Sample Size Consideration for Equivalent Test of Tier-1 Quality Attributes for Analytical Biosimilarity Assessment

Abstract

FDA recommends a stepwise approach for obtaining the totality-of-the-evidence for assessing biosimilarity between a proposed biosimilar product and its corresponding reference biologic product being considered (US Food and Drug Administration.: Guidance for industry: scientific considerations in demonstrating biosimilarity to a reference product. US Food and Drug Administration, Silver Spring, 2015 [6]). The stepwise approach starts with analytical studies for assessing similarity in critical quality attributes (CQAs), which are relevant to clinical outcomes. For critical quality attributes that are most relevant to clinical outcomes (Tier 1 CQAs), FDA requires equivalence testing to be performed for similarity assessment, based on an equivalence acceptance criteria. In practice, the number of Tier 1 CQAs might be greater than one, and should be no more than four. The number of biosimilar lots is often recommended to be no less than 10, and the ratio between the reference product sample size and biosimilar product sample size is recommended within the range from $2/3$ to $3/2$ (US Food and Drug Administration.: Guidance for industry: Statistical Approaches to Evaluate Analytical Similarity. US Food and Drug Administration, Silver Spring, 2017 [7]). Accordingly, we derive the formulas for the power calculation for the sample size for analytical similarity assessment based on the equivalence testing currently used in analytical biosimilar assessment (Tsong et al. J Biopharm Stat 27:197–205, (2017)[10]).

Appendix

• 1. The proof of Theorem 1.

Proof: Because the power function is

$$ {\text{P}}_{1} = {\text{P}}\left( {{\text{T}}_{1} = \frac{{\left( {\overline{\text{X}}_{\text{T}} - \overline{\text{X}}_{\text{R}} } \right) + 1.5\upsigma_{\text{R}} }}{{\sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }} > {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} , {\text{T}}_{2} = \frac{{\left( {\overline{\text{X}}_{\text{T}} - \overline{\text{X}}_{\text{R}} } \right) - 1.5\upsigma_{\text{R}} }}{{\sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }} < - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} } \right) $$

(A.1)

given $\upmu_{\text{T}} -\upmu_{\text{R}} =\uptheta \in \left( { - 1.5\upsigma_{\text{R}} , + 1.5\upsigma_{\text{R}} } \right),\upsigma_{\text{T}} ,\upsigma_{\text{R}}$. The inequailities in (A.1) is equivalent to

$$ - 1.5\upsigma_{\text{R}} + {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} < \left( {\overline{\text{X}}_{\text{T}} - \overline{\text{X}}_{\text{R}} } \right) < 1.5\upsigma_{\text{R}} - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} $$

(A.2)

Because ${\text{Z}} = \frac{{\left( {\overline{\text{X}}_{\text{T}} - \overline{\text{X}}_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }}\sim {\text{N}}\left( {0, 1} \right)$, the inequailities in (A.2) is equivalent to

$$ \frac{{\left( { - 1.5\upsigma_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} + \frac{{{\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} < Z < \frac{{\left( {1.5\upsigma_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} - \frac{{{\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} $$

(A.3)

Let ${\text{x}}_{1} = \frac{{\left( {{\text{n}}_{\text{T}} - 1} \right){\text{S}}_{\text{T}}^{2} }}{{\upsigma_{\text{T}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{T}} - 1}}^{2}$ and ${\text{x}}_{2} = \frac{{\left( {{\text{n}}_{\text{R}} - 1} \right){\text{S}}_{\text{R}}^{2} }}{{\upsigma_{\text{R}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{R}} - 1}}^{2}$, then ${\text{S}}_{\text{T}}^{2} = \frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{\left( {{\text{n}}_{\text{T}} - 1} \right)}}$ and ${\text{S}}_{\text{R}}^{2} = \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{\left( {{\text{n}}_{\text{R}} - 1} \right)}}$, the inequalities in (A.3) is

$$ \frac{{\left( { - 1.5\upsigma_{\text{R}} - {\varvec{\theta}}} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} + {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \frac{{\sqrt {\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} < Z < \frac{{\left( {1.5\upsigma_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \frac{{\sqrt {\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} $$

(A.4)

