How to prove that max{$X_1$,...,$X_n$} is a sufficient statistic for $b$ in the Uniform distribution on$ [a,b]$ I am having a bit of difficulty with the following: I have the Uniform distribution on [a,b] where a is known and b is unknown and $b>a$, I'd like to show that the $T=max$($X_1$,...,$X_n$) is a sufficient statistic for $b$. 
Here is my proof:
Since for any $x_i$$<0$, $f_n$($\vec{x}$,$b$)= $0$, we only look at the case where $x_i$$\geq$$0$:
We have:
$f_n$($\vec{x}$,$b$)= $(1/(b-a))^n$ for $t$$\leq$$b$ and 
0 otherwise, 
As a result, 
let $h(t,b)$$=$$f_n$($\vec{x}$,$b$)= $(1/(b-a))^n$ for $t$$\leq$$b$ and 
0 otherwise, 
Then, let $u(x)$$=$$1$.
Since we can write $f_n$($\vec{x}$,$b$) = $h(t,b)$$u(x)$, then T is a sufficient statistic. However, I have seen in other places that the right answer is to let write the likelihood as:
$f_n$($\vec{x}$,$b$) = $\frac{h(t,b)}{(b-a)^n}$ where here I have no idea what $h(t,b)$ should be. 
Thank you!
 A: $h(t,b)=1_{\{t<b\}}$ is the indicator function of $t<b$. I believe, where $t=\max (x_1,...x_n)$
To see why this is the case:
the pdf of uniform distributions on $(a,b)$ is given by
$$f(x)=\frac{1}{b-a}1_{x\in [a,b]} $$
This is exactly what you said, but using indicator function makes it clearer. $1_{x\in [a,b]}$ takes value 1 when $x\in [a,b]$ and 0 otherwise.
The likelihood function
$$f_n(x)=\frac{1}{(b-a)^n}1_{\{x_1,x_2,...x_n\in [a,b]\}} $$
Now note $\{x_1,x_2,...x_n\in (a,b)\}$ is the same as $\min(x_1,...x_n)\geq a$ and $\max(x_1,...x_n)\leq b$, so
$$f_n(x)=\frac{1}{(b-a)^n}1_{\{\min(x_1,x_2,...x_n)\geq a\}}1_{\{\max(x_1,x_2,...x_n)\leq b\}} $$ 
Then, by factorisation criterion, this is a sufficient statistics.
Now observe: to make $f_n$ as big as possible, you need to make the $(b-a)$ as small as possible (so that $1/(b-a)^n$ is large) while $1_{\{\max(x_1,x_2,...x_n)\geq b\}}=1$, (otherwise it is 0), which means $\max(x_1,x_2,...x_n)\geq b$.
Combining these two things together and the fact $b>a$, you want $b$ as small as possible while $\max(x_1,x_2,...x_n)\geq b$ still holds. This means
$f_n$ is maximised when $\max(x_1,x_2,...x_n)=\hat{b}$
Note: it is obvious to see if $a$ is unknown in addition to/instead of $b$ is unkown, then the sufficient statistics/MLE of $a$ is $\min(x_1,x_2,...x_n)= \hat{a}$
