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

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.

• 1. Your likelihood function is incomplete - you've forgotten to include the fact that $\theta \ge x_i$ for each $i$. The complete function should be $\mathcal{L}_\theta \cdot \prod [\theta \ge x_i]$, where $\mathcal{L}_\theta$ is the function above and $[\cdot]$ is the indicator function. 2. In the range $\theta \ge \max x_i$, the corrected function and the original function agree, and you've determined that the likelihood function is positive and strictly decreasing. In the range $\theta < \max x_i$, it is zero once you include the correction. Ergo, the maximiser is $\theta^* = \max x_i$. – stochasticboy321 Oct 30 '16 at 21:03
• math.stackexchange.com/q/383587/321264 – StubbornAtom Feb 6 '20 at 19:46

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