 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-44/1/g/solution

Let

A_{n}

denote the event that at least one study is positive. Conditional independence gives

P (A_{n} ∣ T) = 1 - β^{n}

and

P (A_{n} ∣ F) = 1 - (1 - α)^{n}

. Updating the prior odds by this event's Bayes factor gives

\frac{P ( T ∣ A _{n} )}{P ( F ∣ A _{n} )} = R \frac{1 - β ^{n}}{1 - ( 1 - α ) ^{n}} .

These are the odds under at-least-one-positive selection, not the odds using every study's actual result.

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