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I have another problem and the density is something like this: $$f(x,\theta)=\frac{1}{\theta} \text{if }0\le x\le\theta \text{ for some }\theta>0$$ $$f(x,\theta)=0 \text{ otherwise}$$

Given the sample $X_1,...,X_n$ be iid.

I found the MLE to be $X_n$ which is the the maximum of the sample.

I'm supposed to find the exact distribution but I'm not sure what do they want? Do I derive the density and CDF of a sample maximum?

Lastly I'm supposed to find the exact distribution and asymptotic distribution of

$$n(\theta-\hat{\theta})$$

I know about Lindeberg Levy-Central Limit Theorem but I'm not sure how to apply that into this case or deriving the exact distribution.

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  • $\begingroup$ Yes you need to calculate the distribution of the sample maximum, which, since the $X$'s are Uniform $(0,\theta)$, is easy. Start with the cdf. for the second question apply the change-of-variables formula to see what happens. $\endgroup$ – Alecos Papadopoulos May 3 '14 at 17:19
  • $\begingroup$ I know the change of variables formula but does that mean I set thetabar=n(theta-thetabar)?. $\endgroup$ – user42668 May 3 '14 at 23:03
  • $\begingroup$ You define a new random variable, say $Z = n(\theta-\hat \theta)$ and find the distribution of $Z$ wich is a function of $\hat \theta$. $\endgroup$ – Alecos Papadopoulos May 3 '14 at 23:17
  • $\begingroup$ ...and you won't need the CLT for the asymptotic distribution. Think $e$, when the time comes. $\endgroup$ – Alecos Papadopoulos May 3 '14 at 23:24
  • $\begingroup$ e as in the exponential right?. $\endgroup$ – user42668 May 3 '14 at 23:33
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The OP correctly found the MLE of $\theta$ to be the maximum order statistic. The $X$- r.v.'s here are $U(0,\theta)$. The distribution function and the density function of the maximum of an i.i.d. sample is here

$$F(\hat \theta) = [F_X(\theta)]^n = \left(\frac {\hat \theta}{\theta}\right)^n,\qquad f_{\hat \theta}(\hat \theta) = n\frac {\hat \theta ^{n-1}}{\theta^n}$$

Define the random variable $Z = n(\theta-\hat \theta)$ (note that $Z\ge 0$, since $\hat \theta$ never overestimates $\theta$). Then applying the change-of-variable formula we have $$\hat \theta = \theta - \frac 1n Z, \qquad \left|\frac {d\hat \theta}{dZ}\right| =\frac 1n $$

So

$$f_Z(z) = \frac 1n n\frac {(\theta - \frac 1n z)^{n-1}}{\theta^n} = \frac 1{\theta}\left(1+\frac {(-z/\theta)}{n}\right)^{n-1}$$

Asymptotically we have

$$\lim_{n\rightarrow \infty}f_Z(z) = \lim_{n\rightarrow \infty}\frac 1{\theta}\left(1+\frac {(-z/\theta)}{n}\right)^{n-1} = \lim_{n\rightarrow \infty}\frac 1{\theta}\left(1+\frac {(-z/\theta)}{n}\right)^{n}\cdot \left(1+\frac {(-z/\theta)}{n}\right)^{-1}$$

The rightmost term goes to unity, and using the limit representation of the base of the natural logarithm, we arrive at

$$\lim_{n\rightarrow \infty}f_Z(z) = \frac 1{\theta}e^{-z/\theta}$$

which is an exponential distribution.

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