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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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1 Answers1

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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 $(\mu,\sigma)$, $$ L(\mu,\sigma)\leqslant L(\mu^*,\sigma^*)=a(\sigma^*). $$

Did
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  • Exercise: adapt these considerations to the simpler previous question. – Did Jun 14 '13 at 07:23
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    Well explained, +1. – Stefan Hansen Jun 14 '13 at 07:27
  • @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 08:26
  • 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 (http://math.stackexchange.com/questions/419287/likelihood-function-for-the-uniform-density/453823#453823) has been answered – Alecos Papadopoulos Jul 31 '13 at 20:08
  • @AlecosPapadopoulos I read your answer and I even left a comment... – Did Jul 31 '13 at 20:47