MLE of $\theta$ when $X_1,\ldots,X_n$ is a random sample from $U(2\theta,5\theta)$ Let $X_1, X_2, \ldots , X_n$ be a random sample from uniform $(2\theta, 5\theta), \,\theta>0$.
Let $X_{(n)} = \max \{X_1, X_2, \ldots, X_n\}$ and $X_{(1)} = \min \{X_1, X_2, \ldots, X_n\}$. Find the Maximum Likelihood estimator of $\theta$.

I think the answer is
MLE of $\theta$ is $\frac {X_{(n)}}{5}$.  Is it correct?
 A: Suppose $X_{1},X_{2},\ldots, X_{n}$ are iid and $X\sim U(a,b)$ where $x\in[a,b]$.
$$f_{X}(x)=\begin{cases}
1, \,\,\,a\leq x\leq b \\
0, \,\text{   otherwise}
\end{cases}$$
The likelihood function is:
$$L(a,b; x)=\bigg(\frac{1}{b-a}\bigg)^{n}\prod_{i=1}^{n}1_{\{a\leq x_{i}\leq b\}}$$
Now we can't use calculus to solve for a maximum. However, we know that:
$$\prod_{i=1}^{n}1_{\{a\leq x_{i}\leq b\}}=\begin{cases}
1, \,\,\,a\leq x_{i}\leq b, \forall i \\
0, \,\text{   otherwise}
\end{cases}$$
Therefore, to maximize, we need $b\geq x_{(n)}$ and $a\leq x_{(1)}$ (otherwise the product goes to zero).
Similarly, we know $\Big(\frac{1}{b-a}\Big)^{n}$ is a decreasing function of $b$ and increasing function of $a$. Thus, we need the smallest value of $b$ that is at least as large as $x_{(n)}$ and the largest value of $a$ that is smaller than or equal to $x_{(1)}$.
Hence we arrive at:
$$\hat{b}^{\text{MLE}}=x_{(n)}$$
$$\hat{a}^{\text{MLE}}=x_{(1)}$$
With regards to your question, given your solution, I can only guess you meant for your samples to come from $X\sim U(a,5\theta)$, which in the case of the solution provided would be $b=5\theta$ and $\hat{\theta}^{\text{MLE}}=x_{(n)}/5$.

For the case of dependent bounds i.e. $X\sim U(c\theta,d\theta)$, the derivation is as follows:
Given, $0\leq c < d$:
$$L=\bigg(\frac{1}{\theta(d-c)}\bigg)^{n}\prod_{i=1}^{n}1_{\{c\theta\leq x_{i} \leq d\theta\}}$$
and
$$\prod_{i=1}^{n}1_{\{c\theta\leq x_{i} \leq d\theta\}}=\begin{cases}
1, \,\,\,c\theta\leq x_{i}\leq d\theta, \forall i \\
0, \,\text{   otherwise}
\end{cases}$$
Or equivalently,
$$\prod_{i=1}^{n}1_{\{c\theta\leq x_{i} \leq d\theta\}}=\begin{cases}
1, \,\,\,x_{i}/d\leq \theta\leq x_{i}/c, \forall i \\
0, \,\text{   otherwise}
\end{cases}$$
i.e. $c\theta\leq x_{(1)}$ and $d\theta\geq x_{(n)}$ giving $x_{(n)}/d\leq \theta \leq x_{(1)}/c$.
This gives, $\theta\in [x_{(n)}/d,\,x_{(1)}/c]$.
Thus, for any value of $\theta$ in that range, the product is maximized at 1. Given that we know $\Big(\frac{1}{\theta(d-c)}\Big)^{n}$ is a decreasing function of $\theta$, we need to choose the smallest value of $\theta$ in that interval.
Therefore, we arrive at $\hat{\theta}^{\text{MLE}}=x_{(n)}/d$. Interestingly, the MLE is independent of $c$.
