Adaptive designs with arbitrary dependence structure based on Fisher’s combination test

Abstract

Adaptive designs were originally developed for independent and uniformly distributed $p$-values. However, in general the type I error rate of a given adaptive design depends on the true dependence structure between the stage-wise $p$-values. Since there are settings, where the $p$-values of the stages might be dependent with even unknown dependence structure, it is of interest to consider the most adverse dependence structure maximizing the type I error rate of a given adaptive design (worst case). In this paper, we explicitly study the type I error rate in the worst case for adaptive designs without futility stop based on Fisher’s combination test. Potential inflation of the type I error rate is studied if the dependence structure between the $p$-values of the stages is not taken into account adequately. It turns out that considerable inflation of the type I error rate can occur. This emphasizes that the examination of the true dependence structure between the stage-wise $p$-values and an adequate choice of the conditional error function is crucial when adaptive designs are used.

Appendix

1.1 Equivalence of Eqs. (8) and (9) for admissible $c$ and $\gamma$

As a consequence of Lemma 1 below, the following statements are equivalent:

$$\begin{aligned} \alpha _{w}&= 2 - 2 F \left( \frac{c}{\sqrt{2}} \right) \qquad \text {and} \quad c \le \sqrt{2} \gamma \\ \Longleftrightarrow \quad \alpha _{w}&= 2 - 2 F \left( \frac{c}{\sqrt{2}} \right) \qquad \text {and} \quad \alpha _{w} \ge 2 \alpha _1 \\ \Longleftrightarrow \quad c&= \sqrt{2} F^{-1}\left( 1 - \frac{\alpha _{w}}{2} \right) \qquad \text {and} \quad \alpha _{w} \ge 2 \alpha _1 \end{aligned}$$

Likewise, as a consequence of Lemma 2 below, the following statements are equivalent:

$$\begin{aligned} \alpha _{w}&= 2 - F \left( \gamma \right) - F \left( \sqrt{2} c - \gamma \right) \quad \text {and} \quad c > \sqrt{2} \gamma \\ \Longleftrightarrow \quad \alpha _{w}&= 2 - F \left( \gamma \right) - F \left( \sqrt{2} c - \gamma \right) \quad \text {and} \quad \alpha _{w} < 2 \alpha _1 \\ \Longleftrightarrow \quad c&= \frac{1}{\sqrt{2}} \left( F^{-1}(1 - \alpha _1) + F^{-1}(1 + \alpha _1 - \alpha _{w}) \right) \quad \text {and} \quad \alpha _{w} < 2 \alpha _1 \end{aligned}$$

A pair of parameters $c$ and $\gamma$ is admissible if $0 < \alpha _{w}(c,\gamma ) < 1$ with $\alpha _{w}(c,\gamma )$ from (8). For admissible $c$ and $\gamma$, we have

$$\begin{aligned} \alpha _{w}&= {\left\{ \begin{array}{ll} 2 \left( 1- F\left( \frac{c}{\sqrt{2}}\right) \right) , &{} \text { if } \quad c \le \sqrt{2} \gamma ,\\ 2- F(\gamma ) - F( \sqrt{2}c - \gamma ), &{} \text { if } \quad \sqrt{2} \gamma \le c. \end{array}\right. } \end{aligned}$$

Thus the above shows that Eqs. (8) and (9) are equivalent for admissible $c$ and $\gamma$.

Lemma 1

For $0 \le \alpha _1 \le \alpha _{w} \le 1$, let $\gamma := F^{-1}(1-\alpha _1)$, and let $c \in \mathbb {R}$ such that $\alpha _{w}=2-2 F \left( \frac{c}{\sqrt{2}}\right)$. Then:

$$\begin{aligned} c \le \sqrt{2} \gamma \Leftrightarrow \alpha _{w} \ge 2 \alpha _1. \end{aligned}$$

(16)

Proof

$c \le \sqrt{2} \gamma$ $\Leftrightarrow$ $F\left( \frac{c}{\sqrt{2}}\right) \le F(\gamma )$ $\Leftrightarrow$ $1-\frac{\alpha _{w}}{2} \le 1 - \alpha _1$ $\Leftrightarrow$ $\alpha _{w} \ge 2 \alpha _1$.$\square$

Lemma 2

For $0 \le \alpha _1 \le \alpha _{w} \le 1$, let $\gamma := F^{-1}(1-\alpha _1)$, and let $c \in \mathbb {R}$ such that $\alpha _{w}= 2- F(\gamma ) - F(\sqrt{2} c -\gamma )$. Then:

$$\begin{aligned} c \le \sqrt{2} \gamma \Leftrightarrow \alpha _{w} \ge 2 \alpha _1. \end{aligned}$$

(17)

Proof

First notice that by assumption $\sqrt{2} c= \gamma + F^{-1}(1 + \alpha _1 - \alpha _{w})$. Thus, $c \le \sqrt{2} \gamma$ $\Leftrightarrow$ $F^{-1}(1 + \alpha _1 - \alpha _{w}) \le \gamma$ $\Leftrightarrow$ $1 + \alpha _1 - \alpha _{w} \le 1- \alpha _1$ $\Leftrightarrow$ $2 \alpha _1 \le \alpha _{w}$.$\square$

