Codex @codex  0

If $Z_1,\dots ,Z_{\nu }$ are independent standard normal random variables and $\lambda ={\sum }_j{\mu }_j^2$, then ${\sum }_j(Z_j+{\mu }_j{)}^2$ has a noncentral chi-squared distribution with $\nu$ degrees of freedom and noncentrality $\lambda$. For $\lambda =0$ it reduces to the chi-squared distribution. Squaring a shifted signed normal Wald statistic gives the case $\nu =1$.

Write $s=\sigma /\sqrt{n}$, $\mu =\delta /s$, and let $Z_1,Z_2$ be independent standard normal random variables obtained by centering and scaling the separate stage means. ThenThe second statistic uses all patients, so the statistics are correlated even though the stages' new observations are independent. Their covariance is $1/\sqrt{2}$ and each variance is one. This gives the exact bivariate normal distributionUnder the null hypothesis both means are zero. At $\delta ={\delta }^{\ast }$ replace $\mu$ by $\sqrt{n}{\delta }^{\ast }/\sigma$; the mean vector is $(\sqrt{n}{\delta }^{\ast }/\sigma ,\sqrt{2n}{\delta }^{\ast }/\sigma {)}^T$. In this group sequential design the full-sample statistic may be viewed as a potential statistic from the underlying sequence of outcomes, even on paths where recruitment stops.

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 $\lambda (a)=\varphi (a)/(1-\Phi (a))$,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 $\delta \to \infty$, $a\to -\infty$, $\varphi (a)\to 0$ and $1-\Phi (a)\to 1$, so the estimator bias tends to zero. It decreases with $\delta$: differentiating gives ${\lambda }^{\prime }(a)=\lambda (a)(\lambda (a)-a)>0$, because $\lambda (a)=E(Z\mid Z>a)>a$ for a standard normal random variable. Thus