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Let the random variable $X$ have a uniform density given by $$ f(x;\mu,\sigma)=\frac{1}{2\sqrt 3\sigma}I_{[\mu-\sqrt 3\sigma,\mu+\sqrt 3\sigma]}(x) $$

where $-\infty\lt\mu\lt\infty$ and $\sigma\gt 0$

Find the maximum-likelihood-estimator [MLE] of $\mu$ and $\sigma$.

the previous question Likelihood Function for the Uniform Density. had only one parameter $\theta$ . So it was easy to change the range with respect to $\theta$ and find the MLE of $\theta$.Also, there i have not asked to find the MLE of $\theta$ rather the questions were different in category. But in the present question, there are two parameters $\mu$ and $\sigma$. So it's not easy to me to change the range.

I started to solve it

$L(\mu,\sigma)=\prod_{i=1}^n f(x_i;\mu,\sigma)=\prod_{i=1}^n\frac{1}{2\sqrt 3\sigma}I_{[\mu-\sqrt 3\sigma,\mu+\sqrt 3\sigma]}(x_i)=[\frac{1}{2\sqrt 3\sigma}]^n \prod_{i=1}^n I_{[\mu-\sqrt 3\sigma,\mu+\sqrt 3\sigma]}(x_i)$

Then i don't know how to proceed.

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Very closely related to this previous question – Dilip Sarwate Jun 13 '13 at 16:05
Changing notations is not going to make people answer this question if they haven't answered your previous one already. If you want to bring more attention to your question, here are the proper ways. But considering that you asked your previous version less than 6 hours ago, I would suggest you to just be patient and wait. For the time being I'll mark this as a duplicate of your previous question. – Willie Wong Jun 13 '13 at 17:18
up vote 2 down vote accepted

My feeling is that your problem in this question as well as in a few recent similar ones is that the likelihood functions you consider involve indicator functions. By nature, these are not differentiable hence one has to find ways to maximize a function without differentiation.

One efficient tool is to imagine what can make the likelihood large. In the present case, you showed that the likelihood $L$ is defined by $$ L(\mu,\sigma)=a(\sigma)\mathbf 1_{(\mu,\sigma)\in A}, $$ for some function $a$ and some domain $A$. How to maximize $L(\mu,\sigma)$? Obviously, by making $a(\sigma)$ as large as possible while $(\mu,\sigma)$ is in $A$.

To be more specific, note that the function $a$ is decreasing hence to maximize $a(\sigma)$ is to minimize $\sigma$, and that $(\mu,\sigma)$ is in $A$ if and only if $\mu-\sqrt3\sigma\leqslant\min(x_i)$ and $\mu+\sqrt3\sigma\geqslant\max(x_i)$. In particular the condition $\max(x_i)-\min(x_i)\leqslant(\mu+\sqrt3\sigma)-(\mu-\sqrt3\sigma)=2\sqrt3\sigma$ must hold for $(\mu,\sigma)$ to belong to $A$. Thus, if $(\mu,\sigma)$ is in $A$ then $\sigma\geqslant\sigma^*$, with $$ \sigma^*=\frac{\max(x_i)-\min(x_i)}{2\sqrt3}. $$ To conclude that the optimal $(\mu,\sigma)$ (or at least some of them) is (are) such that $\sigma=\sigma^*$, one needs that $(\mu,\sigma^*)$ is in $A$ for at least one value of $\mu$ (otherwise one should abandon the optimal value $\sigma^*$ and replace it by a greater value). Once again, the conditions for that are that $\mu\leqslant\min(x_i)+\sqrt3\sigma^*$ and $\mu\geqslant\max(x_i)-\sqrt3\sigma^*$. Since $\min(x_i)+\sqrt3\sigma^*=\max(x_i)-\sqrt3\sigma^*$, this yields exactly one value of $\mu$ such that $(\mu,\sigma^*)$ is in $A$, namely, $\mu=\mu^*$, with $$ \mu^*=\frac{\max(x_i)+\min(x_i)}2. $$ Finally, for every $(\mu,\sigma)$, $$ L(\mu,\sigma)\leqslant L(\mu^*,\sigma^*)=a(\sigma^*). $$

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Exercise: adapt these considerations to the simpler previous question. – Did Jun 14 '13 at 7:23
Well explained, +1. – Stefan Hansen Jun 14 '13 at 7:27
@Did I have not understood some points.(1) How this inequality $\max(x_i)-\min(x_i)\leqslant m+\sqrt3s-(m-\sqrt3s)$ prevail?(2)How $\max(x_i)-\min(x_i)\leqslant m+\sqrt3s-(m-\sqrt3s)$ follows $\sigma^*=\frac{\max(x_i)-\min(x_i)}{2\sqrt3}.$?(3)Why $\min(x_i)+\sqrt3\sigma^*=\max(x_i)-\sqrt3\sigma^*$? – time Jun 14 '13 at 8:26
If $\min\geqslant(\ast)$ and $\max\leqslant(\ast\ast)$ then $\max-\min\leqslant(\ast\ast)-(\ast)$. – Did Jun 14 '13 at 10:07
The maximum likelihood estimation for the uniform density in another question (…) has been answered – Alecos Papadopoulos Jul 31 '13 at 20:08

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