Let ${\text{A}}_{1} = \frac{{1.5\upsigma_{\text{R}} -\uptheta}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}}}}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}}}}$ and ${\text{B}}_{1} = \frac{{ - 1.5\upsigma_{\text{R}} -\uptheta}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} + {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}}}}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}}}}$,

From (A.4) since ${\text{B}}_{1} < {\text{A}}_{1}$, we can get that $\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}} < \frac{{9\upsigma_{\text{R}}^{2} }}{{4{\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}}^{2} }}$, then the power function (A.1) is

$$ {\text{E}}_{{{\text{x}}_{1} ,{\text{x}}_{2} }} \left\{ {{\text{P}}({\text{B}}_{1} < Z < {\text{A}}_{1} )} \right\} $$

Given ${\text{x}}_{1} \sim\upchi_{{{\text{n}}_{\text{T}} - 1}}^{2}$ and ${\text{x}}_{2} \sim\upchi_{{{\text{n}}_{\text{R}} - 1}}^{2}$, the power function is

$$ {\text{P}}_{1} = \mathop \smallint \limits_{0}^{ + \infty } \mathop \smallint \limits_{0}^{ + \infty } \left\{ {\left[ {\Phi \left( {{\text{A}}_{1} } \right) - \Phi \left( {{\text{B}}_{1} } \right)} \right] \times {\text{f}}\left( {{\text{x}}_{1} , {\text{x}}_{2} } \right)} \right\} \times {\mathbf{I}}\left\{ {\frac{{{\text{x}}_{1}\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} \left( {{\text{n}}_{\text{T}} - 1} \right)}} + \frac{{{\text{x}}_{2}\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} \left( {{\text{n}}_{\text{R}} - 1} \right)}} \le \frac{{9\upsigma_{\text{R}}^{2} }}{{4{\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}}^{2} }}} \right\}{\text{dx}}_{1} {\text{dx}}_{2} $$

where ${\text{f}}\left( {{\text{x}}_{1} , {\text{x}}_{2} } \right) = {\text{f}}\left( {{\text{x}}_{1} } \right){\text{f}}\left( {{\text{x}}_{2} } \right)$ is the density function of two independent Chi-square distributions with ${\text{x}}_{1} \sim\upchi_{{{\text{n}}_{\text{T}} - 1}}^{2}$ and ${\text{x}}_{2} \sim\upchi_{{{\text{n}}_{\text{R}} - 1}}^{2}$. ${\mathbf{I}}\{ .\}$ is the indication function to restrict the triangle area formed by ${\text{x}}_{1}$ and ${\text{x}}_{2}$.

• 2. The proof of Theorem 2.

Proof: By using the Satterthwaite approximation

$$ \frac{{{\text{df}}^{ *} \left( {\frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} \right)}}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}}}\sim\upchi_{{{\text{df}}^{ *} }}^{2} $$

(A.5)

Where ${\text{df}}^{ *} = \left( {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} \right)^{2} /\left( {\frac{{\upsigma_{\text{T}}^{4} }}{{\left( {{\text{n}}_{\text{T}}^{ *} } \right)^{2} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{\text{R}}^{4} }}{{\left( {{\text{n}}_{\text{R}}^{ *} } \right)^{2} ({\text{n}}_{\text{R}} - 1)}}} \right)$, let ${\text{S}}^{2} = \frac{{{\text{S}}_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}$ and ${\text{x}} = \frac{{{\text{df}}^{ *} {\text{S}}^{2} }}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}}}\sim\upchi_{{{\text{df}}^{ *} }}^{2}$, then the inequalities (A.3) is

$$ \frac{{\left( { - 1.5\upsigma_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} + {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{\text{x}}{{{\text{df}}^{ *} }}} \frac{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} < Z < \frac{{\left( {1.5\upsigma_{\text{R}} -\uptheta} \right)}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{\text{x}}{{{\text{df}}^{ *} }}} \frac{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} }}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} $$

(A.6)

