p value and posterior probability under uniform priors I'm reading a paper by Andrew Gelman. On page 3, it says,

Let θ be the true (population) difference in sex ratios
  of attractive and less attractive parents. Then the data under
  discussion (with a two-sided P value of 0.2), combined with a
  uniform prior on θ, yield a 90% posterior probability that θ is
  positive.

I don't quite get the 90% part. I know it's 1-0.2/2, but why? 
 A: Not sure if it makes sense to post a response 4 months after the original question was asked; but I stumbled across this post searching for something else, and thought an answer might be useful to others. Please feel free to delete if not deemed suitable.
The key concept here is the difference between the $p$-value (as a probability conditional on the null $H_0$), and the posterior probability making a statement about the alternative hypothesis $H_1$.
The logic is as follows:


*

*A two-sided $p$-value means that we are assessing evidence in support of either $\theta > 0$ or $\theta < 0$ (or in short, $|\theta| > 0$).

*The one-sided "greater than" $p$-value is then just half the two-sided $p$-value.

*Now, the one-sided $p$-value is $p = Pr(X \geq x | H_0)$. Assuming uniform priors, we have $Pr(X \geq x | H_0) = Pr(H_0 | X \geq x)$.

*Then, $Pr(H_1 | X \geq x) = 1 - Pr(H_0 | X \geq x)$, so the posterior probability for $\theta > 0$ is $Pr(H_1 | X \geq x) = 1 - 0.5 \times 0.2 = 0.9$.


So the posterior probability that $\theta > 0$ is 90%.
