MLE Estimator of a uniform distribution with two parameters. I'm currently stuck on a problem to find the maximum likelihood estimator of a uniform distribution. The problem is the following:
"For the uniform distribution unif $[2 \theta_{1} \theta_{2},\theta_{1}^2+\theta_{2}^2]$ find the estimators for $\theta_{1}$ and $\theta_{2}$ where ($\theta_{2}>\theta_{1}>0$) using the Maximum Likelihood estimator."
I've seen many question about the topic. so far I've computed the likelihood and the log likelihood function:
$Likelihood function = (\theta_{1}+\theta_{2})^{-2}$
$LogLikelihoodfunction = -2log(\theta_{1}+\theta_{2})$
from this how should I continue?
I'm really looking forward to some help!  Thank you
 A: Your likelihood (and thus log-likelihood) is incorrect.
The density of a single observation from this distribution is $$f_X(x) = \frac{1}{(\theta_1^2 + \theta_2^2) - 2\theta_1\theta_2} \mathbb{1}(2\theta_1 \theta_2 \le x \le \theta_1^2 + \theta_2^2).$$  The denominator is the square of the difference:  $$f_X(x) = (\theta_2 - \theta_1)^{-2} \mathbb{1}(2\theta_1 \theta_2 \le x \le \theta_1^2 + \theta_2^2).$$  Therefore, the likelihood given a sample $\boldsymbol x = (x_1, \ldots, x_n)$ is $$\mathcal{L}(\theta_1, \theta_2 \mid \boldsymbol x) = \prod_{i=1}^n f_X(x_i) = (\theta_2 - \theta_1)^{-2n} \mathbb{1}(2\theta_1 \theta_2 \le x_{(1)} \le x_{(n)} \le \theta_1^2 + \theta_2^2).$$  The log-likelihood is $$\ell(\theta_1, \theta_2 \mid \boldsymbol x) = -2n \log (\theta_2 - \theta_1), \quad 2\theta_1 \theta_2 \le x_{(1)} \le x_{(n)} \le \theta_1^2 + \theta_2^2.$$  The partial derivatives with respect to $\theta_1, \theta_2$ are $$\frac{\partial \ell}{\partial \theta_1} = \frac{2n}{\theta_2 - \theta_1}, \quad \frac{\partial \ell}{\partial \theta_2} = \frac{2n}{\theta_1 - \theta_2}.$$  Note that for any choice of $\theta_1, \theta_2$ such that $\theta_2 > \theta_1 > 0$, neither partial derivative has a critical point.  Therefore, the log-likelihood is maximized for a choice of $(\theta_1, \theta_2)$ that minimizes the difference $\theta_2 - \theta_1$ subject to the constraint $$2\theta_1 \theta_2 \le x_{(1)} \le x_{(n)} \le \theta_1^2 + \theta_2^2.$$  Consider these conditions separately:  $$2\theta_1 \theta_2 = x_{(1)}$$ describes a hyperbola in the $(\theta_1, \theta_2)$ coordinate system, and similarly, $$\theta_1^2 + \theta_2^2 = x_{(n)}$$ describes a circle of radius $\sqrt{x_{(n)}}$.  The region of the plane that satisfies both inequalities is therefore a region in the first quadrant such that $\theta_1 < \theta_2$, below the hyperbola and above (outside) the circle.  Over this region, it is clear that the point with minimal $\theta_2 - \theta_1$ is located at the intersection of the two curves.  Therefore, the MLE is the solution to the system $$2\theta_1 \theta_2 = x_{(1)}, \quad \theta_1^2 + \theta_2^2 = x_{(n)}.$$  Note: the above is not a completely rigorous argument; it is only intended as motivation.  You will need to prove it rigorously using other methods.  I leave it to you to explicitly solve this system in terms of the minimum and maximum order statistics and fill in the details of the proof.
