Solution
= Solution
The trial uses $N_1$ observations per arm if it stops at the first analysis and $N_2$ if it continues. Since $Z_1\sim N(\delta\sqrt{I_1},1)$,
$$
\mathbb P_\delta(\text{continue})
=\Phi(e_1-\delta\sqrt{I_1})-\Phi(f_1-\delta\sqrt{I_1}).
$$
Therefore the expected per-arm sample size is
$$
\boxed{
\mathbb E_\delta N
=N_1+(N_2-N_1)
\left\{\Phi(e_1-\delta\sqrt{I_1})-\Phi(f_1-\delta\sqrt{I_1})\right\}}.
$$