A New Overall-Subgroup Simultaneous Test for Optimal Inference in Biomarker-Targeted Confirmatory Trials

Abstract

We propose a joint hypothesis test for simultaneous confirmatory inference in the overall population and a pre-defined marker-positive subgroup under the assumption that the treatment effect in the marker-positive subgroup is larger than that in the overall population. The proposed confirmatory overall-subgroup simultaneous test (COSST) is based on partitioning the sample space of the test statistics in the marker-positive and marker-negative subgroups. We define two rejection regions in the joint sample space of the two test statistics: (1) efficacy in the marker-positive subgroup only; (2) efficacy in the overall population. COSST achieves higher statistical power to detect the overall and subgroup efficacy than most sequential procedures while controlling the family-wise type I error rate. COSST also takes into account the potentially harmful effect in the subgroups in the decision. The optimal rejection regions depend on the specific alternative hypothesis and the sample size. COSST can be useful for Phase III clinical trials with tailoring objectives.

Appendix

Let $\mu _{T+} ,\mu _{C+} ,\mu _{T-} ,\mu _{C-}$ and $\sigma _{T+}^2 ,\sigma _{C+}^2 ,\sigma _{T-}^2 ,\sigma _{C-}^2$be the means and variances in the populations defined by their subscripts, respectively; $\bar{X}_{T+}$, $\bar{X}_{C+}$, $\bar{X}_{T-}$, and $\bar{X}_{C-}$ and $s_{T+}^2$, $s_{T-}^2$,$s_{C+}^2$ and $s_{C-}^2$be the corresponding sample means and variances, and $N_{T+}$,$N_{T-}$, $N_{C+}$, $N_{C-}$ be the corresponding sample sizes. Let $p_+$be the biomarker prevalence rate in the study, such that

$$\begin{aligned} p_+ =\frac{N_+ }{N}=\frac{N_{T+} }{N_T }=\frac{N_{C+} }{N_C } \end{aligned}$$

and

$$\begin{aligned} p_- =\frac{N_- }{N}=\frac{N_{T-} }{N_T }=\frac{N_{C-} }{N_C }=1-p_+ , \end{aligned}$$

where $N_T =N_{T+} +N_{T-}$ and $N_C =N_{C+} +N_{C-}$ are the sample sizes in the treatment and control arms, respectively, and $N_+ =N_{T+} +N_{C+}$ and $N_- =N_{T-} +N_{C-}$ are the numbers of the marker-positive and negative subjects, respectively. Let $N=N_+ +N_- =N_T +N_C$ be the total sample size (Table 4).

Table 4 Means and variances by subgroup

1.1 Randomization ratio

Randomization ensures that the proportion of subjects randomized to treatment are the same in each of the subgroups as well as in the overall sample, i.e., $p_T =\frac{N_T }{N}=\frac{N_{T+} }{N_+ }=\frac{N_{T-} }{N_- }$ and $p_C =\frac{N_C }{N}=\frac{N_{C+} }{N_+ }=\frac{N_{C-} }{N_- }=1-p_T$, because the assignment is independent of subgroup status.

Let $R=\frac{p_T }{p_C }=\frac{N_T }{N_C }=\frac{N_{T+} }{N_{C+} }=\frac{N_{T-} }{N_{C-} }$ be the randomization ratio. Note that

$$\begin{aligned} N_T =\frac{NR}{1+R}\,\mathrm{and}N_C =\frac{N}{1+R}. \end{aligned}$$

1.2 Relationships between test statistics and population parameters

Let $S_+^2 =V\left( {\bar{X}_{T+} -\bar{X}_{C+} } \right) =\frac{\sigma _{T+}^2 }{N_{T+} }+\frac{\sigma _{C+}^2 }{N_{C+} }$ and $S_-^2 =V(\bar{X}_{T-} -\bar{X}_{C-} )=\frac{\sigma _{T-}^2 }{N_{T-} }+\frac{\sigma _{C-}^2 }{N_{C-} }$.