Let ${\text{A}}_{1}^{ '} = \frac{{1.5\upsigma_{\text{R}} -\uptheta}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} - {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{\text{x}}{{{\text{df}}^{ *} }}} \sqrt {\frac{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}}}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}}}}$ and ${\text{B}}_{1}^{ '} = \frac{{ - 1.5\upsigma_{\text{R}} -\uptheta}}{{\sqrt {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}} }} + {\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}} \sqrt {\frac{\text{x}}{{{\text{df}}^{ *} }}} \sqrt {\frac{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}}}{{\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}} }}}}}$, using ${\text{B}}_{1}^{ '} < {\text{A}}_{1}^{ '}$, we can get ${\text{x}} < L = \frac{{9\upsigma_{\text{R}}^{2} {\text{df}}^{ *} }}{{4{\text{t}}_{{{\text{df}}^{ *} , 1 -\upalpha}}^{2} \left( {\frac{{\upsigma_{\text{T}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{\text{R}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} \right)}}$, the power function (A.1) is ${\text{E}}_{\text{x }} \left\{ {{\text{P}}({\text{B}}_{1}^{ '} < Z < {\text{A}}_{1}^{ '} )} \right\},$ which is equivalent to

$$ {\text{P}}_{1}^{ '} = \mathop \smallint \limits_{0}^{\text{L}} \left\{ {\left[ {\Phi \left( {{\text{A}}_{1}^{ '} } \right) - \Phi \left( {{\text{B}}_{1}^{ '} } \right)} \right] \times \frac{1}{{2^{{\frac{{{\text{df}}^{ *} }}{2}}} \varGamma \left( {\frac{{{\text{df}}^{ *} }}{2}} \right)}}{\text{x}}^{{\frac{{{\text{df}}^{ *} }}{2} - 1}} {\text{e}}^{{ - \frac{\text{x}}{2}}} } \right\}{\text{dx}} $$

• 3. The proof of Theorem 4.

Proof: If there are 2 quality attributes that are correlated, assume the 2 quality attributes in the reference product ${\text{X}}_{{1,{\text{R}}}}$ and ${\text{X}}_{{2, {\text{R}}}}$ have the correlation value $\uprho$; the same as the 2 quality attributes in the test product ${\text{X}}_{{1,{\text{T}}}}$ and ${\text{X}}_{{2, {\text{T}}}}$

$$ \left[ {\begin{array}{*{20}c} {{\text{X}}_{{1,{\text{R}}}} } \\ {{\text{X}}_{{2, {\text{R}}}} } \\ \end{array} } \right]\sim {\text{MVN}}\left( {\left[ {\begin{array}{*{20}c} {\upmu_{{1,{\text{R}}}} } \\ {\upmu_{{2, {\text{R}}}} } \\ \end{array} } \right], \left[ {\begin{array}{*{20}c} {\upsigma_{{1,{\text{R}}}}^{2} } & {\uprho \upsigma _{{1,{\text{R}}}}\upsigma_{{2,{\text{R}}}} } \\ {\uprho \upsigma _{{1,{\text{R}}}}\upsigma_{{2,{\text{R}}}} } & {\upsigma_{{2,{\text{R}}}}^{2} } \\ \end{array} } \right]} \right) $$

$$ \left[ {\begin{array}{*{20}c} {{\text{X}}_{{1,{\text{T}}}} } \\ {{\text{X}}_{{2, {\text{T}}}} } \\ \end{array} } \right]\sim {\text{MVN}}\left( {\left[ {\begin{array}{*{20}c} {\upmu_{{1,{\text{T}}}} } \\ {\upmu_{{2, {\text{T}}}} } \\ \end{array} } \right], \left[ {\begin{array}{*{20}c} {\upsigma_{{1,{\text{T}}}}^{2} } & {\uprho \upsigma _{{1,{\text{T}}}}\upsigma_{{2,{\text{T}}}} } \\ {\uprho \upsigma _{{1,{\text{T}}}}\upsigma_{{2,{\text{T}}}} } & {\upsigma_{{2,{\text{T}}}}^{2} } \\ \end{array} } \right]} \right) $$

