Maximum likelihood estimator for $\theta$ when $f_{\theta}(x)=\frac{4x^3}{\theta ^4}1_{0\le x\le \theta}$

Question

Let $X_1, \ldots , X_n$ be i.i.d. sample from the absolute continuous distribution given by the p.d.f. $$f_{\theta}(x) = \begin{cases} \frac{4x^3}{\theta ^4},\qquad \text{if } 0\leq x \leq \theta\\ 0, \qquad \text{ otherwise } \end{cases}\quad,$$ where $\theta > 0$.

Find the maximum likelihood estimator of $\theta$.

So far I got: $$\mathcal{L_\theta}(x)=\prod\limits^n_{i=1}\frac{4x_i^3}{\theta ^4}=\theta^{-4n}4^n \left(\prod\limits^n_{i=1}x_i \right)^3.$$ $$\ln[\mathcal{L_\theta}(x)] \equiv -4n \ln (\theta) + n\ln 4 + 3\sum\limits^n_{i=1}\ln(x_i)$$ \begin{align}{\partial L(\theta) \over \partial \theta} =\frac{-4n}{\theta}=0\end{align} What can be the Maximum likelihood estimator for $\theta$ ? Thanks.

Answer 1

If $\exists i \in \left\{ 1, \ldots, n\right\}, x_i > \theta, \mathcal{L_θ}(x)=0$

If $\forall i \in \left\{ 1, \ldots, n\right\}, x_i \leq \theta, $

$$\mathcal{L_θ}(x)=\prod\limits^n_{i=1}\frac{4x_i^3}{θ ^4}=θ^{-4n}4^n (\prod\limits^n_{i=1}x_i)^3\geq0.$$

We want $\mathcal{L_θ}(x)$ to be as large as possible and hence we have $\theta \geq \max_i x_i$. Also, When $\theta \geq \max(x_i)$, it is a decreasing function.

Hence $\theta_{ML}=\max(x_i)$