Then $S^{2}=V\left( {\bar{X}_T -\bar{X}_C } \right) =V\left( {p_+ \bar{X}_+ +p_- \bar{X}_- } \right) =p_+^2 S_+^2 +p_-^2 S_-^2$ (Lemma 1). Furthermore,

$Z_+ =\frac{\bar{X}_{T+} -\bar{X}_{C+} }{S_+ }\sim N\left( {E\left( {Z_+ } \right) ,1} \right)$ and $E\left( {Z_+ } \right) =\frac{\mu _+ }{S_+ }$,

$Z_- =\frac{\bar{X}_{T-} -\bar{X}_{C-} }{S_- }\sim N\left( {E\left( {Z_- } \right) ,1} \right)$ and $E\left( {Z_- } \right) =\frac{\mu _- }{S_- }$

$Z=\frac{\bar{X}_T -\bar{X}_C }{S}=p_+ \frac{S_+ }{S}Z_+ +p_- \frac{S_- }{S}Z_- =\sqrt{K_+ }Z_+ +\sqrt{K_- }Z_-$ (Lemma 2).

$\mu =p_+ \mu _+ +p_- \mu _-$ (Lemma 4)

Lemma 1

The following relationship holds:

$$\begin{aligned} S^{2}=p_+^2 S_+^2 +p_-^2 S_-^2 , \end{aligned}$$

where $S^{2}=V\left( {\bar{X}_T -\bar{X}_C } \right) ,S_+^2 =V\left( {\bar{X}_{T+} -\bar{X}_{C+} } \right)$ and $S_-^2 =V\left( {\bar{X}_{T-} -\bar{X}_{C-} } \right) .$

Proof

$$\begin{aligned}&\bar{X}_T -\bar{X}_C =\frac{N_{T+} \bar{X}_{T+} +N_{T-} \bar{X}_{T-} }{N_T }-\frac{N_{C+} \bar{X}_{C+} +N_{C-} \bar{X}_{C-} }{N_C }\\&\qquad \qquad \quad =\frac{N_{T+} }{N_T }\bar{X}_{T+} +\frac{N_{T-} }{N_T }\bar{X}_{T-} -\frac{N_{C+} }{N_C }\bar{X}_{C+} -\frac{N_{C-} }{N_C }\bar{X}_{C-}\\&\qquad \quad \qquad =p_+ \left( {\bar{X}_{T+} -\bar{X}_{C+} } \right) +p_- \left( {\bar{X}_{T-} -\bar{X}_{C-} } \right) \end{aligned}$$

$$\begin{aligned}&S^{2}=Var(\bar{X}_T -\bar{X}_C )=Var\left( {p_+ \left( {\bar{X}_{T+} -\bar{X}_{C+} } \right) +p_- \left( {\bar{X}_{T-} -\bar{X}_{C-} } \right) } \right) \\&\qquad =p_+^2 Var\left( {\bar{X}_{T+} -\bar{X}_{C+} } \right) +p_-^2 Var\left( {\bar{X}_{T-} -\bar{X}_{C-} } \right) =p_+^2 S_+^2 +p_-^2 S_-^2 . \end{aligned}$$

$\square$

Lemma 2

The overall study sample test statistic Z can be expressed in terms of the marker-positive and negative test statistics, $Z_+$ and $Z_-$, as follows:

$$\begin{aligned} Z=\sqrt{K_+ }Z_+ +\sqrt{K_- }Z_- , \end{aligned}$$

where $K_+ =p_+^2 S_+^2 /S^{2}$ and $K_- =p_-^2 S_-^2 /S^{2}$.

Proof

$$\begin{aligned} Z= & {} \frac{\bar{X}_T -\bar{X}_C }{\sqrt{Var\left( {\bar{X}_T -\bar{X}_C } \right) }}=\frac{p_+ \left( {\bar{X}_{T+} -\bar{X}_{C+} } \right) +p_- \left( {\bar{X}_{T-} -\bar{X}_{C-} } \right) }{S}\\= & {} \frac{p_+ S_+ Z_+ +p_- S_- Z_- }{S}\\ \qquad= & {} \frac{p_+ S_+ }{S}Z_+ +\frac{p_- S_- }{S}Z_- =\sqrt{K_+ }Z_+ +\sqrt{K_- }Z_- . \end{aligned}$$

$\square$

Lemma 3

When the subgroup variances are equal, i.e., $\sigma _{T+}^2 =\sigma _{C+}^2 =\sigma _{T-}^2 =\sigma _{C-}^2$, then $Z=\sqrt{p_+ }Z_+ +\sqrt{p_- }Z_-$.

