Some thoughts on uniform prior probabilities when estimating P values and confidence intervals

The fundamental problem both @EvZ and I have is that, as far as he or I can tell, your formula conflicts with well established results from the algebra of random variables.

Specifically, for normal distribution, the variance of a sum (or difference) 2 independent, normally distributed random variables is N(E(X) \pm E(Y), var(x) + var(y)).

The difference between 2 independent random variables entails subtracting their expectations, but adding their variances. (see first link above)

I think both EvZ and I are assuming both estimates come from a Normal distribution. If that is so, then the difference is distributed as a normal N(0, 2). We can predict the probability of any difference by returning to the standard normal N(0,1) by dividing the observed difference by \sqrt{2}.

None of the attempts at formalizing your argument mathematically up to this point, agree with this.