0
$\begingroup$

I am reading an article from Statistics in Medicine ("Tutorial in biostatistics: multiple hypothesis testing in genomics"). For my question it is only relevant to know that $V$ denote the number of falsely rejected hypotheses in a set of $m$ hypotheses, $R$ the number of all rejected hypotheses and $Q:=V/R$ the false discovery proportion (for $R>0$ and $0$ else).

The authors state that since $0 \le Q \le 1$, $$E(Q) \le P(Q>0) = P(V>0).$$

That is, the false discovery rate ($E(Q)$) is always smaller or equal the family wise error rate ($P(Q>0)$). Why is this a sufficient proof?

$\endgroup$
  • 1
    $\begingroup$ see stats.stackexchange.com/questions/59681/… $\endgroup$ – Christoph Hanck Jun 28 '16 at 14:23
  • $\begingroup$ Interesting but somewhat abstract reasoning. It would make more sense to me to evaluate $E(V/R)$ directly and show it is smaller $1-P(V=0)$. I wonder whether that is possible at all. $\endgroup$ – tomka Jun 28 '16 at 14:47