1.2 Proof of Eq. (14)

In order to prove Eq. (14), we distinguish three cases: (i) $0 < \alpha _1 \le \frac{\alpha _{w}^2}{4}$, (ii) $\frac{\alpha _{w}^2}{4} \le \alpha _1 \le \frac{\alpha _{w}}{2}$, (iii) $\alpha _1 \ge \frac{\alpha _{w}}{2}$. In situation (i), $\alpha _1 \le \frac{\alpha _{w}}{2}$ since $0 < \alpha _{w} < 1$. We find from (13) that $c_{ BK}= \frac{\alpha _{w}^2}{4}$. So, $\alpha _1 \le c_{ BK}$ and $\alpha _{ind} = c_{ BK} - \log (c_{ BK}) c_{ BK} = \frac{\alpha _{w}^2}{4} \left( 1- \log \left( \frac{\alpha _{w}^2}{4}\right) \right)$ according to (12). In situation (ii), $c_{ BK}= \frac{\alpha _{w}^2}{4}$ by (13). This implies $\alpha _1 \ge c_{ BK}$ and consequently $\alpha _{ind} = \alpha _1 - \log (\alpha _1) \frac{\alpha _{w}^2}{4}$. Finally, in situation (iii), we find from (13) that $c_{ BK}= \alpha _1(\alpha _{w}-\alpha _1)$. Notice that we necessarily have $\alpha _{w} > \alpha _1$, since both $c_{ BK} > 0$ and $\alpha _1 > 0$. So, $0 < \alpha _1 < \alpha _{w} < 1$, which yields $c_{ BK}=\alpha _1(\alpha _{w}-\alpha _1) \le \alpha _1$. This yields $\alpha _{ind} = \alpha _1 - \log (\alpha _1) \alpha _1(\alpha _{w}-\alpha _1)$ and the proof is finished.

1.3 Proof of Eq. (15)

For the choice $\alpha _1 = c_{ BK}$ the assertion follows immediately from (14), since $\alpha _1 = c_{ BK} \ne 0$ implies $\alpha _1 = \frac{\alpha _{w}^2}{4}$. It remains to consider the case $\alpha _1 = \frac{\alpha _{ind}}{2}$ and to show that any solution of (14) for given $0 < \alpha _{w} < 1$ with $\alpha _{ind}=2 \alpha _1$ has the property $\frac{\alpha _{w}^2}{4} \le \alpha _1 \le \frac{\alpha _{w}}{2}$. So, let us first assume that we had a solution of (14) with $\alpha _1 \ge \frac{\alpha _{w}}{2}$ and $2 \alpha _1 = \alpha _{ind}$. Then $\alpha _{ind} = \alpha _1 - \log (\alpha _1) \alpha _1(\alpha _{w}-\alpha _1)$, and it follows that $-\log (\alpha _1) (\alpha _{w}- \alpha _1) = 1$. Since $-\log (x) < 1/x$ for all $x > 0$, we obtain $\alpha _{w}= \alpha _1 - \frac{1}{\log (\alpha _1)} > 2 \alpha _1$ in contradiction to $\alpha _{w} \le 2 \alpha _1$. If we assume a solution of (14) with $\alpha _1 \le \frac{\alpha _{w}^2}{4}$ and $2 \alpha _1 = \alpha _{ind}$, then $\alpha _1 = \frac{\alpha _{ind}}{2} = \frac{\alpha _{w}^2}{4} \frac{1- \log \left( \frac{\alpha _{w}^2}{4}\right) }{2}$. Thus, $-\log \left( \frac{\alpha _{w}^2}{4}\right) \le 1$, i.e. $\alpha _{w}^2 \ge \frac{4}{e}$ with $e$ denoting Euler’s number. This is in contradiction to $\alpha _{w} \in (0,1)$. It remains to show that there is an $\alpha _1$ with $\frac{\alpha _{w}^2}{4} \le \alpha _1 \le \frac{\alpha _{w}}{2}$ which is a solution of the equation $2 \alpha _1 = \alpha _{ind} = \alpha _1 - \log (\alpha _1) \frac{\alpha _{w}^2}{4}$. To prove this, let

$$\begin{aligned} \vartheta (\alpha _1) := \sqrt{\frac{4 \alpha _1}{- \log (\alpha _1)}}. \end{aligned}$$

(18)

We claim that $(0,1] \subset \vartheta ((0, \frac{1}{e}])$. This follows from the continuity of $\vartheta$, since $\vartheta (\frac{1}{e})= \frac{2}{\sqrt{e}} \ge 1$ and $\lim _{x \rightarrow 0} \frac{x}{- \log (x)} = 0$. Thus, for all $\alpha _{w} \in (0,1)$ there is a unique $\alpha _1 \in (0,\frac{1}{e})$ with $\vartheta (\alpha _1) = \alpha _{w}$. Uniqueness of $\alpha _1$ follows, because $\vartheta$ is strictly increasing on $(0,e)$. Therefore, let us choose $\alpha _1 \in (0, \frac{1}{e}]$ such that $\vartheta (\alpha _1) = \alpha _{w}$. Then $\alpha _1 \ge \frac{\alpha _{w}^2}{4}$, because $\alpha _1 \le \frac{1}{e} \Leftrightarrow -\frac{1}{\log (\alpha _1)} \le 1$ and thus $\alpha _{w}^2=\frac{4 \alpha _1}{ - \log (\alpha _1)} \le 4 \alpha _1$. Moreover, $\alpha _1 \le \frac{\alpha _{w}}{2}$, because $\alpha _1^2 \le \frac{\alpha _1}{-\log (\alpha _1)} = \frac{\alpha _{w}^2}{4}$ in view of $-\log (x) \le 1/x$ for all $x > 0$. Finally notice that $\vartheta (\alpha _1)= \alpha _{w} \Leftrightarrow \alpha _{ind}= \alpha _1 - \log (\alpha _1) \frac{\alpha _{w}^2}{4}$ if $\alpha _{ind}= 2 \alpha _1$. All in all we have proven the following result: Let $\alpha _{ind}=2 \alpha _1$. Then for all $\alpha _{w} \in (0,1)$ there is an unique $\alpha _1 \in (0,\frac{1}{e})$ such that $\alpha _{ind}= \alpha _1 - \log (\alpha _1) \frac{\alpha _{w}^2}{4}$ and $\frac{\alpha _{w}^2}{4} \le \alpha _1 \le \frac{\alpha _{w}}{2}$. So, if $\alpha _1= \alpha _{ind}/2$ and if the significance level $\alpha _{ind}$ is exactly preserved for independent and uniformly distributed $p$-values, then the type I error rate in the worst case is $\alpha _{w} = \sqrt{{2 \alpha _{ind}}/\left( {- \log \left( \frac{\alpha _{ind}}{2}\right) }\right) }$.

