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Closely related to: Calculate the confidence interval for the mean of a beta distribution

I am fitting a a $\operatorname{Beta}(\alpha, \beta)$ distribution to some data using Matlab's betafit function which has produced confidence intervals for $\alpha$ and $\beta$. I am wondering whether it is possible to derive confidence intervals for the implied mean parameter $\mu = \frac{\alpha}{\alpha + \beta}$ from these.

Note that my question is different to the one I linked previously as that one explicitly relies on parametrizing the Beta log-likelihood function in terms of $\mu$ and the nuisance parameter $\sigma$, and then calculates the CI for $\mu$ using profile likelihood.

In addition, I do not understand the Matlab documentation when it says:

The confidence intervals are based on a normal approximation for the distribution of the logs of the parameter estimates.

What distribution does the log of the parameter estimates take?

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  • $\begingroup$ This question requires the self study tag. $\endgroup$ – Michael Chernick Feb 15 '17 at 13:18
  • $\begingroup$ don't see why it should have a self study tag as it is not "A routine question from a textbook, course, or test used for a class or self-study." $\endgroup$ – Alex Feb 15 '17 at 22:42
  • $\begingroup$ You just explained why it does need the self study tag. $\endgroup$ – Michael Chernick Feb 15 '17 at 22:45
  • $\begingroup$ Please point me to where this is a routine question and I will look up the relevant resource and see whether I can solve it from there. $\endgroup$ – Alex Feb 15 '17 at 22:46
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According to the document, Matlab estimates the parameters for Beta distribution with the maximum likelihood. In this case, I am afraid that the sampling distributions for the estimators are not available. Let $\{X_i\}_{i=1}^n$ be a set of i.i.d. observations from standard Beta distribution $B(\alpha,\beta)$, obtaining the MLEs $\hat{\alpha}$ and $\hat{\beta}$ requires one to solve simultaneous equations \begin{align}\psi(\hat{\alpha})-\psi(\hat{\alpha}+\hat{\beta})&=\frac{1}{n}\sum_{i=1}^n\ln (X_i),\\\psi(\hat{\beta})-\psi(\hat{\alpha}+\hat{\beta})&=\frac{1}{n}\sum_{i=1}^n\ln (1-X_i),\end{align} where $\psi$ denotes the digamma function (see e.g. Gnanadesikan, Pinkham, and Hughes' Maximum Likelihood Estimation of the Parameters of the Beta Distribution from Smallest Order Statistics for more details). As far as I know, no general analytical solutions are available, let alone the exact distributions of $\hat{\alpha}$ and $\hat{\beta}$. It is possible to solve them in special cases however, e.g. when $\beta$ are known to be exactly $1$, by properties of digamma function the first equation can be reduced to $$\hat{\alpha}=-\frac{n}{\sum_{i=1}^n\ln (X_i)},$$ where one can derive that $-\sum_{i=1}^n\ln (X_i)\sim\varGamma(n,1/\alpha)$, which implies $\hat{\alpha}$ has scaled inverse-Gamma distribution. Other than that, one has to resort to the asymptotic normality of MLEs to make inferences on $\hat{\alpha}$ and $\hat{\beta}$, also see e.g. Lau and Lau's Effective procedures for estimating beta distribution's parameters and Their confidence intervals for procedure using Pearson distribution. In addition, one may utilize the normality to derive confidence intervals for the mean, since $$\hat{\mu}=\frac{\hat{\alpha}}{\hat{\alpha}+\hat{\beta}}$$ will have correlated normal ratio distribution if both $\hat{\alpha}$ and $\hat{\beta}$ are normal.

For your second question, I take that logs of estimates are used most likely to cope with the problem of parameter space. Note that both $\alpha$ and $\beta$ must be in $(0,\infty)$. But if your $\hat{\alpha}=1$ and you use normal approximation directly, you may end up with confidence interval such as $(-0.5,2.5)$. You can get around this by mapping to log-scale and then back, though I am not sure how accurate the result will be then.

As an interesting side note, the book (Hahn and Shapiro's Statistical Models in Engineering) cited in the documentation actually talked about method of moments estimate instead of MLE, it is not clear to me why it is cited here.

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  • $\begingroup$ thanks for the detailed analysis of each part of my question. $\endgroup$ – Alex Feb 16 '17 at 21:22

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