We further assume the sample size of either quality attribute is the same in the reference product or in the test product, that is ${\text{n}}_{{1,{\text{R}}}} = {\text{n}}_{{2,{\text{R}}}} = {\text{n}}_{\text{R}}$ and ${\text{n}}_{{1,{\text{T}}}} = {\text{n}}_{{2,{\text{T}}}} = {\text{n}}_{\text{T}}$, given the hypotheses in (6) and test statistics in (7), the power of passing both equivalence testings is equivalent to calculate

$$ {\text{P}}\left( {\left[ {\begin{array}{*{20}c} { - 1.5\upsigma_{{1, {\text{R}}}} + {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} } \\ { - 1.5\upsigma_{{2, {\text{R}}}} + {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} } \\ \end{array} } \right] \le \left[ {\begin{array}{*{20}c} {\overline{\text{X}}_{{1.{\text{T}}}} - \overline{\text{X}}_{{1.{\text{R}}}} } \\ {\overline{\text{X}}_{{2.{\text{T}}}} - \overline{\text{X}}_{{2.{\text{R}}}} } \\ \end{array} } \right] \le \left[ {\begin{array}{*{20}c} { + 1.5\upsigma_{{1, {\text{R}}}} - {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} } \\ { + 1.5\upsigma_{{2, {\text{R}}}} - {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\frac{{{\text{S}}_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{{\text{S}}_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} } \\ \end{array} } \right]} \right) $$

(A.9)

The degrees of freedom ${\text{df}}_{1}^{ *} = \left( {\frac{{\upsigma_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} \right)^{2} /\left( {\frac{{\upsigma_{{1, {\text{T}}}}^{4} }}{{\left( {{\text{n}}_{\text{T}}^{ *} } \right)^{2} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{4} }}{{\left( {{\text{n}}_{\text{R}}^{ *} } \right)^{2} ({\text{n}}_{\text{R}} - 1)}}} \right)$ and ${\text{df}}_{2}^{ *} = \left( {\frac{{\upsigma_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}}^{ *} }} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}}^{ *} }}} \right)^{2} /\left( {\frac{{\upsigma_{{1, {\text{T}}}}^{4} }}{{\left( {{\text{n}}_{\text{T}}^{ *} } \right)^{2} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{4} }}{{\left( {{\text{n}}_{\text{R}}^{ *} } \right)^{2} ({\text{n}}_{\text{R}} - 1)}}} \right)$. Let ${\text{u}}_{\text{T}} = \frac{{({\text{n}}_{\text{T}} - 1){\text{S}}_{{1, {\text{T}}}}^{2} }}{{\upsigma_{{1, {\text{T}}}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{T}} - 1}}^{2}$, ${\text{v}}_{\text{T}} = \frac{{({\text{n}}_{\text{T}} - 1){\text{S}}_{{2, {\text{T}}}}^{2} }}{{\upsigma_{{2, {\text{T}}}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{T}} - 1}}^{2}$ and ${\text{u}}_{\text{R}} = \frac{{({\text{n}}_{\text{R}} - 1){\text{S}}_{{1, {\text{R}}}}^{2} }}{{\upsigma_{{1, {\text{R}}}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{R}} - 1}}^{2}$, ${\text{v}}_{\text{R}} = \frac{{({\text{n}}_{\text{R}} - 1){\text{S}}_{{2, {\text{R}}}}^{2} }}{{\upsigma_{{2, {\text{R}}}}^{2} }}\sim\upchi_{{{\text{n}}_{\text{R}} - 1}}^{2}$, and $\uptheta_{1} =\upmu_{{1,{\text{T}}}} -\upmu_{{1,{\text{R}}}}$, $\uptheta_{2} =\upmu_{{2,{\text{T}}}} -\upmu_{{2,{\text{R}}}}$. The inequailities in (A.9) is equivalent to

$$\begin{aligned} &\left[ {\begin{array}{*{20}c} { - 1.5\upsigma_{{1, {\text{R}}}} -\uptheta_{1} + {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\upsigma_{{1, {\text{T}}}}^{2} {\text{u}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} {\text{u}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}}} } \\ { - 1.5\upsigma_{{2, {\text{R}}}} -\uptheta_{2} + {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\upsigma_{{2, {\text{T}}}}^{2} {\text{v}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} {\text{v}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}}} } \\ \end{array} } \right] \le {\mathbf{B}} \\ & \quad \le \left[ {\begin{array}{*{20}c} { + 1.5\upsigma_{{1, {\text{R}}}} -\uptheta_{1} - {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\upsigma_{{1, {\text{T}}}}^{2} {\text{u}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} {\text{u}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}}} } \\ { + 1.5\upsigma_{{2, {\text{R}}}} -\uptheta_{2} - {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\frac{{\upsigma_{{2, {\text{T}}}}^{2} {\text{v}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} {\text{v}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}}} } \\ \end{array} } \right] \end{aligned}$$

