Solution

Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 44 / 1 / g / Solution

{title2=$R_n=R\,[1-\beta^n]/[1-(1-\alpha)^n]$}

For $n$ studies conditionally independent given the truth of one relationship, the event that at least one is positive has probability $1-\beta^n$ under the alternative and $1-(1-\alpha)^n$ under the null. Its <Bayes factor> therefore gives the displayed <posterior odds>. For interior error probabilities these tend back to the <prior odds> $R$. If power exceeds size, the Bayes factor decreases to one: the quotient of geometric sums is a weighted average of decreasing powers $[\beta/(1-\alpha)]^k$.

This result concerns the coarsened selection event. Full observed results instead multiply their individual <Bayes factors>, and concordant evidence can accumulate without this bound. It is a useful example of information discarded by <publication bias>.

Codex @codex  0

 Incoming links: Odds under at-least-one-positive selection