Proof

When sample variances are equal among groups, note that

$$\begin{aligned}&\frac{S_-^2 }{S_+^2 }=\frac{\frac{\sigma ^{2}}{N_{T-} }+\frac{\sigma ^{2}}{N_{C-} }}{\frac{\sigma ^{2}}{N_{T+} }+\frac{\sigma ^{2}}{N_{C+} }}=\frac{\frac{1}{N_{T-} }+\frac{1}{N_{C-} }}{\frac{1}{N_{T+} }+\frac{1}{N_{C+} }}=\frac{N_{T-} +N_{C-} }{N_{T-} N_{C-} }\frac{N_{T+} N_{C+} }{N_{T+} +N_{C+} }=\frac{N_- N_{T+} N_{C+} }{N_{T-} N_{C-} N_+ }\\&\qquad =\frac{p_- Np_+ N_T p_+ N_C }{p_- N_T p_- N_C p_+ N}=\frac{p_+ }{p_- } \end{aligned}$$

From Lemma 2, we observe that

$$\begin{aligned} Z= & {} \frac{p_+ S_+ }{S}Z_+ +\frac{p_- S_- }{S}Z_- =\frac{p_+ S_+ Z_+ +p_- S_- Z_- }{\sqrt{p_+^2 S_+^2 +p_-^2 S_-^2 }}=\frac{p_+ Z_+ +p_- \frac{S_- }{S_+ }Z_- }{\sqrt{p_+^2 +p_-^2 \frac{S_-^2 }{S_+^2 }}}\\= & {} \frac{p_+ Z_+ +p_- \sqrt{\frac{p_+ }{p_- }}Z_- }{\sqrt{p_+^2 +p_-^2 \frac{p_+ }{p_- }}}\\= & {} \frac{\sqrt{p_+ }Z_+ +\sqrt{p_- }Z_- }{\sqrt{p_+ +p_- }}=\sqrt{p_+ }Z_+ +\sqrt{p_- }Z_- . \end{aligned}$$

$\square$

Lemma 4

The following relationship holds for population parameters $\mu$, $\mu _+$ and $\mu _-$:

$$\begin{aligned} \mu =p_+ \mu _+ +p_- \mu _- , \end{aligned}$$

where $p_+ =1-p_-$ is the biomarker prevalence rate in the study.

Proof

Note that

$$\begin{aligned} E\left( Z \right) =E\left[ {\frac{\bar{X}_T -\bar{X}_C }{\sqrt{\frac{\sigma _T^2 }{N_T }+\frac{\sigma _C^2 }{N_C }}}} \right] =\frac{\mu _T -\mu _C }{\sqrt{\frac{\sigma _T^2 }{N_T }+\frac{\sigma _C^2 }{N_C }}}=\frac{\mu }{\sqrt{\frac{\sigma _T^2 }{N_T }+\frac{\sigma _C^2 }{N_C }}}=\frac{\mu }{S}. \end{aligned}$$

Similarly,

$$\begin{aligned} E\left( {Z_+ } \right) =\frac{\mu _+ }{\sqrt{\frac{\sigma _{T_+ }^2 }{N_{T_+ } }+\frac{\sigma _{C_+ }^2 }{N_{C_+ } }}}=\frac{\mu _+ }{S_+ }\,\mathrm{and}\,E\left( {Z_- } \right) =\frac{\mu _- }{\sqrt{\frac{\sigma _{T_- }^2 }{N_{T_- } }+\frac{\sigma _{C_- }^2 }{N_{C_- } }}}=\frac{\mu _- }{S_- }. \end{aligned}$$

Using Lemma 2, we observe that

$$\begin{aligned}&\frac{\mu }{S}=E\left( Z \right) =E\left( {\sqrt{K_+ }Z_+ +\sqrt{K_- }Z_- } \right) =\sqrt{K_+ }E\left( {Z_+ } \right) +\sqrt{K_- }E\left( {Z_- } \right) \\&=\frac{p_+ S_+ }{S}\frac{\mu _+ }{S_+ }+\frac{p_- S_- }{S}\frac{\mu _- }{S_- }=\frac{p_+ }{S}\mu _+ +\frac{p_- }{S}\mu _-. \end{aligned}$$

Therefore, $\mu =p_+ \mu _+ +p_- \mu _-$. $\square$

1.2.1 Safety boundary

We use the standardized safety boundary SB of $-2$ for the marker-negative group in all numeric studies. That is, if $Z_-$ falls below $-2$, we will not reject the overall null hypothesis. The safety boundary, $SB_\mu$, on the $\mu$ scale can be easily calculated using the corresponding scaling factor:

$$\begin{aligned} SB_\mu =SB*S_{-} =SB\sqrt{\frac{\sigma _{T_- }^2 }{N_{T_- } }+\frac{\sigma _{C_- }^2 }{N_{C_- } }}. \end{aligned}$$