(A.10)

where ${\mathbf{B}}$ is a bi-variate normal distribution with mean $0$ and variance-covariance matrix

$$ {\varvec{\Sigma}} = \left[ {\begin{array}{*{20}c} {\frac{{\upsigma_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}} & {\frac{{\uprho \upsigma _{{1,{\text{T}}}}\upsigma_{{2, {\text{T}}}} }}{{{\text{n}}_{\text{T}} }} + \frac{{\uprho \upsigma _{{1,{\text{R}}}}\upsigma_{{2, {\text{R}}}} }}{{{\text{n}}_{\text{R}} }}} \\ {\frac{{\uprho \upsigma _{{1,{\text{T}}}}\upsigma_{{2, {\text{T}}}} }}{{{\text{n}}_{\text{T}} }} + \frac{{\uprho \upsigma _{{1,{\text{R}}}}\upsigma_{{2, {\text{R}}}} }}{{{\text{n}}_{\text{R}} }}} & {\frac{{\upsigma_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}} \\ \end{array} } \right] $$

let $\uptau_{1} = \sqrt {\frac{{\upsigma_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}}$; $\uptau_{2} = \sqrt {\frac{{\upsigma_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}}$; $\uprho^{ *} = \frac{{\uprho \upsigma _{{1,{\text{T}}}}\upsigma_{{2, {\text{T}}}} }}{{{\text{n}}_{\text{T}} }} + \frac{{\uprho \upsigma _{{1,{\text{R}}}}\upsigma_{{2, {\text{R}}}} }}{{{\text{n}}_{\text{R}} }}/\sqrt {\left( {\frac{{\upsigma_{{1, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}} \right)\left( {\frac{{\upsigma_{{2, {\text{T}}}}^{2} }}{{{\text{n}}_{\text{T}} }} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} }}{{{\text{n}}_{\text{R}} }}} \right)}$, then ${\mathbf{B}}$ is a bi-variate normal distribution with mean $0$ and variance-covariance matrix $\left[ {\begin{array}{*{20}c} {\uptau_{1}^{2} } & {\uprho^{ *}\uptau_{1}\uptau_{2} } \\ {\uprho^{ *}\uptau_{1}\uptau_{2} } & {\uptau_{2}^{2} } \\ \end{array} } \right]$. The power function would be

$$ {\text{P}}_{2} = {\text{E}}_{{{\text{u}}_{\text{T}} , {\text{v}}_{\text{T}} , {\text{u}}_{\text{R}} , {\text{v}}_{\text{R}} }} \left\{ {{\text{P(}}{\mathbf{Inequality}} ({\text{A}}.\,10)\,{\text{Holds | u}}_{\text{T}} , {\text{v}}_{\text{T}} , {\text{u}}_{\text{R}} , {\text{v}}_{\text{R}} )} \right\} $$

Let ${\text{U}} = \frac{{\upsigma_{{1, {\text{T}}}}^{2} {\text{u}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} {\text{u}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}} \le \frac{{9\upsigma_{{1.{\text{R}}}}^{2} }}{{4{\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}}^{2} }}$ and ${\text{V}} = \frac{{\upsigma_{{2, {\text{T}}}}^{2} {\text{v}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} {\text{v}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}} \le \frac{{9\upsigma_{{2.{\text{R}}}}^{2} }}{{4{\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}}^{2} }}$ by the inequalities (A.10), the limits ${\text{LL}}_{1} = - 1.5\upsigma_{{1, {\text{R}}}} -\uptheta_{1} + {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\text{U}}$, ${\text{LL}}_{2} = - 1.5\upsigma_{{2, {\text{R}}}} -\uptheta_{2} + {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\text{V}}$ and ${\text{UL}}_{1} = 1.5\upsigma_{{1, {\text{R}}}} -\uptheta_{1} - {\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}} \sqrt {\text{U}}$, ${\text{UL}}_{2} = 1.5\upsigma_{{2, {\text{R}}}} -\uptheta_{2} - {\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}} \sqrt {\text{V}}$; then

