$X_1,\ldots,X_n$ be a random sample drawn form the uniform distribution on interval $[0,\theta]$

$T_n=\text{max}(X_1, \ldots ,X_n)$

show that $T_n$ is a function of the sufficient statistic for $\theta$.

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Since this is a direct consequence of the definition and of Fisher's factorization theorem (the result in the domain), you could add some indications about what causes you trouble here. – Did Jun 15 '12 at 7:20

A sufficient statistic is a function of sample data that gives as much information about the parameter as exists in the entire set of sample data.

If you're dealing with the $U(0,\theta)$ distribution, then all information about $\theta$ is contained in $X_{(n)}$

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The joint probability density for $X_1,\ldots,X_n$ is $$f_\theta (x_1,\ldots,x_n)=\prod_{j=1}^{n} \frac{1}{\theta} \mathbf{1}_{\{0\leq x_j\leq\theta\}} = \frac{1}{\theta^n} \prod_{j=1}^{n} \mathbf{1}_{\{0\leq x_j\leq\theta\}} = \mathbf{1}_{\{0\leq \min x_j \}} \frac{\mathbf{1}_{\{\max x_j \leq\theta\}} }{\theta^n}$$

so the the density can be factored into a product such that one factor does not depend on $\theta$, and the other factor which does depend on $\theta$, depends on the $x_j$ only through $\max x_j$. So $T_n = \max X_j$ is a sufficient statistic.

Since $$\frac{ f_\theta (x_1,\ldots,x_n)}{f_\theta (y_1,\ldots,y_n)} = \frac{\mathbf{1}_{\{0\leq \min x_j \}} }{\mathbf{1}_{\{0\leq \min y_j \}} } \cdot \frac{\mathbf{1}_{\{\max x_j \leq\theta\}}}{\mathbf{1}_{\{\max y_j \leq\theta\}}}$$ depends on $\theta$ only through the maxima, $T_n$ is a minimum sufficient statistic and must be a function of any sufficient statistic.

You can read more on Wikipedia.

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