1.2.2 Rejection regions

The proposed simultaneous test is based on the partition of the two-dimensional sample space of $Z_-$ and $Z_+$ into two rejection regions and one acceptance region (Fig. 1). Based on the observed $(Z_- ,Z_+ )$, the proposed test will

• (1) Reject only $H_{0+} ,$ if $\left( {Z_- ,Z_+ } \right) \in R_1 \cup {\Delta }_1$ (rejection region 1),

• (2) Reject both $H_{0O}$ and $H_{0+}$, if $\left( {Z_- ,Z_+ } \right) \in R_2$ (rejection region 2),

• (3) Reject neither $H_{0O}$ nor $H_{0+}$, if $\left( {Z_- ,Z_+ } \right) \notin R_1 \cup {\Delta }_1 \cup R_2 .$

Note that $H_{0C}$ is rejected in any of the two rejection regions, that is, $H_{0C}$ is rejected if $\left( {Z_- ,Z_+ } \right) \in R_1 \cup {\Delta }_1 \cup R_2$.

Provided below are calculations for the probability of the rejection regions. To ease the notation, we drop the subscript for standardized safety boundary and denote it SB.

$$\begin{aligned} P(R_1 )=P(Z_+ >C_{Z+} |\mu _+ )P(Z_- <SB|\mu _- ) \end{aligned}$$

$$\begin{aligned}&P(\Delta _1 )=\mathop \int \limits _{C_{Z+} }^{\frac{C_Z }{\sqrt{K_+ }}-SB\sqrt{K_- /K_+ }} \mathop \int \limits _{SB}^{\frac{C_Z }{\sqrt{K_- }}-Z_+ \sqrt{K_+ /K_- }} f\left( {z_+ |\mu _+ } \right) f\left( {z_- |\mu _- } \right) dz_- dz_+ \end{aligned}$$

$$\begin{aligned} \quad ~~~~~~~~~~~\,=\mathop \int \limits _{C_{Z+} }^{\frac{C_Z }{\sqrt{K_+ }}-SB\sqrt{K_- /K_+ }} f\left( {z_+ |\mu _+ } \right) \left[ {\Phi \left( {\frac{C_Z }{\sqrt{K_- }}-Z_+ \sqrt{K_+ /K_- }|\mu _- } \right) -\Phi \left( {SB|\mu _- } \right) } \right] dz_+ \end{aligned}$$

$$\begin{aligned}&=\Phi \left( {SB|\mu _- } \right) \left[ {\Phi \left( {C_{Z+} |\mu _+ } \right) -\Phi \left( {\frac{C_Z }{\sqrt{K_+ }}-SB\sqrt{K_- /K_+ }|\mu _+ } \right) } \right] \end{aligned}$$

$$\begin{aligned} \qquad \quad \,+\mathop \int \limits _{C_{Z+} }^{\frac{C_Z }{\sqrt{K_+ }}-SB\sqrt{K_- /K_+ }} f\left( {z_+ |\mu _+ } \right) \Phi \left( {\frac{C_Z }{\sqrt{K_- }}-Z_+ \sqrt{K_+ /K_- }|\mu _- } \right) dz_+ \end{aligned}$$

$$\begin{aligned} P(R_2 )= & {} P\left( {Z\ge C_Z \,\mathrm{and}\,Z_- \ge SB|\mu ,\mu _+ } \right) \\= & {} \mathop \int \nolimits _{SB}^\infty f\left( {z_- |\mu _- } \right) dz_- \mathop \int \nolimits _{\frac{C_Z }{\sqrt{K_+ }}-\frac{\sqrt{K_- }}{\sqrt{K_+ }}Z_- }^\infty f\left( {z_+ |\mu _+ } \right) dz_+ \end{aligned}$$

Lemma 5

$\mathop {\max }\limits _{\mu \le 0<\mu _+ } \alpha _{32} =\alpha _{31} (\mu =\mu _+ =0)\le \alpha _C (\mu =\mu _+ =0)$

