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Two-look O’Brien–Fleming-type spending

Example

Assume AC, take K=2, t1=1/2, t2=1, and α=0.05. Let Φ(x)=P0(Z1≤x) be the standard normal CDF, and write zp for the unique real number with Φ(zp)=p. Use the two-sided O’Brien–Fleming-type spending curve AOF(0)=0,AOF(t)=2[1−Φ(z0.975t)](0<t≤1). Its first spend is a1=2[1−Φ(z0.9752)]. The first symmetric two-sided boundary is c1=z1−a1/2. The second boundary is the unique finite c2≥0 satisfying P0(∣Z1∣≥c1 or ∣Z2∣≥c2)=0.05, where Corr⁡0(Z1,Z2)=1/2. The two-sided boundaries come from this absolute-value crossing equation, not from doubling the one-sided boundaries.

Facts & Assumptions

Given: AC, K=2, t1=1/2,t2=1, α=0.05, the canonical Gaussian null law, and the displayed O’Brien–Fleming-type spending curve. Verify the spending conditions and identify the two symmetric boundaries.

[F1]

Canonical group-sequential alpha-spending design: An admissible spending function is nondecreasing from [0,1] into [0,α], with endpoint values 0 and α; the symmetric two-sided crossing event at look j is {∣Zj∣≥cj}.

[F2]

Recursive group-sequential boundary calibration: At a positive increment the symmetric two-sided boundary solves P0(Rk−1∪{∣Zk∣≥ck})=A(tk).

[F3]

Group-sequential boundaries exist uniquely and spend alpha exactly: Every positive spending increment has a unique finite boundary and P0(Rk)=A(tk) at each look.

[F4]

Standard normal and normal laws: The standard normal law is the probability measure with density ϕ(x)=e−x2/2/2π.

[F5]

The standard normal density has total mass one: The density ϕ is positive and has integral one.

[F6]

Continuity from below for measures: For an increasing sequence of measurable sets, the measure of their union is the supremum of their measures.

[F7]
[A1]

The Axiom of Choice: Assume the Axiom of Choice.

Verification

1.1F4F5A1algebra

Define Φ(x)=∫−∞xϕ(s) ds. For x<y, positivity in [F5] gives Φ(y)−Φ(x)>0, while ϕ(s)≤1/2π gives Φ(y)−Φ(x)≤(y−x)/2π. Thus Φ is strictly increasing and continuous.

2.1F4F5F6F7step 1.1algebra

The density ϕ is even, so Φ(0)=1/2. Applying [F6] to the increasing upper half-lines (−n,∞) gives Φ(x)→0 as x→−∞; applying it to the increasing lower half-lines (−∞,n] gives Φ(x)→1 as x→+∞. Hence [F7] gives a unique quantile zp for each p∈(0,1), and z0.975>0.

3.1givenF1step 1.1step 2.1algebra

For 0<t≤1, z0.975/t decreases as t increases; since Φ increases, AOF(t) is nondecreasing. Its values are positive and at most 2[1−Φ(z0.975)]=0.05, and its limit at 0 is 0 by the upper-tail limit in step 2.1. Thus the declared AOF(0)=0 makes it admissible. At t1=1/2, its value is a1=2[1−Φ(z0.9752)], with 0<a1<0.05.

4.1givenF1F2F3F4step 2.1step 3.1algebra∎

By symmetry of the density in [F4], P0(∣Z1∣≥c)=2[1−Φ(c)] for c≥0. Therefore c1=z1−a1/2 has first-look crossing probability a1. The canonical covariance matrix is Σ=(11/21/21), so its determinant is 1/2>0 and the correlation is 1/2. Applying [F2]–[F3] at the second look gives the unique finite c2 satisfying the displayed cumulative two-sided crossing equation, with total probability AOF(1)=0.05.

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