Find minimal sufficient statistics and ML estimator 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?
 A: 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.