Proof

$$\begin{aligned} \alpha _{32} =P\,\mathrm{(}R_2 |\mu ,\mu _+ :\mu \le 0<\mu _+ )=P\left( {Z\ge C_Z ,Z_- \ge SB|\mu \le 0<\mu _+ } \right) \end{aligned}$$

$$\begin{aligned} =\mathop \int \nolimits _{SB}^\infty f\left( {z_- |E\left( {Z_- } \right) } \right) dz_- \mathop \int \nolimits _{\frac{C_Z }{\sqrt{K_+ }}-\frac{\sqrt{K_- }}{\sqrt{K_+ }}Z_- }^\infty f\left( {z_+ |E\left( {Z_+ } \right) } \right) dz_+ \end{aligned}$$

$$\begin{aligned} \quad \!=\mathop \int \nolimits _{SB}^\infty f\left( {z_- |\frac{\mu _- }{S_- }} \right) \left\{ {1-{\Phi }\left( {\frac{C_Z }{\sqrt{K_+ }}-\frac{\sqrt{K_- }}{\sqrt{K_+ }}z_- -\frac{\mu _+ }{S_+ }} \right) } \right\} dz_- \end{aligned}$$

$$\begin{aligned} \qquad \!=\mathop \int \nolimits _{SB}^\infty \phi \left( {z_- -\frac{\mu _- }{S_- }} \right) \left\{ {1-{\Phi }\left( {\frac{C_Z }{\sqrt{K_+ }}-\frac{\sqrt{K_- }}{\sqrt{K_+ }}z_- -\frac{\mu _+ }{S_+ }} \right) } \right\} dz_- \end{aligned}$$

$$\begin{aligned} =\mathop \int \nolimits _{SB}^\infty \phi \left( {z_- -\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }} \right) \left\{ {1-{\Phi }\left( {\frac{C_Z }{\sqrt{K_+ }}-\frac{\sqrt{K_- }}{\sqrt{K_+ }}z_- -\frac{\mu _+ }{S_+ }} \right) } \right\} dz_- (\,\mathrm{Lemma}\, 4) \end{aligned}$$

Let $y=z_- -\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }$. Then $z_- =y+\frac{\mu }{p_- S_- }-\frac{p_+ \mu _+ }{p_- S_- }.$

Note that $\frac{\sqrt{K_- }}{\sqrt{K_+ }}z_- =\frac{\sqrt{K_- }}{\sqrt{K_+ }}\left( {y+\frac{\mu }{p_- S_- }-\frac{p_+ \mu _+ }{p_- S_- }} \right) =\frac{\sqrt{K_- }}{\sqrt{K_+ }}y+\frac{\sqrt{K_- }}{\sqrt{K_+ }}\frac{\mu }{p_- S_- }-\frac{\sqrt{K_- }}{\sqrt{K_+ }}\frac{p_+ \mu _+ }{p_- S_- }$

$$\begin{aligned} =\frac{\sqrt{K_- }}{\sqrt{K_+ }}y+\frac{p_- S_- }{p_+ S_+ }\frac{\mu }{p_- S_- }-\frac{p_- S_- }{p_+ S_+ }\frac{p_+ \mu _+ }{p_- S_- }(\,\mathrm{Lemma}\, l2) \end{aligned}$$

$$\begin{aligned} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\frac{\sqrt{K_- }}{\sqrt{K_+ }}y+\frac{\mu }{p_+ S_+ }-\frac{\mu _+ }{S_+ } \end{aligned}$$

Therefore,

$$\begin{aligned}&\alpha _{32} =\mathop \int \nolimits _{SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }}^\infty \phi \left( y \right) \left\{ {1-{\Phi }\left( {\frac{c_z }{\sqrt{K_+ }}-\frac{\mu }{p_+ S_+ }-\frac{\sqrt{K_- }}{\sqrt{K_+ }}y} \right) } \right\} dy\\&\frac{\partial \alpha _{32} }{\partial \mu _+ }=-\frac{p_+ }{p_- S_- }\phi \left( {SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }} \right) \\&\left\{ {1-{\Phi }\left( {\frac{C_z }{\sqrt{K_+ }}-\frac{\mu }{p_- S_- }-\frac{\sqrt{K_- }}{\sqrt{K_+ }}\left( {SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }} \right) } \right) } \right\} \\&\le 0 \end{aligned}$$

$$\begin{aligned}&\frac{\partial \alpha _{32} }{\partial \upmu }=\frac{1}{p_- S_- }\phi \left( {SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }} \right) \\&\left\{ {1-{\Phi } \left( {\frac{C_z }{\sqrt{K_+ }}-\frac{\mu }{p_+ S_+ }-\frac{\sqrt{K_- }}{\sqrt{K_+ }}\left( {SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }} \right) } \right) } \right\} \\&+\mathop \int \nolimits _{SB-\frac{\mu }{p_- S_- }+\frac{p_+ \mu _+ }{p_- S_- }}^\infty \phi \left( y \right) \left\{ {1+\frac{1}{p_+ S_+ }\phi \left( {\frac{C_z }{\sqrt{K_+ }}-\frac{\mu }{p_+ S_+ }-\frac{\sqrt{K_- }}{\sqrt{K_+ }}y} \right) } \right\} dy\\&>0 \end{aligned}$$