1.4 The choice $\alpha _1= \alpha _{ind}/n$ in the limit $n \rightarrow 1$

Let $\alpha _1, \alpha _{w}$ and $\alpha _{ind}$ be a solution of (14) with $0 < \alpha _1 < \alpha _{w} < 1$, and let $\alpha _1= \frac{\alpha _{ind}}{n}$ for some $n > 1$. We are going to prove that $\alpha _{ind} \rightarrow \alpha _{w}$ as $n \rightarrow 1$. To see this, notice that according to (12) and by definition of $\psi$ we have: $n \alpha _1 = \alpha _{ind} \ge \alpha _1 - \log (\psi ) c_{ BK}$, i.e. $- \log (\psi ) c_{ BK} \le (n-1) \alpha _1$. In the limit $n \rightarrow 1$, we thus obtain $- \log (\psi ) c_{ BK} \rightarrow 0$, since $0 \le \alpha _1, c_{ BK} \le 1$. So, we have $- \log (\psi ) \rightarrow 0$ or $c_{ BK} \rightarrow 0$. Now let $0 < \alpha _{ind} < 1$. We claim that $- \log (\psi ) \rightarrow 0$, i.e. $\psi \rightarrow 1$, is impossible. To see this, assume that $\psi =\max (\alpha _1,c_{ BK}) \rightarrow 1$ as $n \rightarrow 1$. We obtain $c_{ BK} \rightarrow 1$, because $\alpha _1$ converges by assumption to the fixed constant $\alpha _{ind} < 1$ as $n \rightarrow 1$. Now, notice that $\alpha _1(\alpha _{w}-\alpha _1) \le \frac{\alpha _{w}^2}{4}$, because the function $x \mapsto x (\alpha _{w}-x)$ has a global maximum in $x=\frac{\alpha _{w}}{2}$. Thus we conclude from (13) that $\frac{\alpha _{w}^2}{4} \ge c_{ BK} \rightarrow 1$ as $n \rightarrow 1$ in contradiction to $\alpha _{w} < 1$. Thus $c_{ BK} \rightarrow 0$. This implies together with (13) that $\alpha _{1}(\alpha _{w}-\alpha _1) \rightarrow 0$. Indeed, if $2 \alpha _1 > \alpha _{w}$, this is clear, and if $2 \alpha _1 \le \alpha _{w}$, we obtain $0 \le \alpha _1 (\alpha _{w} - \alpha _1) \le \frac{\alpha _{w}^2}{4} = c_{ BK} \rightarrow 0$. Since $\alpha _1 > 0$, we obtain $\alpha _{w} \rightarrow \alpha _1$ and thus $\alpha _{w} \rightarrow \alpha _{ind}$ as claimed, because $\alpha _1 \rightarrow \alpha _{ind}$ as $n \rightarrow 1$.

1.5 Proof of Eq. (8)

Consider the adaptive design defined by (5). Unless otherwise specified, let $z:= \sqrt{2} c$ and $\gamma :=F^{-1}(1-\alpha _1)$ below. In order to determine the type I error rate in the worst case for design (5), we see from (7) that it suffices to derive $\inf _C \mathbb {P}_{H_0}(Z_1 + Z_2 < z, Z_1 < \gamma )$, where the infimum is taken over all copulas $C$. For this purpose, we first assemble some essential mathematical results in Sect. 6.5.1 of Appendix. The strategy consists in deriving an upper and a lower bound for $\inf _C \mathbb {P}_{H_0}(Z_1 + Z_2 < z, Z_1 < \gamma )$ and in showing that both the upper and lower bound coincide. In Sect. 6.5.2 of Appendix, a lower bound is derived. A suitable upper bound is derived in Sect. 6.5.3 of Appendix and is shown to coincide with the lower bound mentioned before. On this basis, the type I error rate $\alpha _{w}$ for uniformly distributed but arbitrarily dependent $p$-values stated in Eq. (8) is finally derived in Sect. 6.5.4 of Appendix.

1.5.1 Bounds in the worst case scenario

Let $X$ and $Y$ be real-valued random variables. Then according to Schmidt et al. (2014), Corollary A.1, we have

$$\begin{aligned} \sup _{x+y=z, \ x < \gamma } W(\mathbb {P}(X < x),\mathbb {P}(Y < y)) \le \mathbb {P}_{}(X + Y < z, X < \gamma ). \end{aligned}$$

(19)

with $W(u,v):= \max \{ u+v-1,0 \}$ denoting Fréchet-Hoeffding’s lower bound. In particular, the left side of Eq. (19) is a lower bound for $\inf _C \mathbb {P}_{}^C(X + Y < z, X < \gamma )$, and thus yields an upper bound for the type I error rate in the worst case. Thereby recall that $\mathbb {P}^C(A)$ denotes the probability of any event $A \subset \mathbb {R}^2$ given that $C$ is the copula for $X$ and $Y$. To find an upper bound for $\inf _C \mathbb {P}_{}^C(X + Y < z, X < \gamma )$, we will make use of the following result. Let $\Delta$ denote the set of all functions $G:[- \infty , \infty ] \rightarrow [0,1]$ such that $G$ is left-continuous and non-decreasing on $\mathbb {R}$, $G(+\infty )=1$ and $G(- \infty )=0= \lim _{x \rightarrow -\infty } G(x)$. Then, according to Frank et al. (1987), Theorem 3.1 and Theorem 3.2, for any pair $X$, $Y$ of random variables such that $\mathbb {P}(X<x)$, $\mathbb {P}(Y<y) \in \Delta$, we have:

$$\begin{aligned} \inf _C \mathbb {P}^C(X+Y < z)&= \sup _{x+y=z} W(\mathbb {P}(X<x),\mathbb {P}(Y<y)). \end{aligned}$$

(20)

Notice that Eqs. (19) and (20) are not restricted to continuous random variables $X$ and $Y$, but actually hold for any pair $X$, $Y$ of random variables such that $\mathbb {P}(X<x)$, $\mathbb {P}(Y<y) \in \Delta$. Equation (20) will be used to derive the desired upper bound.