$$ {\text{P}}_{2} = {\text{E}}_{{{\text{U}}, {\text{V}}}} \left\{ {\mathop \smallint \limits_{{{\text{LL}}_{1} }}^{{{\text{UL}}_{1} }} \mathop \smallint \limits_{{{\text{LL}}_{2} }}^{{{\text{UL}}_{2} }} \frac{1}{{2\uppi \uptau _{1}\uptau_{2} \sqrt {1 -\uprho^{ * 2} } }}{ \exp }\left\{ { - \frac{{{\text{x}}_{1}^{2} }}{{2\uptau_{1}^{2} \left( {1 -\uprho^{ * 2} } \right)}} + \frac{{\uprho^{ *} {\text{x}}_{1} {\text{x}}_{2} }}{{\uptau_{1}\uptau_{2} \left( {1 -\uprho^{ * 2} } \right)}} - \frac{{{\text{x}}_{2}^{2} }}{{2\uptau_{2}^{2} \left( {1 -\uprho^{ * 2} } \right)}}} \right\}{\text{dx}}_{1} {\text{dx}}_{2} } \right\} $$

The problem remains to find the joint distribution of U and V. Please note that from theorem 4, ${\text{F}}\left( { {\text{u}}_{\text{T}} , {\text{u}}_{\text{R}} ,{\text{v}}_{\text{T}} , {\text{v}}_{\text{R}} } \right) = {\text{f}}_{2} \left( {{\text{u}}_{\text{T}} , {\text{v}}_{\text{T}} } \right) * {\text{f}}_{2} \left( {{\text{u}}_{\text{R}} , {\text{v}}_{\text{R}} } \right)$ where ${\text{f}}_{2}$ is the joint density of a bivariate chi-square distribution. Let $\frac{{\upsigma_{{1, {\text{T}}}}^{2} {\text{u}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{1, {\text{R}}}}^{2} {\text{u}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}} =\uplambda_{1} {\text{u}}_{\text{T}} +\upmu_{1} {\text{u}}_{\text{R}} = {\text{U}}$ and $\frac{{\upsigma_{{2, {\text{T}}}}^{2} {\text{v}}_{\text{T}} }}{{{\text{n}}_{\text{T}}^{ *} ({\text{n}}_{\text{T}} - 1)}} + \frac{{\upsigma_{{2, {\text{R}}}}^{2} {\text{v}}_{\text{R}} }}{{{\text{n}}_{\text{R}}^{ *} ({\text{n}}_{\text{R}} - 1)}} =\uplambda_{2} {\text{v}}_{\text{T}} +\upmu_{2} {\text{v}}_{\text{R}} = {\text{V}}$, and we want to get the joint distribution of ${\text{U }}$ and ${\text{V}}$. The Jacobian transformation matrix is

$$ {\text{Jacobian}}\left[ {\left( {{\text{u}}_{\text{T}} = \frac{{{\text{U}} -\upmu_{1} {\text{u}}_{\text{R}} }}{{\uplambda_{1} }}, {\text{u}}_{\text{R}} , {\text{v}}_{\text{T}} , {\text{v}}_{\text{R}} = \frac{{{\text{V}} -\uplambda_{2} {\text{v}}_{\text{T}} }}{{\upmu_{2} }}} \right) \to \left( {{\text{U}}, {\text{u}}_{\text{R}} ,{\text{v}}_{\text{T}} , {\text{V}}} \right)} \right] = \frac{1}{{\uplambda_{1}\upmu_{2} }} $$

