How are group sequential analysis, random walks, and Brownian motion related? Assume that I am planning a clinical trial comparing two groups using a binary outcome.  I will do the $\chi^2$ test after 3 equal enrollment intervals: interim test #1 after $m_1$ enrollments in group 1 and $m_2$ enrollments in group 2; interim test #2 after $2m_1$ and $2m_2$ enrollments; and the final test after $3m_1$ and $3m_2$ enrollments.  On the nth test (n = 1 ,2,3), I will compare the $\chi^2$ statistic to $\frac{3}{n} \times c$, where $c$ is the O'Brien Fleming constant.  For $N=3$ and $\alpha = 0.05$, $c = 3.940$. (Note that this is not much different from 1.96^2 = 3.84.)
For each study interval, we see $y_{1n}$ outcomes in group 1 and $y_{2n}$ in group 2.  Let $\pi_1$ and $\pi_2$ be the true outcome probabilities.  The hypothesis being tested (H0) is that $\pi_1 = \pi_2 = \pi$.  For each of the 3 intervals, we create a standard normal random variable $U_n$:
$$ U_n = \frac{y_{1n}/m_1 - y_{2n}/m_2}{\sqrt{\pi(1-\pi)(1/m_1+1/m_2)}}$$
Let $S_n$ be the sum of the $U_n$s, so
\begin{align*}
S_1 &= U_1 \quad \quad \quad \quad \sim N(0,1)\\
S_2 &= U_1 +  U_2  \quad \quad \sim N(0,2)\\
S_3 &= U_1 + U_2 +U_3 \sim N(0,3) 
\end{align*}
This is a 3-step Gaussian random walk starting at 0 with $U_n$ as the steps and $S_n$ as the location after $n$ steps.  The steps $U_1$, $U_2$, and $U_3$ are independent but the locations $S_1$, $S_2$, and $S_3$ are not.
O'Brien and Fleming (1979) define $Z_n^*$ and $T_n$ as follows:
$$Z_n^* = S_n/\sqrt{n}  \sim N(0,1)$$
$$T_n = S_n^2/3 = (Z_n^*)^2\big(\frac{n}{3}\big) $$
$(Z_n^*)^2$ is the random variable corresponding to my $\chi^2$ statistics, which I am going to compare to $\frac{3}{n} \times c$.  To get $c$, O'Brien and Fleming say we need to know the distribution of $max\{T_n\}$ (the maximum of $T_1$, $T_2$, and $T_3$) which they say has the same distribution as $max\{[W(n/3)]^2\}$ where $\{W(t)|0 \le t \le1\}$ "represents Brownian motion".
Q1: Are they correct that $max\{T_n\}$ for $n = 1,2,3$ has the same distribution as $max\{[W(n/3)]^2\}$ where $\{W(t)|0 \le t \le1\}$ "represents Brownian motion"?
They say that $c = 3.940$ is the 95th percentile value for $max\{T_n\}$ = $max\{[W(n/3)]^2\}$.
Q2: Is this correct?  How does one get this value? Could I use this formula?

Since posting, I have partially answered my own questions.
Q1: Are O'Brien and Fleming correct when they say that the distribution of $max\{T_n\}$ has the same distribution as $max \{ [W(n/3)]^2 \}$ where $\{ W(t)|0 \le t \le 1 \}$ "represents Brownian motion"?
A1: Yes.  Also, "Brownian motion" and "Wiener process" are used synonymously.
Q2: How does one get the value of $c = 3.940$ for $N=3$ "looks" and $\alpha = 0.05$?
A2: I was able to get close to this value using simulation in R.  In fact, I was able to approximately reproduce their Table 1 of $c$ values for all their combinations of $N$ and $\alpha$.  I was hoping that there was a better way than brute force simulation.
