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Let $X_1, X_2, \ldots, X_n$ be i.i.d. uniform $(a-b, a+b)$, where $b>0$ and let $\theta = (a,b)$. Find a minimal sufficient statistics $t$ for $\theta$. Find the ML estimator $\hat{\theta}$ of $\theta$.

So, what I have done is for minimal sufficient statistics $t$ for $\theta$, by Fisher Neyman theorem, showed that $T(X_i) = (\min(X_i), \max(X_i))$. Is this correct?

And for the ML estimator, I just can understand the distribution would be $p = 1/2b$. Then can I just take a log function and calculate ML?

Michael Hardy
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ssandi
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  • I don't think the way of using factorization theorem to find minimal sufficient statistic is correct. Factorization theorem only states a way of finding sufficient statistic but not minimal sufficient statistic. – Tan Jan 07 '21 at 19:34

1 Answers1

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The density for a single observation is $$ f(x) = \begin{cases} \dfrac{1}{2b} & \text{for }a-b<x<a+b, \\[12pt] 0 & \text{for }x>a+b\text{ or }x<a-b. \end{cases} $$ and for $n$ independent observations it is $$ f(x_1,\ldots,x_n)=\begin{cases} \dfrac{1}{(2b)^n} & \text{if }a-b<\min\{x_1,\ldots,x_n\}\text{ and }\max\{x_1,\ldots,x_n\}<a+b, \\[12pt] 0 & \text{otherwise}. \end{cases} $$ The likelihood is therefore $$ L(a,b) = \begin{cases} \dfrac{1}{(2b)^n} & \text{if }(a,b)\in\{(a,b) : a-b<\min\ \&\ a+b>\max\}, \\[12pt] 0 & \text{otherwise}. \end{cases} $$ If we let $c=a+b$ and $d=a-b$ we then have $$ L(a,b) = K(c,d) = \begin{cases} \dfrac{1}{(c-d)^n} & \text{if }d<\min\ \&\ \max<c, \\[12pt] 0 & \text{otherwise}, \end{cases} $$ so $K$ is the likelihood as a function of $c$ and $d$.

Notice now that you can make $(c-d)^n$ as small as possible by choosing $c,d$ to make the distance $c-d$ as small as possible subject to the constraints that $d\le\min$ and $\max\le c$, and that means making $d=\min$ and $c=\max$. So the MLE for $a=(c+d)/2$ is $(\max+\min)/2$ and that for $b=(c-d)/2$ is $(\max-\min)/2$.

Your use of the factorization theorem to find the minimal sufficient statistic is correct.

Michael Hardy
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  • When specifying a density function one doesn't worry about boundaries, so one my write $f(x)=\left.\begin{cases} \text{something} & \text{if }x<k, \ \text{something} & \text{if }x>k.\end{cases}\right}$ But in finding MLEs, working with functions of a parameter, if increasing the likelihood means pushing it toward a point on the boundary, one takes that boundary point to be the MLE. – Michael Hardy Nov 27 '13 at 20:23
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    I don't think the way of using factorization theorem to find minimal sufficient statistic is correct. Factorization theorem only states a way of finding sufficient statistic but not minimal sufficient statistic. – Tan Jan 07 '21 at 19:33