Better Students Ask More Questions.
Bayesian analysis Symmetric 90% and 95% confidence intervals for a random...
Topics: Math, Statistics
Symmetric 90% and 95% confidence intervals for a random variable `p` (where `p` is the probability of an event) are given. If a subjective probability `p_0` an effective number of observations `n_0` upon which it is based are also given, how does one revise the subjective probability in the light of the data? (Assume a conjugate prior distribution)
1 Answer | add yours
The confidence intervals given will be of the form
`hat(p) pm z_(1-alpha/2)sqrt((hat(p)(1-hat(p)))/n)`
where `z_(1-alpha/2)` is the 100(alpha/2)th percentile of the standard Normal distribution. For symmetric 90% intervals` ` `alpha = 0.1` and for symmetric 95% intervals `alpha = 0.05` so that the corresponding percentiles would be
`z(0.95) = 1.64 ` , `z(0.975) = 1.96`
The center of the interval will give you `hat(p)`` ` and the distance between the center and each end will give you `z_(1-alpha/2)sqrt((hat(p)(1-hat(p)))/n)` . Knowing `z_(1-alpha/2)` and an estimate of `p`, `hat(p)`, one can then obtain `n`.
The 'subjective' probability defines what the prior distribution for the probability `p` (a random variable in the Bayesian paradigm) is. If that distribution is assumed to be a Beta distribution (so that it is a 'conjugate prior') then it will be of the form Beta(`alpha`,`beta`) where `alpha-1` is the number of hypothetical successes and `beta -1` the hypothetical number of failures. If the subjective 'expected' probability is `p_0` and is worth `n_0` hypothetical observations (a larger `n_0` implies a stronger belief in `p_0`) then
`alpha + beta - 2 = n_0` and `alpha - 1 = n_0p_0`
Therefore `alpha = (n_0p_0) + 1` and `beta = n_0 + 2 - (n_0p_0) - 1`
`= n_0(1-p_0) + 1`
To update this prior in the light of the data add the number of successes to `alpha` and the number of failures to `beta`. The number of successes is `nhat(p)` (`n` and `hat(p)` obtained earlier from the confidence intervals) and the number of failures `n(1-hat(p))`. So the posterior distribution after updating the prior in the light of the data is Beta(`alpha',beta'`) where
`alpha' = alpha + nhat(p) = (n_0p_0) + 1 + nhat(p) `
`beta' = beta + n(1-hat(p)) = n_0(1-p_0) + 1 + n(1-hat(p))`
Answer: update the Beta prior in the light of the (binary) data to give a Beta posterior
Posted by mathsworkmusic on August 26, 2013 at 6:45 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.