SarahNibs comments on Confidence levels inside and outside an argument

SarahNibs 17 Dec 2010 4:50 UTC
1 point
I let L(B|E) be uniform from x-s/2 to x+s/2 and got that P(B|E) =
$\\frac\{1\}\{s\} \\ln\{\\frac\{A e^\{\\frac\{s\}\{2\}\}\+1\}\{A e^\{\-\\frac\{s\}\{2\}\}\+1\}\}$
where A is the odds if L(B|E)=x. In the limit as s goes to infinity, it looks like the interesting pieces are a term that’s the log of the prior probability dropping off as s grows linearly, plus a term that eventually looks like (1/s)*ln(e^(s/2))=1/2 which means we approach ¹⁄₂.