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i don't quite understand how to show that T is the mle.

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Please do not post images with the question in it. Retype the question using the built-in $\LaTeX$ rendering engine. That will make the question more self-contained. Please also add any thoughts you have and, if this is homework, tagged it as such. Cheers. –  cardinal Mar 12 '12 at 8:46
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1 Answer

The likelihood of $X_i$ is proportional to $\dfrac{1}{\theta} I[0\le X_i \le \theta]$ so multiply the likelihoods together and then find the value of $\theta$ which maximises the product, namely $\dfrac{1}{\theta^n}$ times the product of the indicator functions.

Note that $\prod_i I[0\le X_i \le \theta] = I[0\le \min X_i \le \max X_i \le \theta]$ and that if all the $X_i$ are non-negative this becomes $I[T \le \theta]$ or $I[\theta \ge T]$.

Added:

$\dfrac{1}{\theta^n}$ is a positive but decreasing funtion of $\theta$ so $\dfrac{1}{\theta^n} I[\theta \ge T]$ is zero for $\theta \le T$, jumps up to $\dfrac{1}{T^n}$ when $\theta = T$ and then is lower for $\theta \gt T$ . So the maximum likelihood estimator is $\hat{\theta}=T$.

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how do you know that T is sufficient for $\theta$? the indicator function is $0\leq X_i \leq \theta$ instead of $max{Xi}$ so i think we need to include $min{X_i}$ right? –  Mathematics Mar 12 '12 at 8:54
    
@Mathematics: If you did not know from the question the lower end of the support then you are correct, you would want the minimum too, for example if the distribution was uniform on $[\theta_1, \theta_2]$. But here you are told that the distribution is on $[0, \theta]$, so the minimum tells you nothing new; if $\theta \ge \max X_i$ then you automatically have $\theta \ge \min X_i$. –  Henry Mar 12 '12 at 9:00
    
o i see. But why you can arrive the ans without finding any extremum point of a log-likelihood function? I mean to show that T is mle –  Mathematics Mar 12 '12 at 9:04
    
@Mathematics: I have added something –  Henry Mar 12 '12 at 12:40
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