting but incorrect alternative is to report the unconditional mean $\theta$ as the expected estimate among retained arms, on the grounds that $\hat{\theta}$ is unbiased. Unbiasedness means $E(\hat{\theta})=\theta$ before conditioning; it does not imply $E(\hat{\theta}\mid A)=\theta$. The formula quantifies precisely the difference. Nor should this result be applied unchanged when retention depends on multiple correlated contrasts, a random standard error, or a sequence of decisions. Those situations require their own selection event and probability model.

Common pitfall: Treating the retained estimates as an unselected sample and using $E(\hat{\theta})=\theta$ after the threshold has been applied. The correction is to condition on the actual event: under the stated upper-truncation model, add $\sigma\lambda((c-\theta)/\sigma)$ to $\theta$. The adjustment is positive and depends on both the threshold’s standardized distance and the standard error.

Check Your Understanding

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

An estimate has $\theta=5$ units and $\sigma=2$ units, and it is retained above $c=7$ units. Find $a$, the retention probability, and the conditional mean.
If the threshold equals the true effect, what is the conditional bias in units of $\sigma$?
Why can an unbiased estimate have a positive conditional bias after an upper-threshold rule?
A calculation gives a conditional mean below the retention threshold. What should be checked first?

Solutions. (1) $a=(7-5)/2=1$. The retention probability is $1-\Phi(1)\approx0.158655$. With $\lambda(1)\approx1.525135$, the conditional mean is $5+2(1.525135)=8.05027$ units, approximately $8.050$ units. (2) When $c=\theta$, $a=0$, so the bias is $\sigma\lambda(0)=\sigma\sqrt{2/\pi}\approx0.7979\sigma$. (3) Unbiasedness is an unconditional property. Conditioning on $\hat{\theta}>c$ removes estimates below the cutoff, shifting the retained distribution upward. (4) Check the event direction, standardization $a=(c-\theta)/\sigma$, and use of the upper-tail probability $1-\Phi(a)$. Under this model every retained value exceeds $c$, so its conditional mean must also exceed $c$.