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Let $X_1, X_2, ..., X_n$ be independent and identically distributed uniform $U(0, \theta)$ distribution where $0 < \theta < \infty $. Show that $T(X)=max X_i$ is a complete statistics.

My biggest problem here is how to find the uniform distribution? Is it either:

a) The pdf is $f(x; \theta )$=$ \frac{1}{ \theta -0 }$=$ \frac{1}{ \theta }$

or

b) $f(x; \theta )$=$ \frac{1}{ \theta_2 - \theta_1 }$=$ \frac{1}{ \sqrt{3 \theta} + \sqrt{3 \theta } }$ = $ \frac{1}{ 2 \sqrt{3 \theta} }$ because the mean is $0$ and variance $ \theta $, where I find the value of $ \theta_1 $ and $ \theta_2 $ from it.

Then to find the complete statistics, I need to find the likelihood function and because it's need to show the maximum value, so I need to find it using order statistics $f_{Y_n}$

But, I can't continue since I'm not sure with my pdf of Uniform distribution.

Really appreciated if anyone could clear up the way to find the pdf of the Uniform distribution?

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  • $\begingroup$ The parameters of Uniform distribution $0, \theta$. They are the end points of your interval not mean and variance. So (a) is the correct pdf. Now, the reason you have to use ordered statistics is because your parameter $\theta$ depends on your support. Therefore, taking the derivative of your Log Likehood function to find the maximum does not work. $\endgroup$ – Harry Apr 13 '18 at 7:13
  • $\begingroup$ Also, putting the indicator into the density will help solve the problem. The density $f(x|\theta) = \frac{1}{\theta}I_{(x\in(0,\theta))}$ with the indicator shows that the distribution gives zero mass to values outside of $(0,\theta)$ $\endgroup$ – Ryan Warnick Apr 15 '18 at 14:53

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