MLE of uniform distribution with dependent bounds The normal derivation of the MLE for the uniform is as follows:
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:
$L(a,b; x)=\Big(\frac{1}{b-a}\Big)^{n}\prod_{i=1}^{n}1_{\{a\leq x_{i}\leq b\}}$
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:
$\hat{b}^{\text{MLE}}=x_{(n)}$
$\hat{a}^{\text{MLE}}=x_{(1)}$.

Now, let $X\sim U(c\theta,d\theta)$ where $c,d$ are constants with $c<d$. So the upper and lower bounds of the uniform are now dependent or related.
Considering the above derivation, this would imply our MLE for $\theta$ is:
Lower bound: $\hat{\theta}^{\text{MLE}}=x_{(1)}/c$
Upper bound: $\hat{\theta}^{\text{MLE}}=x_{(n)}/d$
So, what is $\hat{\theta}^{\text{MLE}}$? Is it an interval? Does the derivation have to jointly consider both bounds?
 A: I think the problem assumes $c$ and $d$ are known constants, and $\theta$ is the only unknown.
The likelihood is $(d-c)^{-n} \theta^{-n} \prod_{i=1}^n 1_{\{\theta c \le x_i \le \theta d\}}$.
Note that each indicator can be rewritten as $1_{\{x_i/d\le \theta \le x_i/c\}}$, so we need to maximize the likelihood over $\theta \in [x_{(n)}/d, x_{(1)}/c]$. Since $\theta^{-n}$ is decreasing in $\theta$, the MLE is $x_{(n)}/d$. The result is intuitive: the smaller the $\theta$, the smaller the interval $[\theta c, \theta d]$, so the density is larger on this interval. The MLE chooses the smallest $\theta$ such that all the points $x_i$ lie in the interval.
A: The MLE is not well defined in your parametrization of the distribution, because if $c$, $d$, and $\theta$ are unknown such that $c < d$ and $\theta > 0$, then the system $$c\theta = x_{(1)}, \quad d\theta = x_{(n)}$$ is underdetermined:  you can choose any $\hat \theta > 0$, and there is a corresponding choice of $\hat c, \hat d$ that results in $$(\hat d - \hat c) \hat \theta = x_{(n)} - x_{(1)}.$$  Put another way, your parametrization does not permit you to distinguish whether your sample was generated, for example, from a $\operatorname{Uniform}(c = 1, d = 3, \theta = 2)$ distribution, or a $\operatorname{Uniform}(c = 2, d = 6, \theta = 1)$ distribution, because both are uniform on $[2,6]$.
