= Solution
Let $X$ and $Y$ denote the first- and second-stage <sample means>, let $s=\sigma/\sqrt n$, and write $S=(X+Y)/2$. Continuation is the selection event $\mathcal C=\{X\ge sf\}$. The second-stage <sample mean> stays independent of $\mathcal C$, so $E(Y\mid\mathcal C)=\delta$. The first-stage <sample mean> has a <truncated normal distribution>. With $a=f-\delta/s$ and the upper-tail <Inverse Mills ratio> $\lambda(a)=\phi(a)/(1-\Phi(a))$,
$$
E(X\mid\mathcal C)=\delta+s\lambda(a),\qquad
\boxed{E(S\mid\mathcal C)-\delta=\frac{s}{2}\lambda\left(f-\frac{\delta}{s}\right)>0.}
$$
The positive <conditional selection bias after futility continuation> comes from selecting unusually large first-stage outcomes. The unconditional <sample mean> of a fixed $2n$ observations would be unbiased; that is a different sampling distribution from the one restricted to continued trials.
As $\delta\to\infty$, $a\to-\infty$, $\phi(a)\to0$ and $1-\Phi(a)\to1$, so the <estimator bias> tends to zero. It decreases with $\delta$: differentiating gives $\lambda'(a)=\lambda(a)(\lambda(a)-a)>0$, because $\lambda(a)=E(Z\mid Z>a)>a$ for a <standard normal random variable>. Thus
$$
\boxed{\text{the bias is upward, decreases as the effect increases, and tends to }0.}
$$
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