1.5.2 Derivation of the lower bound

Given the two-stage adaptive design (5) with stage-wise $p$-values $P_1$ and $P_2$, let $z:= \sqrt{2} c$ and $\gamma :=F^{-1}(1-\alpha _1)$. Moreover, let $Z_i:=F^{-1}(1-P_i)$, $i=1,2$. Since $F(0)=0$, $Z_i$ is a random variable with distribution function

$$\begin{aligned} F_{Z_i}(z_i):= \mathbb {P}(Z_i \le z_i) = {\left\{ \begin{array}{ll} 0, \quad &{} z_i \le 0, \\ F(z_i), &{} z_i \ge 0, \end{array}\right. } \end{aligned}$$

(21)

under $H_0$. In particular, $Z_i$ has a decreasing density function almost everywhere. The lower bound (19) for $\inf _C \mathbb {P}_{}^C(Z_1 + Z_2 < z, Z_1 < \gamma )$ is given by the following theorem.

Theorem 1

Let $\gamma , z \ge 0$, and let $Z_1$, $Z_2$ be two random variables with distribution functions $F_{Z_1}, F_{Z_2}$ as stated in (21). Then

$$\begin{aligned} \sup _{x+y=z, x < \gamma } W(F_{Z_1}(x),F_{Z_2}(y)) = {\left\{ \begin{array}{ll} \max \{ 2 F\left( \frac{z}{2}\right) -1 , 0 \}, &{} \text { if } \quad z \le 2 \gamma , \\ \max \{ F(\gamma ) + F(z - \gamma ) -1 , 0 \}, &{} \text { if } \quad z \ge 2 \gamma , \\ \end{array}\right. } \end{aligned}$$

(22)

where $W(u,v):= \max \{ u+v-1,0 \}$ denotes the Fréchet–Hoeffding lower bound.

Proof

Let $\phi :=\sup _{x < \gamma } \left( F_{Z_1}(x)+ F_{Z_2}(z-x) \right)$. It suffices to show that $\phi = \max \{2 F\left( \frac{z}{2}\right) ,1\}$ if $z \le 2 \gamma$ and $\phi = \max \{ F(\gamma ) + F(z - \gamma ), 1 \}$ for $z \ge 2 \gamma$. By (21), $F_{Z_1}(x)=0$ for $x \le 0$ and $F_{Z_1}(x)=F(x)$ for $x \ge 0$. Similarly, $F_{Z_2}(z-x)=0$ for $x \ge z$ and $F_{Z_2}(z-x)=F(z-x)$ for $x \le z$. We now distinguish two cases: (i) $z \le 2 \gamma$, (ii) $z \ge 2 \gamma$.

• (i) If $z \le 2 \gamma$, we find that

  $$\begin{aligned} F_{Z_1}(x)+F_{Z_2}(z-x) = {\left\{ \begin{array}{ll} F(z-x), &{} \text { if } \quad x \le 0 \le z \le 2 \gamma , \\ F(x) + F(z-x), &{} \text { if } \quad 0 \le x \le z \le 2 \gamma , \\ F(x), &{} \text { if } \quad 0 \le z \le x \le 2 \gamma . \end{array}\right. } \end{aligned}$$

  (23)

  Consequently, $\phi$ is the maximum of $\sup _{x \le 0} F(z-x) = 1$, $\sup _{0 \le x < \min (z, \gamma )} ( F(x) + F(z-x) )$ and $\sup _{\min (z, \gamma ) \le x < \gamma } F(x)$. If the interval $[\min (z, \gamma ), \gamma )$ is non-empty, then $\sup _{\min (z, \gamma ) \le x < \gamma } F(x) = F(\gamma ) \le 1$. According to Lemma 3, the supremum of $F(x)+ F(z-x)$ over $x$ is taken in $x_{max}=\frac{z}{2} \in [0, \min (z, \gamma )]$. So, $\phi = \max \{2 F\left( \frac{z}{2}\right) ,1 \}$.

• (ii) If $z \ge 2 \gamma$, we find that

  $$\begin{aligned} F_{Z_1}(x)+F_{Z_2}(z-x) = {\left\{ \begin{array}{ll} F(z-x), &{} \text { if } \quad x \le 0 \le 2 \gamma \le z, \\ F(x) + F(z-x), &{} \text { if } \quad 0 \le x \le 2 \gamma \le z, \\ F(x) + F(z-x), &{} \text { if } \quad 0 \le 2 \gamma \le x \le z, \\ F(x), &{} \text { if } \quad 0 \le 2 \gamma \le z \le x. \end{array}\right. } \end{aligned}$$

  (24)

  Consequently, $\phi$ is the maximum of $\sup _{x \le 0} F(z-x) = 1$ and $\sup _{0 \le x < \gamma } (F(x) + F(z-x))$. The maximum of $F(x)+F(z-x)$ is located at $\frac{z}{2}$ and $F(x)+F(z-x)$ is increasing on $(-\infty , \frac{z}{2})$. By assumption on $z$, we have $\frac{z}{2} \ge \gamma$. Thus $\sup _{0 \le x < \gamma } (F(x)+F(z-x)) = F(\gamma )+F(z- \gamma )$, i.e. $\phi$ is the maximum of $1$ and $F(\gamma )+F(z- \gamma )$. $\square$

Lemma 3

Let $z \ge 0$. Let $F:\mathbb {R} \rightarrow \mathbb {R}$ be a twice continuously differentiable function with $F''(x)<0$ for all $x \in \mathbb {R}$. Then $F(x) + F(z-x)$ has a unique maximum in $x=\frac{z}{2}$.

Proof

Let $\varPsi (x):=F(x) + F(z-x)$, let $f(x):=F'(x)$. Then $\varPsi '(x)=f(x)-f(z-x)$. $\varPsi '(x)=0$ implies $f(x)=f(z-x)$, which yields a unique extremum in $x=\frac{z}{2}$ by injectivity of $f$. Since, $f'(x) < 0$ for all $x \in \mathbb {R}$, $x=\frac{z}{2}$ is a maximum. $\square$

Lemma 4

Let $z, \gamma \ge 0$. Let $F:\mathbb {R} \rightarrow \mathbb {R}$ be a twice continuously differentiable function with $F''(x)<0$ for all $x \in \mathbb {R}$. Then $2 F\left( \frac{z}{2}\right) \ge F(\gamma ) + F(z-\gamma )$.

