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This question already has an answer here:

I have elementary knowledge in statistics. I'm trying to estimate the confidence interval for mean of a beta distribution as specified in this article using log likelihood estimation given alpha, beta and sample size. This is closely related to this question. The main difference is I have the alpha, beta and the sample size estimated using log likelihood from the data.

Lets suppose I have estimated $(\alpha,\beta)$ for two different data with different sample sizes:

  1. ($\alpha,\beta$) = (0.688,3.806) using a sample size $n=500$ and
  2. ($\alpha,\beta$) = (0.704,1.182) using a sample size $n=1500$

    The mean of a beta distribution is $$E(X) = \frac{\alpha}{(\alpha+\beta)}$$ and the variance is given by $$Variance(x) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$$

the mean and variance for 2 data are as follows:

  1. mean = 0.15 and variance = 0.0236
  2. mean = 0.37 and variance = 0.0810

Edit:

Using Bootstrap we could estimate the confidence interval of the beta distribution, This is based on the answer provided earlier to a similar question:

# Sample size
n = 50

# Parameters of the beta distribution
alpha = 0.688 
beta = 3.806 

# Simulate some data
set.seed(1)
x = rbeta(n, alpha, beta)

curve(dbeta(x,alpha,beta))


library(simpleboot)

x.boot = one.boot(x, mean, R=10^4)
hist(x.boot)               
boot.ci(x.boot, type="bca")

Below is the confidence interval for beta distribution when $\alpha = 0.688$ and $\beta = 3.806$ with a mean of 0.15.

Intervals : 
Level       BCa          
95%   ( 0.1159,  0.2082 )  
Calculations and Intervals on Original Scale

My question is around using the alpha, beta, and estimated mean and variance to estimate the CI of the mean as opposed to using bootstrap estimates. For normal distribution you could estimate CI given a mean, variance and sample size. How do I do the same for Beta distribution ?

How do I estimate the confidence interval for the mean, given the alpha, beta, and estimated mean and variance?

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marked as duplicate by whuber May 11 '15 at 19:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ If alpha and beta are estimated from a sample, that's not "known", but "estimated". $\endgroup$ – Glen_b May 11 '15 at 15:14
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    $\begingroup$ I cannot see any difference between this question and the one you say is only "closely related." They are identical in that they ask how to find a CI for the mean based on ML estimates of the parameters. $\endgroup$ – whuber May 11 '15 at 19:35
  • $\begingroup$ @whuber, Given the mean and variance of the beta distribution, can you please help me calculate the CI ? Thanks $\endgroup$ – forecaster May 11 '15 at 20:06
  • $\begingroup$ I cannot, nor can anyone else, because you don't supply enough information. Have you applied the approaches and code offered in this answer to the duplicate? $\endgroup$ – whuber May 11 '15 at 20:43
  • $\begingroup$ Yes, I did using bootstrap test outlined in the duplicate.Forgive my naivety, I was under the assumption that mean, variance and sample size is sufficient for calculating confidence interval. If more information is required, can you please let me know what those information would be. The article uses probability mixture models - beta and geometric to estimate alpha and bet parameters for beta distribution. Thanks $\endgroup$ – forecaster May 11 '15 at 20:48