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If $(x_1,\cdots,x_n)$ is a sample from a $\mathcal{U}(\theta_1,\theta_2)$ (uniform) distribution with $$\Omega=\{(\theta_1,\theta_2)\in\mathbb{R}^2:\theta_1<\theta_2\}$$ determine the MLE of $(\theta_1,\theta_2)$ without using calculus.

OK, I know that, $$f(x|(\theta_1,\theta_2))=\frac{1}{(\theta_2-\theta_1)}$$ and

$$L(\theta_1,\theta_2)=\prod_{i=1}^n\frac{1}{(\theta_2-\theta_1)}=\frac{1}{(\theta_2-\theta_1)^n}$$

Now I need to maximize the $\theta_2$ and minimize the $\theta_1$ values in order to miximize the likelihood. I know how to find those value using LL (Log likelihood function) and partial derivatives. But how to find them without calculus? I have no idea where to start.

falamiw
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2 Answers2

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The MLE is when $\theta_2=x_{(n)},\theta_1=x_{(1)}$. This is roughly because $\frac 1 {(\theta_2-\theta_1)^n}\le\frac 1 {(X_{(n)}-X_{(1)})^n}$ for all possible other choices of thetas. To see this, consider that the difference between the max and min values is 5. If you choose the max to be theta_2, and the min to be theta_1, then the likelihood is 1/5^n. Anything beyond these "boundaries" of data and you will get a fraction "smaller" than 1/5, so the fraction to the power of n will be "smaller" than 1/5^n, and you would get a "smaller" likelihood. Intuitively, these data are the most likely when the uniform distribution has boundaries of the smallest data point and the largest data point.

You should put some indicator variables to be more correct, i.e. for an individual data point, $f(x|\theta_1,\theta_2)=\frac{1}{\theta_2-\theta_1}1_{\theta_1\le x\le\theta_2}$

Vons
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There is so much you are missing that it is difficult for me to know where to begin addressing your issues.

First of all, if your density $$f(x \mid \theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1}$$ is not actually a function of $x$, then you are saying that any value of $x$ is equally likely, even if it is not in the interval $[\theta_1, \theta_2]$. Is that really what you want to say?

Second, do you know anything about sufficient statistics? Trying to maximize a likelihood function without understanding data reduction and sufficiency is like trying to do calculus without knowing basic algebra.

Third, your likelihood function is nonsense. It doesn't even depend on the sample, so what function of the sample will produce a maximum?

heropup
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  • To be honest my confusion get stronger when I see the likelihood function doesn't depend on the sample/data. I know a bit of sufficient statistics which is a function of sample/data whose value contains all the information needed to compute any estimate of the parameter. But I truly not understand that very much. Like if $X_i \sim N(\mu,\sigma^2)$ then the sufficient statistic I found is $T(X)=x_1+\cdots+x_n$ for unknown parameter $\mu$. What does it actually mean? And the problem I post is the exercise $6.2.24$ of Probability and Statistics by Evans M., Rosenthal J.. – falamiw Feb 02 '21 at 08:11
  • I got one of your answer where you clear my doubt, Thanks @heropup – falamiw Feb 02 '21 at 17:28