Proof

Let $\varPsi _{\gamma }(z):= F(\gamma ) + F(z-\gamma ) - 2 F\left( \frac{z}{2}\right)$. Let $f(z):=F'(z)$ with prime denoting partial differentiation with respect to $z$. Then $\varPsi '_{\gamma }(z) = f(z-\gamma ) - f\left( \frac{z}{2}\right)$. Since $f$ is strictly decreasing, $\varPsi '_{\gamma }(z) = 0$ implies $z=2 \gamma$. Moreover, $\varPsi ''_{\gamma }(z) = f'(z-\gamma ) - \frac{1}{2} f'\left( \frac{z}{2}\right)$, i.e. $\varPsi ''_{\gamma }(2\gamma ) = \frac{1}{2} f'(\gamma ) < 0$. All in all, $\varPsi _{\gamma }$ has a unique extremum at $z= 2\gamma$, which is a maximum. At the maximum, $\varPsi _{\gamma }(2\gamma ) = 0$. $\square$

1.5.3 Derivation of the upper bound

Given the two-stage adaptive design (5), let $z:= \sqrt{2} c$ and $\gamma :=F^{-1}(1-\alpha _1)$. For $r \ge 1$, let $f_r:\mathbb {R} \rightarrow \mathbb {R}$ denote the continuous, decreasing function

$$\begin{aligned} f_r(z_1):= {\left\{ \begin{array}{ll} -z_1, &{} \text { if } \quad z_1 \le \gamma , \\ -r(z_1 - \gamma ) -\gamma , &{} \text { if } \quad z_1 \ge \gamma . \end{array}\right. } \end{aligned}$$

(25)

For $z_1 \in (-\infty ,\gamma ]$, $f_r$ is a straight line through the origin with slope $-1$. At $z_1 = \gamma$, the slope switches to $-r \le -1$. Notice that $\{ z_1 + z_2 < z, z_1 < \gamma \} \subset \{z_2-f_r(z_1) < z \} \subset \mathbb {R}^2$ for all $r \ge 1$ (see Fig. 2). In particular $\inf _C \mathbb {P}_{H_0}^{\, C}(Z_1 + Z_2 < z, Z_1 < \gamma ) \le \inf _C \mathbb {P}_{H_0}^{\, C}(X + Y < z)$ for all $r \ge 1$ with $X:=-f_r(Z_1)$ and $Y:=Z_2$. The left side of the latter inequality does not depend on $r$. Consequently, we conclude

$$\begin{aligned} \inf _C \mathbb {P}_{H_0}^{\, C}(Z_1 + Z_2 < z, Z_1 < \gamma ) \le \lim _{r \rightarrow \infty } \inf _C \mathbb {P}_{H_0}^{\, C}(X + Y < z). \end{aligned}$$

(26)

Fig. 2

Probability of the events $\{ Z_1 + Z_2 < z, Z_1 < \gamma \} \subseteq \{ Z_2 < f_r(Z_1) + z \}$ for $z > \gamma$. $\{ Z_1 + Z_2 < z, Z_1 < \gamma \}$ is highlighted in dark grey and $\{ Z_2 < f_r(Z_1) + z \}$ \$\{ Z_1 + Z_2 < z, Z_1 < \gamma \}$ is highlighted in grey

We derive the right side of Eq. (26) as follows. According to (20),

$$\begin{aligned} \inf _C \mathbb {P}_{H_0}^C(X+Y < z)&= \sup _{x+y=z} \max \{ F_X(x)+F_Y(y)-1,0 \} \nonumber \\&= \max \{ \sup _{x+y=z}(F_X(x)+F_Y(y))-1,0 \} \nonumber \\&= \max \{ \sup _{x \in \mathbb {R}}(F_X(x)+F_Y(z-x))-1,0 \}. \end{aligned}$$

(27)

By (21), $F_Y(y)=0$ for $y \le 0$, and $F_Y(y)=F(y)$ for $y \ge 0$. The distribution function $F_X(x)$ of $X$ is given as follows:

Lemma 5

Let $Z$ be a random variable with distribution function $F_Z$ of type (21), i.e. $F_Z(z)=0$ for $z \le 0$ and $F_Z(z)=F(z)$ for all $z \ge 0$. Let $X:=-f_r(Z)$ for some $r \ge 1$ with $f_r$ as defined in (25) with $\gamma \ge 0$. Then

$$\begin{aligned} F_X(x):=\mathbb {P}(X \le x)= {\left\{ \begin{array}{ll} 0, &{} \text { if } \quad x \le 0, \\ F(x), &{} \text { if } \quad 0 \le x \le \gamma ,\\ F\left( \frac{x+(r-1)\gamma }{r}\right) , &{} \text { if } \quad x \ge \gamma . \end{array}\right. } \end{aligned}$$

(28)

Proof

Let $h_r(x):=-f_r(x)$. We claim that the inverse of $h_r$ is given by

$$\begin{aligned} g_r(y):= {\left\{ \begin{array}{ll} y, &{} \text { if } \quad y \le \gamma ,\\ \frac{y - \gamma }{r} + \gamma , &{} \text { if } \quad y \ge \gamma . \end{array}\right. } \end{aligned}$$

(29)

Indeed, let $x \le \gamma$. Then $g_r(h_r(x)))=g_r(x)=x$. Now let $x \ge \gamma$. Then $g_r(h_r(x))) = g_r(r(x-\gamma )+\gamma ) = \frac{\left( r(x-\gamma )+\gamma \right) - \gamma }{r} + \gamma =x$, since $r(x-\gamma )+\gamma \ge \gamma$, i.e. $g_r \circ h_r =$ id. Similarly, $h_r \circ g_r =$ id: For $y \le \gamma$, $h_r(g_r(y)))=h_r(y)=y$. For $y \ge \gamma$, $h_r(g_r(y))) = h_r\left( \frac{y - \gamma }{r} + \gamma \right) = r\left( \left( \frac{y - \gamma }{r} + \gamma \right) -\gamma \right) +\gamma =y$, because $\frac{y - \gamma }{r} + \gamma \ge \gamma$. Since $h_r$ is monotonically increasing, $\mathbb {P}(-f_r(Z) \le x) = \mathbb {P}(Z \le g_r(x))$. By (29), we are finished. $\square$

To maximize $F_X(x)+F_Y(z-x)$, we consider three scenarios: (i) $z \le \gamma$, (ii) $\gamma \le z \le 2 \gamma$, (iii) $z \ge 2 \gamma$.

