Ex. 18.18

Ex. 18.18

Use result (18.53) to show that

$$\text{pFDR}=\frac{{\pi }_0\cdot \{\text{Type I error of }\mathrm{\Gamma }\}}{{\pi }_0\cdot \{\text{Type I error of }\mathrm{\Gamma }\}+{\pi }_1\cdot \{\text{Power of }\mathrm{\Gamma }\}}$$

(Storey, 2003 \cite{storey2003positive}).

Soln. 18.18

From (18.53) in the text we obtain

$$\begin{array}{rcl}\text{pFDR}(\mathrm{\Gamma })& =& \text{Pr}(Z_j=0|t_j\in \mathrm{\Gamma })\\ & =& \frac{{\pi }_0\cdot \text{Pr}(t_j\in \mathrm{\Gamma }|Z_j=0)}{{\pi }_0\cdot \text{Pr}(t_j\in \mathrm{\Gamma }|Z_j=0)+{\pi }_1\cdot \text{Pr}(t_j\in \mathrm{\Gamma }|Z_j=1)}\\ & =& \frac{{\pi }_0\cdot \{\text{Type I error of }\mathrm{\Gamma }\}}{{\pi }_0\cdot \{\text{Type I error of }\mathrm{\Gamma }\}+{\pi }_1\cdot \{\text{Power of }\mathrm{\Gamma }\}}.\end{array}$$