Thus

$$ \begin{aligned} {\text{FF}}\left( {{\text{U}}, {\text{u}}_{\text{R}} , {\text{v}}_{\text{T}} , {\text{V}}} \right) & = {\text{F}}\left( {\frac{{{\text{U}} -\upmu_{1} {\text{u}}_{\text{R}} }}{{\uplambda_{1} }}, {\text{u}}_{\text{R}} , {\text{v}}_{\text{T}} , \frac{{{\text{V}} -\uplambda_{2} {\text{v}}_{\text{T}} }}{{\upmu_{2} }}} \right)\frac{1}{{\uplambda_{1}\upmu_{2} }} \\ & = {\text{f}}_{2} \left( {\frac{{{\text{U}} -\upmu_{1} {\text{u}}_{\text{R}} }}{{\uplambda_{1} }}, {\text{v}}_{\text{T}} |{\text{m}} = {\text{n}}_{\text{T}} - 1} \right) * {\text{f}}_{2} \left( {{\text{u}}_{\text{R}} , \frac{{{\text{V}} -\uplambda_{2} {\text{v}}_{\text{T}} }}{{\upmu_{2} }}|{\text{m}} = {\text{n}}_{\text{R}} - 1} \right)\frac{1}{{\uplambda_{1}\upmu_{2} }} \\ \end{aligned} $$

We then integrate out ${\text{u}}_{\text{R}} , {\text{v}}_{\text{T}}$, we can get the joint distribution of ${\text{U}}$ and ${\text{V}}$

$$ {{\text{F}}}_{2} \left( {{{\text{U}}}, {{\text{V}}}} \right) = \mathop \smallint \limits_{0}^{{\frac{{\text{V}}}{{\uplambda_{2} }}}} \mathop \smallint \limits_{0}^{{\frac{{\text{U}}}{{\upmu_{1} }}}} {{\text{FF}}}\left( {{{\text{U}}}, {{\text{u}}}_{{\text{R}}} , {{\text{v}}}_{{\text{T}}} , {{\text{V}}}} \right){{\text{du}}}_{{\text{R}}} {{\text{dv}}}_{{\text{T}}} $$

We can get the power function as

$$ {\text{P}}_{2} = \mathop \smallint \limits_{0}^{{\frac{{9\upsigma_{{2.{\text{R}}}}^{2} }}{{4{\text{t}}_{{{\text{df}}_{2}^{ *} , 1 -\upalpha}}^{2} }}}} \mathop \smallint \limits_{0}^{{\frac{{9\upsigma_{{1.{\text{R}}}}^{2} }}{{4{\text{t}}_{{{\text{df}}_{1}^{ *} , 1 -\upalpha}}^{2} }}}} \left\{ {\mathop \smallint \limits_{{{\text{LL}}_{1} \left( {\text{U}} \right)}}^{{{\text{UL}}_{1} \left( {\text{U}} \right)}} \mathop \smallint \limits_{{{\text{LL}}_{2} \left( {\text{V}} \right)}}^{{{\text{UL}}_{2} \left( {\text{V}} \right)}} {\text{f}}_{1} \left( {{\text{x}}_{1} , {\text{x}}_{2} } \right){\text{dx}}_{1} {\text{dx}}_{2} } \right\} \times {\text{F}}_{2} \left( {{\text{U}}, {\text{V}}} \right){\text{dUdV }} $$

where ${\text{f}}_{1} \left( {{\text{x}}_{1} , {\text{x}}_{2} } \right) = \frac{1}{{2\uppi \uptau _{1}\uptau_{2} \sqrt {1 -\uprho^{ * 2} } }}{ \exp }\left\{ { - \frac{{{\text{x}}_{1}^{2} }}{{2\uptau_{1}^{2} \left( {1 -\uprho^{ * 2} } \right)}} + \frac{{\uprho^{ *} {\text{x}}_{1} {\text{x}}_{2} }}{{\uptau_{1}\uptau_{2} \left( {1 -\uprho^{ * 2} } \right)}} - \frac{{{\text{x}}_{2}^{2} }}{{2\uptau_{2}^{2} \left( {1 -\uprho^{ * 2} } \right)}}} \right\}$ be the joint distribution of the bi-variate normal distribution with mean zero and variance-covariance matrix $\left[ {\begin{array}{*{20}c} {\uptau_{1}^{2} } & {\uprho^{ *}\uptau_{1}\uptau_{2} } \\ {\uprho^{ *}\uptau_{1}\uptau_{2} } & {\uptau_{2}^{2} } \\ \end{array} } \right]$.