Therefore, $\alpha _{32}$ is a decreasing function of $\mu _+$ and an increasing function of $\mu$. Thus, the maximum value for $\alpha _{32}$ for $\mu \le 0<\mu _+$ is achieved at $\mu _+ =\mu =0$, i.e., $\,\mathrm{max}\left( {\alpha _{32} ,\mu \le 0<\mu _+ } \right) =\alpha _{31} (\mu =\mu _+ =0)\le \alpha _C (\mu =\mu _+ =0)$. $\square$

1.2.3 Effect sizes

We define effect sizes $\delta$, $\delta _+$and $\delta _-$ in the overall, $M_{+}$ and $M_{-}$ subgroups as follows:

$$\begin{aligned} \delta =\frac{EZ}{\sqrt{N}},\delta _+ =\frac{EZ_+ }{\sqrt{N_+ }}\,\mathrm{and}\,\delta _- =\frac{EZ_- }{\sqrt{N_- }}. \end{aligned}$$

Note that our definition of effect size differs from commonly used definitions. It is common to define effect size as the ratio of treatment effect and the pooled estimate of standard deviation. However, since our test statistics do not rely on pooled estimates of standard deviation, but instead use statistics for two independent samples, our effect sizes are defined based on these statistics. Note that

$$\begin{aligned}&\delta =\frac{EZ}{\sqrt{N}}=\frac{1}{\sqrt{N}}E\left[ {\sqrt{N_T }\frac{\bar{X}_T -\bar{X}_C }{\sqrt{\sigma _T^2 +\sigma _C^2 R}}} \right] =\frac{1}{\sqrt{N}}E\left[ {\sqrt{N\frac{R}{1+R}}\frac{\bar{X}_T -\bar{X}_C }{\sqrt{\sigma _T^2 +\sigma _C^2 R}}} \right] \\&=E\left[ {\sqrt{\frac{R}{1+R}}\frac{\bar{X}_T -\bar{X}_C }{\sqrt{\sigma _T^2 +\sigma _C^2 R}}} \right] =\sqrt{\frac{R}{1+R}}\frac{\mu }{\sqrt{\sigma _T^2 +\sigma _C^2 R}}. \end{aligned}$$

Similarly,

$$\begin{aligned} \delta _+ =\sqrt{\frac{R}{1+R}}\frac{\mu _+ }{\sqrt{\sigma _{T+}^2 +\sigma _{C+}^2 R}} \end{aligned}$$

and

$$\begin{aligned} \delta _- =\sqrt{\frac{R}{1+R}}\frac{\mu _- }{\sqrt{\sigma _{T-}^2 +\sigma _{C-}^2 R}}. \end{aligned}$$

Lemma 6

The following relationship among effect sizes $\delta ,\delta _+$and $\delta _-$holds:

$$\begin{aligned} \delta =\frac{p_+ \delta _+ S_+ \sqrt{p_+ }+p_- \delta _- S_- \sqrt{p_- }}{S}. \end{aligned}$$

Proof

Note that $\mu =p_+ \mu _+ +p_- \mu _-$ (Lemma 4). Therefore,

$$\begin{aligned}&\delta =\frac{EZ}{\sqrt{N}}=\frac{\mu }{S\sqrt{N}}=\frac{p_+ \mu _+ +p_- \mu _- }{S\sqrt{N}}=\frac{p_+ \delta _+ S_+ \sqrt{N_+ }+p_- \delta _- S_- \sqrt{N_- }}{S\sqrt{N}}\\&=\frac{p_+ \delta _+ S_+ \sqrt{p_+ }+p_- \delta _- S_- \sqrt{p_- }}{S}. \end{aligned}$$

$\square$

Lemma 7

The following relationship among $EZ,EZ_+$and $EZ_-$holds:

$$\begin{aligned} EZ=\frac{p_+ S_+ }{S}EZ_+ +\frac{p_- S_- }{S}EZ_- \left( 4 \right) ; \end{aligned}$$

Proof

Using Lemma 4, we observe that

$$\begin{aligned} EZ=\frac{\mu }{S}=\frac{p_+ \mu _+ +p_- \mu _- }{S}=\frac{p_+ S_+ EZ_+ +p_- S_- EZ_- }{S}. \end{aligned}$$

$\square$