Sample joint density conditional on sufficient statistic depends on parameter for uniform distribution Lex $X_1, X_2 \sim U(0, \theta)$ i.i.d , then obviously the sufficient statistic would be $T = \max\{X_1, X_2\}$, which can be easily proven by factorization theorem. However, I try to compute the density $f(x_1, x_2\mid t)$, which turns out to depend on the parameter $\theta$. This absolutely contradicts the definition of sufficient statistic:
$$
f(x_1, x_2, t) = \frac 1 {\theta^2}I\{t\leq \theta, \max\{x_1, x_2\}=t\}
$$
$$
f(t) = \frac{2t}{\theta^2}I\{t \leq \theta\}
$$
Thus
$$
f(x_1, x_2\mid t) = \frac 1 {2t} I\{t\leq \theta\}
$$
which depends on $\theta$ through the indicator function $I\{t\leq \theta\}$. 
What's wrong with my reasoning? Thanks a lot!
 A: I think you should have
$$f(x_1,x_2 \mid t) = \frac{1}{2t} I\{0 \le  t \le \theta, \max\{x_1,x_2\} = t\} = \frac{1}{2t} I\{\max\{x_1,x_2\} = t\}.$$
Regarding the indicator arithmetic: if $I\{0\le t \le \theta\} = 0$, then $f(x_1,x_2 \mid t)$ is not even a well-defined probability distribution, so I think you can ignore this case and assume $0 \le t \le \theta$ so that the denominator is nonzero.


*

*if $I\{0 \le  t \le \theta, \max\{x_1,x_2\}=t\}=1$, then $I\{0 \le  t \le \theta\} =1$, so the ratio is $1$.

*otherwise $I\{0 \le  t \le \theta, \max\{x_1,x_2\}=t\}=0$, so the ratio is $0$ regardless of what the denominator is (even when the denominator is zero)

Edit:
$$P(T \le t) = (t/\theta)^2 I\{0 \le t \le \theta\} \implies f(t) = \frac{2t}{\theta^2} I\{0 \le t \le \theta\}$$
$$f(x_1,x_2,t) = \frac{1}{\theta^2} I\{0 \le t \le \theta, \max\{x_1,x_2\}=t\}$$
A: $$f(x_1,x_2\mid t)\overset{\text{Bayes}}{=}\frac{f(x_1,x_2,t)}{f(t)}=\frac{f(x_1,x_2)\mathbb{I}[x_1,x_2\le t]}{\mathbb{P}[x_1,x_2\le t]}=\frac{\frac{1}{\theta^2}\mathbb{I}[x_1,x_2\le t]}{\left(\frac t \theta \right)^2} = \frac{\mathbb{I}[x_1,x_2\le t]}{t^2}$$
