Ex. 18.18

Ex. 18.18

Use result (18.53) to show that

\[\begin{equation} \text{pFDR} = \frac{\pi_0\cdot \{\text{Type I error of } \Gamma\} }{\pi_0\cdot \{\text{Type I error of } \Gamma\} + \pi_1\cdot\{\text{Power of }\Gamma\}}\non \end{equation}\]

(Storey, 2003 \cite{storey2003positive}).

Soln. 18.18

From (18.53) in the text we obtain

\[\begin{eqnarray} \text{pFDR}(\Gamma) &=& \text{Pr}(Z_j=0|t_j\in \Gamma)\non\\ &=&\frac{\pi_0\cdot\text{Pr}(t_j\in \Gamma|Z_j=0)}{\pi_0\cdot\text{Pr}(t_j\in \Gamma|Z_j=0) + \pi_1\cdot\text{Pr}(t_j\in \Gamma|Z_j=1)}\non\\ &=&\frac{\pi_0\cdot \{\text{Type I error of } \Gamma\} }{\pi_0\cdot \{\text{Type I error of } \Gamma\} + \pi_1\cdot\{\text{Power of }\Gamma\}}.\non \end{eqnarray}\]