I was given a question to find the estimator $\theta$ by moments method of the function: $$f_{X}(x)= \begin{Bmatrix} \frac{3x^2}{2\theta^3} & -\theta<x<\theta\\ 0& \text{else} \end{Bmatrix}$$
The first moment is zero, and by the second moment I got that: $$\hat{\theta}= \sqrt{\frac{5S^2_x}{3}}$$
Now if I were to find the estimator with MLE, for the sports, I would find the function $L(\theta)$:
$$L(\theta) = \prod_{i=1}^n \frac{3x_i^2}{2\theta^3} \times{I_{(-\theta<x_i<\theta)}}$$ Now I'll play a little with the condition of the Indicator variable I so that: $$ L(\theta) = \prod_{i=1}^{n}\Bigl(\frac{3x_i^2}{2\theta^3}\Bigr) \times I_{(-\theta<\min(x_i))} \times I_{(\max(x_i)<\theta}$$ Suppose I ignore the constant $ \prod_{i=1}^{n}\Bigl(\frac{3x_i^2}{2}\Bigr)$ part, and I'm only trying to convert the condition on $x$ to a condition of $\theta$, but I'm in a little problem here: $$-\theta<\min(x_i) \rightarrow-\min(x_i)<\theta$$ And $$\max(x_i)<\theta $$ Eventually I can only tell what is smallr than $\theta$, but not what bounds it from above. Virtually I'm looking for a condition such as $$f(x_i) <\theta<g(x_i)$$ where $f$ and $g$ are some functions that probably involve $\max$ or $\min$. Can anyone direct me where can I extract a condition of an upper bound?
\begin{align} \require{cancel} L(\theta) = {} & \prod_{i=1}^n \xcancel{\frac{3x^2}{2\theta^3}} \times{I_{(-\theta<x_i<\theta)}} \\[8pt] L(\theta) = {} & \prod_{i=1}^n \frac{3x_i^2}{2\theta^3} \times{I_{(-\theta<x_i<\theta)}} \\[8pt] = {} & \text{constant} \times \frac 1 {(2\theta^3)^n} \times I_{\theta > \max_i\{x_i\}} \times I_{\theta>-\min_i\{x_i\}} \\[8pt] & \text{(where “constant'' means not depending on $\theta$)} \end{align}
So $L(\theta)$ increases as $\theta>0$ decreases, UNTIL $\theta$ gets as small as $\max\{\pm x_1,\ldots,\pm x_n\}.$