Find maximum-likehood estimator

Question

Say we have $n$ identically distributed random variables with marginal density function given by $$ p_\theta(x)=\dfrac{2\theta}{9^\theta}x^{2\theta-1} $$ for $x\in[0,3]$. It is given that $\theta>0$ (the unknown parameter). I want to find the maximum-likelihood estimtor. I started off with writing down the log likelihood function: $$ \log(\theta;x)=\sum_{i=1}^n\log\left(\dfrac{2\theta}{9^\theta}x_i^{2\theta-1}\right). $$ I differentiated this w.r.t. $\theta$, and setting this to zero we get: $$ \dfrac{\partial}{\partial\theta}\log(\theta;x)=\sum_{i=1}^n\dfrac{9^\theta}{2\theta}\left(\dfrac{2\theta}{9^\theta}\right)^{'}+x_i^{1-2\theta}\cdot(2\theta-1)x^{2\theta-2}=0. $$ I'm not really sure how to proceed from here on. Anyone a hint?

Answer 1

\begin{align} \log(\theta;x) & = \sum_{i=1}^n\log\left(\dfrac{2\theta}{9^\theta}x_i^{2\theta-1}\right) \\ & = \sum_{i=1}^n [\log 2 + \log \theta - \theta \log 9 + (2 \theta - 1) \log x_i] \\ \frac{\partial \log(\theta;x)}{\partial \theta} & = \frac{n}{\theta} - n \log 9 + 2 \sum_{i=1}^n \log x_i = 0. \end{align} Solving for $\theta$ we get $$ \theta_\text{MLE} = \frac{1}{\log 9 - \frac{2}{n} \sum_{i=1}^n \log x_i}. $$