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Two-look Pocock-type symmetric spending

Example

Assume AC, take K=2, t1=1/2, t2=1, and α=1/20. Use the symmetric two-sided Pocock-type spending curve AP(0)=0,AP(t)=αlog⁡(1+(e−1)t)(0<t≤1), where e=exp⁡(1) and log⁡ is the natural logarithm. Its first spend is a1=αlog⁡((1+e)/2). Let Φ(x)=P0(Z1≤x) be the standard normal CDF and let q(u) be the unique real number with Φ(q(u))=u. The first symmetric boundary is c1=q(1−a1/2). The second boundary is the unique finite c2 satisfying P0(∣Z1∣≥c1 or ∣Z2∣≥c2)=α, where the canonical covariance gives Corr⁡0(Z1,Z2)=1/2. The joint probability of continuing past look 1 and crossing at look 2 is therefore α−a1. Conditional on continuing, the second-look crossing probability is (α−a1)/(1−a1). The Pocock-type spending curve does not assert constant standardized boundaries; each boundary is calibrated under the joint law.

Facts & Assumptions

Given: AC, the two planned information fractions t1=1/2,t2=1, α=1/20, the canonical Gaussian null law, and the stated Pocock-type spending curve. Verify the curve is admissible and identify the two two-sided boundaries exactly.

[F1]

Canonical group-sequential alpha-spending design: A design has nondecreasing A:[0,1]→[0,α] with A(0)=0, A(1)=α, and canonical covariance Σjk=tj/tk for j≤k.

[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 increment has a unique finite boundary and P0(Rk)=A(tk) at each look.

[F4]

Standard normal and normal laws: The standard normal law has 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]

The real exponential function and the number e by a power series: The real exponential is defined by its power series and e:=exp⁡(1).

[F7]

The natural logarithm as the inverse of the exponential function: For x>0, log⁡x is the unique real y with exp⁡(y)=x.

[F8]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: log⁡ is continuous and strictly increasing on (0,∞), and log⁡1=0.

[F9]

The exponential function is strictly increasing: The real exponential is strictly increasing on R.

[F10]

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

[F11]

Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b] takes every value between f(a) and f(b): A continuous real function on a closed bounded interval attains every value between its endpoint values.

[A1]

The Axiom of Choice: Assume the Axiom of Choice.

Verification

1.1givenF1F6F7F8F9algebra

Since exp⁡(0)=1 by its power series and exp⁡ is strictly increasing by [F9], e=exp⁡(1)>1. Thus 1+(e−1)t increases from 1 to e as t goes from 0 to 1. By [F7]–[F8], AP(0)=0, AP(1)=αlog⁡e=α, and AP is nondecreasing with values in [0,α]. At t1=1/2 this gives a1=αlog⁡((1+e)/2) and 0<a1<α.

1.2F4F5F10F11A1algebra

The standard normal CDF satisfies Φ(y)−Φ(x)=∫xyϕ(s) ds for x<y. By [F4]–[F5], this difference is positive and at most (y−x)/2π, so Φ is strictly increasing and continuous. Its limits at −∞ and +∞ are 0 and 1: [F10] applied to (−n,∞) gives the lower-tail limit, and [F10] applied to (−∞,n] gives the upper-tail limit; the density has no point masses. Therefore [F11] gives a unique q(u) for every u∈(0,1). In particular c1=q(1−a1/2) is finite, positive, and symmetry and atomlessness give P0(∣Z1∣≥c1)=a1.

2.1givenF1F2F3step 1.1step 1.2algebra∎

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. The event {∣Z1∣≥c1} and the continued-path event {∣Z1∣<c1, ∣Z2∣≥c2} are disjoint and their union is the final rejection event. Hence the continued-path event has joint probability α−a1, and since P0(∣Z1∣<c1)=1−a1>0, its conditional probability given continuation is (α−a1)/(1−a1). The total rejection probability is AP(1)=α.

Depends on

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