Ex. 18.20

Proof of result (18.53). Write

\[\begin{eqnarray} \text{pFDR} &=& E\left(\frac{V}{R}|R > 0\right)\non\\ &=&\sum_{k=1}^ME\left[\frac{V}{R}|R=k\right]\text{Pr}(R=k|R > 0)\non \end{eqnarray}\]

Use the fact that given \(R=k\), \(V\) is a binomial random variable, with \(k\) trials and probability of success \(\text{Pr}(H=0|T\in \Gamma)\), to complete the proof.

Soln. 18.20

Note that given \(R=k\), \(V\) is a binomial random variable, with \(k\) trials and probability of success \(\text{Pr}(H=0|T\in \Gamma)\), we have

\[\begin{equation} E[V|R=k] = k \cdot \text{Pr}(H=0|T\in \Gamma).\non \end{equation}\]

Therefore,

\[\begin{eqnarray} \text{pFDR}(\Gamma) &=&\sum_{k=1}^M\frac{1}{k} \cdot k \cdot \text{Pr}(H=0|T\in \Gamma)\text{Pr}(R=k|R > 0)\non\\ &=&\text{Pr}(H=0|T\in \Gamma)\sum_{k=1}^M\text{Pr}(R=k|R > 0)\non\\ &=&\text{Pr}(H=0|T\in \Gamma).\non \end{eqnarray}\]

Note the notation \(H\) comes from \cite{storey2003positive}, and it should be \(Z\) defined (18.51) in our context.