# MLE for Beta distribution, with $\beta$ = 3

I'm trying to calculate the Maximum-Likelihood Estimator for $$\alpha$$, using the beta distribution with $$\beta = 3$$. I'm kind of stuck at the last bit. Perhaps I've made a mistake somewhere, or this isn't possible.

$$f(x) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)} x^{\alpha -1} (1-x)^{\beta-1} \hspace{0.5cm} \alpha, \beta >0 \hspace{0.5cm} x \in (0,1)$$

Assuming $$\beta=3$$, the PDF simplifies to the following: $$f(x) = \frac{\Gamma(\alpha + 3)}{\Gamma(\alpha) \Gamma(3)} x^{\alpha -1} (1-x)^{3-1} \hspace{0.5cm} \alpha>0 \hspace{0.5cm} x \in (0,1)$$ The Likelihood function for $$\alpha$$, calculated in general as the product of the marginal densities for each observation, given a certain value for the parameter of interest, is calculated as follows: \begin{align*} \mathcal{L}(\alpha |\mathbf{X}) &= \prod_{i=1}^{n} f(x_i| \alpha) \\ & = \frac{\Gamma(\alpha + 3)}{\Gamma(\alpha) \Gamma(3)} x_1^{\alpha-1} (1-x_1)^{3-1} \frac{\Gamma(\alpha + 3)}{\Gamma(\alpha) \Gamma(3)} x_2^{\alpha-1} (1-x_2)^{3-1} \cdots \frac{\Gamma(\alpha + 3)}{\Gamma(\alpha) \Gamma(3)} x_n^{\alpha-1} (1-x_n)^{3-1} \\ & = \left (\frac{\Gamma(\alpha + 3)}{\Gamma(\alpha) \Gamma(3)} \right)^n \prod_{i=1}^{n} x_i^{\alpha-1} \prod_{i=1}^{n} (1-x_i)^{2}\\ \end{align*} The natural logarithim of the Likelihood function is usually used to simplify the maximisation problem. The log-likelihood function is calculated as follows \begin{align*} \text{log}[\mathcal{L}(\alpha|\mathbf{X})] &= \ell(\alpha|\mathbf{X}) \\ &= n \text{log}[\Gamma(\alpha + 3)] - n \text{log}[\Gamma(\alpha)] - n \text{log}[\Gamma(3)] + (\alpha-1)\sum_{i=1}^{n} \text{log}(x_i) + 2 \sum_{i=1}^{n} \text{log}(1-x_i)\\ \end{align*}

The maximum point of $$\ell(\alpha|\mathbf{X})$$ is found by taking the partial derivative of the function with respect to $$\alpha$$; equating the expression to zero, and solving for $$\hat{\alpha}$$, yielding the MLE for $$\alpha$$. \begin{align*} \frac{\partial}{\partial \alpha} \ell(\alpha|\mathbf{X}) &= \frac{n \Gamma '(3+\alpha)}{\Gamma(3+\alpha)} + \sum_{i=1}^{n}\text{log}(1-x_i) = 0 \\ \end{align*}

Where: $$\Gamma '(m) = \frac{d}{dx}\Gamma(m) = \int_{0}^{\infty} t^{m-1}e^{-t}\text{log}(t)dt$$

Any ideas about solving for $$\alpha$$?

• You have to solve this numerically. See related post: stats.stackexchange.com/questions/352161/…. – StubbornAtom Apr 16 at 8:45
• In the second last step, it should be $\sum \log x_i$ – gunes Apr 16 at 8:46
• Thanks a ton! Also, thanks for spotting my error in the second last line :) – Sky Apr 16 at 9:02