• (i) If $z \le \gamma$, we find that

  $$\begin{aligned} F_X(x)+F_Y(z-x)= {\left\{ \begin{array}{ll} F(z-x) , &{} \text { if } \quad x \le 0 \le z \le \gamma ,\\ F(x) + F(z-x), &{} \text { if } \quad 0 \le x \le z \le \gamma ,\\ F(x), &{} \text { if } \quad 0 \le z \le x \le \gamma ,\\ F\left( \frac{x+(r-1)\gamma }{r} \right) , &{} \text { if } \quad 0 \le z \le \gamma \le x. \end{array}\right. } \end{aligned}$$

  (30)

  Thus $\psi := \sup _{x+y=z} (F_X(x)+F_Y(y)) = \sup _{x \in \mathbb {R}} (F_X(x)+F_Y(z-x))$ is the maximum of $\sup _{x \le 0} F(z-x) = 1$, $\sup _{0 \le x \le z} (F(x)+F(z-x))$, $\sup _{z \le x \le \gamma } F(x) = F(\gamma ) \le 1$ and $\sup _{\gamma \le x} F \left( \frac{x+(r-1)\gamma }{r} \right) = 1$. According to Lemma 3, the maximum of $F(x)+F(z-x)$ is taken at $x=\frac{z}{2}$. Thus we find that $\psi = \max \{ 1 , 2F \left( \frac{z}{2} \right) \}$.

• (ii) If $\gamma \le z \le 2 \gamma$,

  $$\begin{aligned} F_X(x)+F_Y(z-x)= {\left\{ \begin{array}{ll} F(z-x), &{} \text { if } \quad x \le 0 \le \gamma \le z \le 2\gamma ,\\ F(x) + F(z-x), &{} \text { if } \quad 0 \le x \le \gamma \le z \le 2\gamma ,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x), &{} \text { if } \quad 0 \le \gamma \le x \le z \le 2\gamma ,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) , &{} \text { if } \quad 0 \le \gamma \le z \le x \le 2\gamma ,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) , &{} \text { if } \quad 0 \le \gamma \le z \le 2\gamma \le x. \end{array}\right. } \end{aligned}$$

  (31)

  Thus $\psi = \sup _{x \in \mathbb {R}} (F_X(x)+F_Y(z-x))$ is the maximum of $\sup _{x \le 0} F(z-x) = 1$, $\sup _{0 \le x \le \gamma } (F(x)+F(z-x))$, $\sup _{\gamma \le x \le z} \left( F\left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \right)$ and $\sup _{z \le x} F \left( \frac{x+(r-1)\gamma }{r} \right) = 1$. The maximum of $F(x)+F(z-x)$ is taken at $\frac{z}{2}$. By assumption, $z \le 2 \gamma$. Thus $\sup _{0 \le x \le \gamma } (F(x)+F(z-x)) = 2F \left( \frac{z}{2} \right)$. All in all, $\psi$ is the maximum of $1$, $2F \left( \frac{z}{2} \right)$ and $\sup _{\gamma \le x \le z} \left( F\left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \right)$.

• (iii) If $z \ge 2 \gamma$,

  $$\begin{aligned} F_X(x)+F_Y(z-x)= {\left\{ \begin{array}{ll} F(z-x), &{} \text { if } \quad x \le 0 \le \gamma \le 2 \gamma \le z,\\ F(x) + F(z-x), &{} \text { if } \quad 0 \le x \le \gamma \le 2\gamma \le z,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x), &{} \text { if } \quad 0 \le \gamma \le x \le 2 \gamma \le z,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x), &{} \text { if } \quad 0 \le \gamma \le 2\gamma \le x \le z,\\ F \left( \frac{x+(r-1)\gamma }{r} \right) , &{} \text { if } \quad 0 \le \gamma \le 2\gamma \le z \le x. \end{array}\right. } \end{aligned}$$

  (32)

  Thus $\psi = \sup _{x \in \mathbb {R}} (F_X(x)+F_Y(z-x))$ is the maximum of $\sup _{x \le 0} F(z-x) = 1$, $\sup _{0 \le x \le \gamma } (F(x)+F(z-x))$, $\sup _{\gamma \le x \le z} \left( F\left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x)\right)$ and $\sup _{z \le x} F \left( \frac{x+(r-1)\gamma }{r} \right) = 1$. The maximum of $F(x)+F(z-x)$ is taken at $\frac{z}{2}$ and $F(x)+F(z-x)$ is increasing on $(-\infty , \frac{z}{2})$. By assumption on $z$, we have $\frac{z}{2} \ge \gamma$. Thus $\sup _{0 \le x \le \gamma } (F(x)+F(z-x)) = F(\gamma )+F(z- \gamma )$. All in all, $\psi$ is the maximum of $1$, $F(\gamma )+F(z- \gamma )$ and $\sup _{\gamma \le x \le z} \left( F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \right)$.

