D optimal designs for three Poisson dose–response models

Abstract

The objective of this paper was to find and investigate the performance of the D optimal designs for three Poisson dose–response models. Phase II dose ranging studies are pivotal in the drug development program, being used to select dose(s) for phase III. Count data is encountered in a number of clinical areas. The Poisson distribution provides an intuitive platform for modelling such data, especially when combined with random effects which allow subjects to differ in their response rates. This work investigated three Poisson dose–response models of increasing complexity. A simple E_max model was used to describe the drug effect, and D optimal designs under a range of different parameter values (scenarios) were found. The relative performances between scenarios were assessed using: the precision of all parameters, the precision of the drug effect parameters, and the percent coefficient of variation (%CV) of the ED₅₀ parameter. The results showed that the D optimal designs were similar across models and scenarios, with the D optimal designs consisting of placebo, the maximum dose, and a dose just below the ED₅₀. However the relative performance of the optimal designs was very different. For example, with 1,000 subjects, the %CV of the ED₅₀ parameter ranged from 1.4 to 91 %. Performance typically improved with higher baseline counts, smaller random effects, and larger E_max. This work introduces a framework for determining and evaluating the performance of D optimal designs for phase II dose ranging studies with count data as the primary endpoint.

Appendix

The individual components on the expected FIM matrix for model 1 were determined analytically. The expected FIM matrix for model 1 is:

$$ {\text{expected}}\;{\text{FIM}} = - {\text{E}}\left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} LL}}{{\partial E_{0}^{2} }}} & {\frac{{\partial^{2} LL}}{{\partial E_{0} ED_{50} }}} & {\frac{{\partial^{2} LL}}{{\partial E_{0} E_{{max} } }}} \\ {\frac{{\partial^{2} LL}}{{\partial E_{0} ED_{50} }}} & {\frac{{\partial^{2} LL}}{{\partial ED_{50}^{2} }}} & {\frac{{\partial^{2} LL}}{{\partial ED_{50} E_{{max} } }}} \\ {\frac{{\partial^{2} LL}}{{\partial E_{0} E_{{max} } }}} & {\frac{{\partial^{2} LL}}{{\partial ED_{50} E_{{max} } }}} & {\frac{{\partial^{2} LL}}{{\partial E_{{max} }^{2} }}} \\ \end{array} } \right] $$

where LL is the log-likelihood. For the Poisson model, the log likelihood is:

$$ LL = x \cdot { \log }\left( \lambda \right) - \lambda - { \log }(x!) $$

For x counts, and for model 1:

$$ { \log }\left( \lambda \right) = E_{0} + \frac{{E_{{max} } \cdot Dose}}{{ED_{50} + Dose}} $$

Thus

$$ \frac{\partial LL}{{\partial E_{0} }} = x - \lambda $$

$$ \frac{\partial LL}{{\partial ED_{50} }} = \frac{{E_{max} \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{2} }}(\lambda - x) $$

$$ \frac{\partial LL}{{\partial E_{{max} } }} = \frac{Dose}{{ED_{50} + Dose}}(x - \lambda ) $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{0}^{2} }} = - \lambda $$

$$ \frac{{\partial^{2} LL}}{{\partial ED_{50}^{2} }} = \frac{{E_{{max} } \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{3} }}\left( {2x - 2\lambda - \lambda \frac{{E_{{max} } \cdot Dose}}{{ED_{50} + Dose}}} \right) $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{{max} }^{2} }} = - \lambda \left( {\frac{Dose}{{ED_{50} + Dose}}} \right)^{2} $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{0} ED_{50} }} = \lambda \frac{{E_{{max} } \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{2} }} $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{0} E_{{max} } }} = - \lambda \frac{Dose}{{ED_{50} + Dose}} $$

$$ \frac{{\partial^{2} LL}}{{\partial ED_{50} E_{{max} } }} = \frac{Dose}{{\left( {ED_{50} + Dose} \right)^{2} }}\left( {\lambda - x + \lambda \frac{{E_{{max} } \cdot Dose}}{{ED_{50} + Dose}}} \right) $$

By noting that E(x) = $\lambda$, the above expressions can be simplified and used to determine the expected FIM. Thus we obtain:

$$ \frac{{\partial^{2} LL}}{{\partial E_{0}^{2} }} = - \lambda $$

$$ \frac{{\partial^{2} LL}}{{\partial ED_{50}^{2} }} = - \frac{{\lambda \left( {E_{{max} } \cdot Dose} \right)^{2} }}{{\left( {ED_{50} + Dose} \right)^{4} }} $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{{max} }^{2} }} = - \lambda \left( {\frac{Dose}{{ED_{50} + Dose}}} \right)^{2} $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{0} ED_{50} }} = \frac{{\lambda \cdot E_{max} \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{2} }} $$

$$ \frac{{\partial^{2} LL}}{{\partial E_{0} E_{{max} } }} = - \frac{\lambda \cdot Dose}{{ED_{50} + Dose}} $$

$$ \frac{{\partial^{2} LL}}{{\partial ED_{50} E_{{max} } }} = \frac{{\lambda \cdot E_{{max} } \cdot Dose^{2} }}{{\left( {ED_{50} + Dose} \right)^{3} }} $$

So finally the expected FIM can be written as:

$$ {\text{expected}}\;{\text{FIM}}\; = \;\lambda \left[ {\begin{array}{*{20}c} 1 & { - \frac{{E_{{max} } \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{2} }}} & {\frac{Dose}{{ED_{50} + Dose}}} \\ { - \frac{{E_{{max} } \cdot Dose}}{{\left( {ED_{50} + Dose} \right)^{2} }}} & {\frac{{\left( {E_{{max} } \cdot Dose} \right)^{2} }}{{\left( {ED_{50} + Dose} \right)^{4} }}} & { - \frac{{E_{{max} } \cdot Dose^{2} }}{{\left( {ED_{50} + Dose} \right)^{3} }}} \\ {\frac{Dose}{{ED_{50} + Dose}}} & { - \frac{{E_{{max} } \cdot Dose^{2} }}{{\left( {ED_{50} + Dose} \right)^{3} }}} & {\left( {\frac{Dose}{{ED_{50} + Dose}}} \right)^{2} } \\ \end{array} } \right] $$