4 ms·
I think the formula should be (n+1)/(n+m+1) which should correspond to the mean of a binomial distribution with a uniform prior. So it's adding 1 to each count
by yobbo 1y ago
I think the formula should be (n+1)/(n+m+1) which should correspond to the mean of a binomial distribution with a uniform prior. So it's adding 1 to each count of observations.
This is probably the formula to memorise and check against.
- kqr 1y agoYou mean (n+1)/(n+m+2) and yes, this is Laplace's rule of succession. It won't give you a confidence interval, but it gives you a posterior point estimate. If you want a rough 95 % confidence interval without complicated maths, the Agresti–Coull interval is useful. It's computed as if the distribution was normal, but pretending there were two more successes and failures instead.
- yobbo 1y agoYep, you're correct. It should be (n+1)/(n+m+2). If you have access to a machine or lookup tables, you might as well plug in the values for the distribution Beta(1+n,1+m) which should correspond to the joint density. (The formula above corresponds to the mean of this distribution, so it's probably right but I haven't work it through myself now ...)
- usgroup 1y agoSee the Jeffrey posterior section here: https://en.m.wikipedia.org/wiki/Binomial_proportion_confidence_interval https://en.m.wikipedia.org/wiki/Binomial_proportion_confiden... The blog post uses a non informative Jeffrey prior.