So, it remains to determine $\sup _{\gamma \le x \le z} \left( F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \right)$. For this purpose, we study the extrema of the function

$$\begin{aligned} \varPsi _z(x):= F\left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x). \end{aligned}$$

(33)

We claim that $\varPsi _z(x)$ has a unique extremum in $x=x_e$, which is actually a maximum. The condition for an extremum $x_{e}$ of (33) is $G:=\frac{\partial }{\partial x} \varPsi _z(x)=0$, i.e.:

$$\begin{aligned} G = f\left( \frac{x_{e}+(r-1)\gamma }{r}\right) \frac{1}{r} - f(z-x_{e}) =0, \end{aligned}$$

(34)

with $f(x):=F'(x)=\frac{\partial F(x)}{\partial x}$. To prove the existence of an extremum, we have to show that for all $z, \gamma \ge 0$, $r \ge 1$ there is an $x_e \in \mathbb {R}$ such that (34) holds. To see this, notice that the maximum $x_{e}$ is connected with $z$ as follows:

$$\begin{aligned} z(x_e)=x_e + f^{-1}\left( \frac{1}{r} f \left( \frac{x_e + (r-1) \gamma }{r} \right) \right) . \end{aligned}$$

(35)

By assumption on $F$, $\lim _{y \rightarrow 0 +} f^{-1}(y) = \infty$ and $\lim _{y \rightarrow \infty } f^{-1}(y) = -\infty$. Since $\lim _{x_e \rightarrow \infty } \frac{1}{r} f \left( \frac{x_e + (r-1) \gamma }{r} \right) = 0$ and $\lim _{x_e \rightarrow -\infty } \frac{1}{r} f \left( \frac{x_e + (r-1) \gamma }{r} \right) = \infty$, we find that $\lim _{x_e \rightarrow \pm \infty } z(x_e) = \pm \infty$. Thus, by continuity of $z(x_e)$ in $x_e$ there exists an $x_e$ such that $z(x_e)=z$ for any given $z, \gamma$ and $r \ge 1$. To prove uniqueness, differentiate the function $F(ax+b) + F(z-x)$, $a, b \in \mathbb {R}$, twice with respect to $x$. This yields $a^2 f'(a x+b)+f'(z-x) < 0$ for all $x \in \mathbb {R}$. In particular, each extremum is a maximum. Thus, since $F$ is continuously differentiable, there can be at most one $x \in \mathbb {R}$ which maximizes $F(ax+b) + F(z-x)$.

We now study the location of the unique maximum $x_{e}$ of $F\left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x)$ in dependence of $z$ and compare it to the cutoff value $\gamma$, where the arm of the function $F_X(x)+F_Y(z-x)$ switches. Recall that $z, \gamma \ge 0$ and $r \ge 1$. By (35), the maximum $x_{e}$ depends on $z$, i.e. $x_{e}=x_{e}(z)$. We claim that $x_{e}(z)$ is a strictly increasing function in $z$. This follows from implicit differentiation:

$$\begin{aligned} \frac{\partial x_{e}(z)}{\partial z} = - \frac{ \frac{\partial G}{\partial z} }{ \frac{\partial G}{\partial x_{e}} } = - \frac{ -f'(z-x_{e}) }{ f'\left( \frac{x_{e}+(r-1)\gamma }{r} \right) \frac{1}{r^2} + f'(z-x_{e}) } > 0, \end{aligned}$$

(36)

because $f'(x)<0$ for all $x \in \mathbb {R}$. We claim that $x_e \le \gamma$ if and only if $z \le \gamma + f^{-1}(f(\gamma )/r)$. To see this, we first determine the value $z_{\gamma }$ for $z$ such that $x_{e}(z_{\gamma })=\gamma$. From (35), we find with $x_e=x_{e}(z_\gamma )=\gamma$ that $z_{\gamma } = \gamma + f^{-1}(f(\gamma )/r)$. Now, according to (36) $x_{e}(z)$ is an increasing function in $z$. In particular, $x_{e} \le \gamma$ if and only if $z \le \gamma + f^{-1}(f(\gamma )/r)$. All in all, we may summarize as follows:

Proposition 1

Let $z, \gamma \ge 0$ and $r \ge 1$. Let $x_{e}$ denote the unique maximum of $F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x)$. Then the following statements hold:

• $x_{e} \le \gamma$ $\Leftrightarrow$ $z \le \gamma + f^{-1}(f(\gamma )/r)$

• $F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x)$ is strictly increasing on $(-\infty , x_{e})$ and strictly decreasing on $(x_{e},\infty )$.

• As $x \rightarrow \pm \infty$, $F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \rightarrow - \infty$.

Now we are in a position to determine $\sup _{\gamma \le x \le z} \left( F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x) \right)$.

Proposition 2

Let $z, \gamma \ge 0$. Then there exists an $r_0 \ge 1$ such that for all $r \ge r_0$

$$\begin{aligned} \sup _{\gamma \le x \le z} \Big (F\left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x)\Big ) = F(\gamma ) + F(z- \gamma ). \end{aligned}$$

(37)

Proof

According to the characterization given in Proposition 1, the function $F \left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x)$ has an unique maximum $x_e$ and is decreasing on $[x_e, \infty )$. Thus it suffices to show that there is an $r_0$ such that $x_e \le \gamma$ for all $r \ge r_0$, which by Proposition 1 is equivalent to showing that $z \le \gamma + f^{-1}(f(\gamma )/r)$ for all $r \ge r_0$. Thus, it suffices to show that $\psi (r):=f^{-1}(f(\gamma )/r)$ is strictly increasing in $r$ with $\lim _{r \rightarrow \infty } \psi (r) = \infty$. Now, this holds true, because $g(r):=f(\gamma )/r$ is strictly decreasing in $r$ with $\lim _{r \rightarrow \infty } g(r)=0$ since $f(\gamma )>0$, and because $f^{-1}(y)$ is strictly decreasing in $y$ with $\lim _{y \rightarrow 0 +} f^{-1}(y)= \infty$. $\square$

