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The Wald statistic has law $Z_1={\hat{\delta }}_1\sqrt{I_1}\sim N(\delta \sqrt{I_1},1)$. Write $C=\{Z_1\ge f_1\}$ for continuation. By the normal cumulative distribution function,The final sample size per arm is $N=n_1+n_2^{\ast }{1}_C$, soIt lies between $n_1$ and $n_2=n_1+n_2^{\ast }$, increasing with the treatment effect: a more promising treatment is more likely to reach full enrollment. Total expected recruitment over both arms is twice this expression. Equality at the futility boundary has probability zero under the normal model.

Set $z=z_{1-\alpha }$. Under the null hypothesis rejection requires both $W_1\ge f$ and $W_2>z$. Since $W_2$ has the standard normal distribution,The bivariate normal distribution in the preceding calculation has a nonsingular covariance matrix and strictly positive statistical probability density everywhere. For every finite futility boundary $f$ and $0<\alpha <1$, the open rectangle $\{W_1<f,W_2>z\}$ has positive probability. ThereforeFor an explicit expression, conditional on $W_1=w$ the full-sample statistic is $N(w/\sqrt{2},1/2)$ under the null hypothesis, yieldingThe type I error is reduced because some otherwise rejecting paths stop for futility. Equality is approached as $f\to -\infty$, but does not hold at a finite futility boundary.