# MLE (Maximum Likelihood Estimator) of Beta Distribution

Let $X_1,\ldots,X_n$ be i.i.d. random variables with a common density function given by:

$f(x\mid\theta)=\theta x^{\theta-1}$

for $x\in[0,1]$ and $\theta>0$.

Clearly this is a $\operatorname{BETA}(\theta,1)$ distribution. Calculate the maximum likelihood estimator of $\theta$.

After going through all the steps with the log likelihood, I end up calculating that the maximum likelihood estimator is $\hat\theta$ below: $$L:=\prod_{i=1}^N\theta x_i^{\theta-1}$$ $$l:=\ln(L)=\ln\left(\prod_{i=1}^N\theta x_i^{\theta-1}\right)=n\ln(\theta)+\sum_{i=1}^n(\theta-1)\ln(x_i)$$ $$\frac{dl}{d\theta}=\frac{n}{\theta}+\sum_{i=1}^n\ln(x_i)$$ $$\hat \theta=\frac{-n}{\sum_{i=1}^n\ln(x_i)}$$

• if you're worried that its negative, i think its okay, the denominator will be non-positive because x is a fraction. The negative in the numerator will cancel. Feb 14, 2018 at 1:50
• Right, but does my method for deriving the MLE look correct? Feb 14, 2018 at 1:52
• More general discussion on Wikipedia article on beta distributions, under MLE. Feb 14, 2018 at 2:09
• @ereHsaWyhsipS : You seem to have correctly found the only critical point of the likelihood function, but being a critical point doesn't always means there's a maximum there. So your argument for the proposition that that's where the absolute maximum occurs is incomplete. Feb 15, 2018 at 3:22
• The MLE is indeed correct. Why doesn't it look right? It is a function of the sufficient statistic $\sum \ln X_i$ after all. Aug 4, 2018 at 17:43

Additional comments: Your answer seems OK. It may be of interest to know that $\hat \theta$ is not unbiased. One can get a rough idea of the distribution of $\hat \theta$ for a particular $\theta$ by simulating many samples of size $n.$ I don't know of a convenient 'unbiasing' constant multiple. The Wikipedia article I linked in my Comment above gives more information.

Here is a simulation for $n = 10$ and $\theta = 5.$

th = 5;  n = 10
th.mle = -n/replicate(10^6, sum(log(rbeta(n, th, 1))))
mean(th.mle)
## 5.555069   # aprx expectation of th.mle > th = 5.
median(th.mle)
## 5.172145


The histogram below shows the simulated distribution of $\hat \theta.$ The vertical red line is at the mean of that distribution, and the green curve is its kernel density estimator (KDE). According to the KDE, its mode is near $4.62.$

den.inf = density(th.mle)
den.inf$x[den.inf$y==max(den.inf$y)] ## 4.624876 hist(th.mle, br=50, prob=T, col="skyblue2", main="") abline(v = mean(th.mle), col="red") lines(density(th.mle), lwd=2, col="darkgreen")  Addendum on Parametric Bootstrap Confidence Interval for$\theta:$In order to find a confidence interval (CI) for$\theta$based on MLE$\hat \theta,$we would like to know the distribution of$V = \frac{\hat \theta}{\theta}.$When that distribution is not readily available, we can use a parametric bootstrap. If we knew the distribution of$V,$then we could find numbers$L$and$U$such that$P(L \le V = \hat\theta/\theta \le U) = 0.95$so that a 95% CI would be of the form$\left(\frac{\hat \theta}{U},\, \frac{\hat\theta}{L}\right).$Because we do not know the distribution of$V$we use a bootstrap procedure to get serviceable approximations$L^*$and$U^*$of$L$and$U.$respectively. To begin, suppose we have a random sample of size$n = 50$from$\mathsf{Beta}(\theta, 1)$where$\theta$is unknown and its observed MLE is$\hat \theta = 6.511.$Entering, the so-called 'bootstrap world'. we take repeated 're-samples of size$n=50$from$\mathsf{Beta}(\hat \theta =6.511, 0),$Then we we find the bootstrap estimate$\hat \theta^*$from each re-sample. Temporarily using the observed MLE$\hat \theta = 6.511$as a proxy for the unknown$\theta,$we find a large number$B$of re-sampled values$V^* = \hat\theta^2/\hat \theta.$Then we use quantiles .02 and .97 of these$V^*$'s as$L^*$and$U^*,$respectively. Returning to the 'real world' the observed MLE$\hat \theta$returns to its original role as an estimator, and the 95% parametric bootstrap CI is$\left(\frac{\hat\theta}{U^*},\, \frac{\hat\theta}{L^*}\right).$The R code, in which re-sampled quantities are denoted by .re instead of$*$, is shown below. For this run with set.seed(213) the 95% CI is$(4.94, 8.69).$Other runs with unspecified seeds using$B=10,000$re-samples of size$n = 50$will give very similar values. [In a real-life application, we would not know whether this CI covers the 'true' value of$\theta.$However, I generated the original 50 observations using parameter value$\theta = 6.5,$so in this demonstration we do know that the CI covers the true parameter value$\theta.$We could have used the probability-symmetric CI with quantiles .025 and .975, but the one shown is a little shorter.] set.seed(213) B = 10000; n = 50; th.mle.obs=6.511 v.re = th.mle.obs/replicate(B, -n/sum(log(rbeta(n,th.mle.obs,1)))) L.re = quantile(v.re, .02); U.re = quantile(v.re, .97) c(th.mle.obs/U.re, th.mle.obs/L.re) ## 98% 3% ## 4.936096 8.691692 ` • Thanks, I appreciate the additional information. But what program did you provide code for that produced that simulation? Feb 14, 2018 at 2:44 • That R statistical software, should have said. Since you're interested in the code, I just appended the code to make the figure. Feb 14, 2018 at 2:45 • Thanks, can't wait to try it out. I appreciate it! Feb 14, 2018 at 2:49 • If you're experimenting with code, you might want to try making a nonparametric bootstrap confidence interval based on your MLE of$\theta.\$ Code is shown in the Addendum. If you have questions about procedure or code, please leave a Comment. Feb 14, 2018 at 7:17