Theorem 2

Let $z, \gamma \ge 0$, and let $X:=-f_r(Z_1)$ and $Y:=Z_2$ with $f_r$ defined in (25). Then there is an $r_0$ such that for all $r \ge r_0$:

$$\begin{aligned} \sup _{x \in \mathbb {R}} \left( F_X(x)+F_Y(z-x) \right) = {\left\{ \begin{array}{ll} \max \left\{ 2 F\left( \frac{z}{2} \right) , 1 \right\} , &{} \text { if } \quad z \le 2 \gamma ,\\ \max \left\{ F(\gamma ) + F(z-\gamma ) , 1 \right\} , &{} \text { if } \quad 2 \gamma \le z. \end{array}\right. } \end{aligned}$$

(38)

Proof

Let $\psi := \sup _{x \in \mathbb {R}} (F_X(x)+F_Y(z-x))$. If $z \le \gamma$, we showed immediately after Eq. (30) that $\psi = \max \{ 1 , 2F\left( \frac{z}{2} \right) \}$. If $\gamma \le z \le 2 \gamma$, we showed immediately after Eq. (31) that $\psi$ is the maximum of $1$, $2F\left( \frac{z}{2} \right)$ and $\sup _{\gamma \le x \le z} \left( F \left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x) \right)$. According to Proposition 2, there is an $r_0$ such that the latter is equal to $F(\gamma ) + F(z - \gamma )$ for all $r \ge r_0$. So choose $r \ge r_0$. Then, in view of Lemma 4, $\psi$ is the maximum of $1$ and $2F\left( \frac{z}{2}\right)$ if $\gamma \le z \le 2 \gamma$. Finally assume that $z \ge 2 \gamma$. Then we showed immediately after Eq. (32) that $\psi$ is the maximum of $1$, $F(\gamma )+F(z- \gamma )$ and $\sup _{\gamma \le x \le z} \left( F\left( \frac{x+(r-1)\gamma }{r} \right) + F(z-x)\right)$. Again, by Proposition 2, there is an $r_0$ such that $\sup _{\gamma \le x \le z} \left( F \left( \frac{x+(r-1)\gamma }{r}\right) + F(z-x) \right)$ is equal to $F(\gamma ) + F(z - \gamma )$ for all $r \ge r_0$. So, $\psi$ is the maximum of $1$ and $F(\gamma )+F(z- \gamma )$ in this situation. $\square$

Thus, for any fixed $z, \gamma \ge 0$, Theorem 2 yields in view of Eq. (27)

$$\begin{aligned} \lim _{r \rightarrow \infty } \inf _C \mathbb {P}_{H_0}^C(X + Y < z) = {\left\{ \begin{array}{ll} \max \left\{ 2 F\left( \frac{z}{2} \right) -1,0 \right\} , &{} \text { if } \quad z \le 2 \gamma ,\\ \max \left\{ F(\gamma ) + F(z-\gamma ) -1 , 0 \right\} , &{} \text { if } \quad 2 \gamma \le z. \end{array}\right. } \end{aligned}$$

(39)

1.5.4 Worst case designs without futility stop

Let $0 < \alpha _1 < \alpha _{w} < 1$, and let $c$ denote the design parameter of a two-stage adaptive design with conditional error function given by (5). The type I error rate $\alpha _{w}$ for uniformly distributed but arbitrarily dependent $p$-values is $\alpha _{w} = 1- \inf _C \mathbb {P}_{H_0}^C(Z_1 + Z_2 < z, Z_1 < \gamma )$ with $z = \sqrt{2} c$, $\gamma = F^{-1}(1- \alpha _1)$, $Z_i:=F^{-1}(1-P_i)$, and with $P_i$ denoting the $p$ value of stage $i$, $i=1,2$ (see also Eq. (7)). According to Eqs. (26) and (19), we have

$$\begin{aligned} \sup _{x+y=z, \ x < \gamma } W(F_{Z_1}(x),F_{Z_2}(y))&\le \inf _C \mathbb {P}_{H_0}^C(Z_1 + Z_2 < z, - \infty \le Z_1 < \gamma ) \nonumber \\&\le \lim _{r \rightarrow \infty } \inf _C \mathbb {P}_{H_0}^C(X + Y < z), \end{aligned}$$

(40)

under $H_0$ with $X:=-f_r(Z_1)$, $f_r$ as defined in (25) and $Y:=Z_2$. Thus it follows from (39) and (22) that

$$\begin{aligned} \inf _C \mathbb {P}_{H_0}^C(Z_1 + Z_2 < z, Z_1 < \gamma ) = {\left\{ \begin{array}{ll} \max \left\{ 2 F \left( \frac{z}{2} \right) -1 , 0 \right\} , &{} \text { if } \quad z \le 2 \gamma ,\\ \max \left\{ F(\gamma ) + F(z-\gamma ) -1, 0 \right\} , &{} \text { if } \quad 2 \gamma \le z. \end{array}\right. } \end{aligned}$$

(41)

Thus, in view of (7), we have

$$\begin{aligned} \alpha _{w}&= {\left\{ \begin{array}{ll} \min \left\{ 2 \left( 1- F \left( \frac{z}{2} \right) \right) , 1 \right\} , &{} \text { if } \quad z \le 2 \gamma ,\\ \min \left\{ 2- F(\gamma ) - F( z - \gamma ) , 1 \right\} , &{} \text { if } \quad 2 \gamma \le z. \end{array}\right. } \end{aligned}$$

(42)

Since $0< \alpha _{w} < 1$ in situations of practical relevance, we restrict attention to those choices for $z$ and $\gamma$ which are admissible in the sense that $0 < 2 \left( 1- F\left( \frac{z}{2}\right) \right) < 1$ if $z \le 2 \gamma$, and $0 < 2- F(\gamma ) - F( z - \gamma ) < 1$ if $z > 2 \gamma$. Other choices for $z$ and $\gamma$ are irrelevant in practical situations. For such admissible $z$ and $\gamma$, we find that

$$\begin{aligned} \alpha _{w}&= {\left\{ \begin{array}{ll} 2 \left( 1- F\left( \frac{z}{2}\right) \right) , &{} \text { if } \quad z \le 2 \gamma ,\\ 2- F(\gamma ) - F( z - \gamma ), &{} \text { if } \quad 2 \gamma \le z. \end{array}\right. } \end{aligned}$$

(43)