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Assume $X_1, X_2, ..., X_n$ is a random sample of a uniform random variable $X$ on the interval $(0,\theta)$. What is the density function for $X$ and the density function for $X_{(n)}$ where $X_{(n)}$ is the $n$-th ordered statistic of $X$.

So, it's pretty trivial that $f_X(x) = \dfrac{1}{\theta}$.
I also know that the density function for an ordered statistic of a uniform random variable on interval $0<x<1$ follows a beta distribution:
$$f_{X_r}(x) = \dfrac{\Gamma{(n+1)}}{\Gamma{(r)}\Gamma{(n-r+1)}}x^{r-1}(1-x)^{n-r}.$$ So, considering that I want to know the the density function for the interval $0<x<\theta$, I feel like I could just divide through by $\theta$ to make the interval $0<\frac{x}{\theta}<1$, and then (considering that I want the density function for the n'th ordered statistic), the distribution function would become:
$$f_{X_{(n)}}{(x)} = n\left(\dfrac{x}{\theta}\right)^{n-1}$$

However, the text I am following along with reports that this value is:
$$f_{X_{(n)}}{(x)} = n\dfrac{x^{n-1}}{\theta^n}.$$

Can anyone help me figure out what I'm missing here? Thanks!

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There are several ways to remedy the issue. The simplest is to recognize that you are performing a linear transformation and hence $$f_{X_{(n)}}(x) = \frac{f_{U(n)}(x/\theta)}{\left|\frac{du}{dx}\right|_{u = x/\theta}} = \frac{1}{\theta}\cdot\frac{\Gamma(n,1)}{\Gamma(n)\Gamma(1)}(x/\theta)^{n-1}(1-x/\theta)^{1-1} = n\cdot\frac{x^{n-1}}{\theta^n}$$ which gives the desired result.

You could also calculate the cdf $$P(X_{(n)}\leq x) = \left(\frac{x}{\theta}\right)^n.$$ This gives that the pdf is $$f_{X_{(n)}}(x) = n\cdot\left(\frac{x}{\theta}\right)^{n-1}\cdot\frac{1}{\theta} = n\cdot\frac{x^{n-1}}{\theta^n}.$$

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  • $\begingroup$ All the answers were very well done, but I'm accepting this one because you not only answered why my equation was coming up incorrect, but also concisely showed how to evaluate the pdf from first principles. $\endgroup$ – lstbl Mar 22 '16 at 21:50
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    $\begingroup$ Yes, I was going to write that you should not expect to just plug stuff in and expect it to work, but I felt like that sounded rude. Haha. Anyway, thanks for noticing. Good luck. $\endgroup$ – Em. Mar 22 '16 at 21:52
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The largest of the $X_i$, say $W$, is less than or equal to $w$ if and only if all the $X_i$ are $\le w$. Thus for $w$ between $0$ and $\theta$ we have $$F_W(w)=\Pr(W\le w)=\left(\frac{w}{\theta}\right)^n.$$ For the density function $f_W(w)$, differentiate. We get $$\frac{n}{\theta^n}w^{n-1}$$ in the interval $0\lt w\lt \theta$. (The density function is $0$ elsewhere.)

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  • $\begingroup$ Very simple and much easier explanation than the one I was working with. Thanks as always Andre. $\endgroup$ – lstbl Mar 22 '16 at 21:46
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    $\begingroup$ You are welcome. It can be important to know the distribution of the $k$-th order statistics, but the max, and to a lesser degree the min, are particularly simple. By the way, I never remember the distribution of the order statistics (have only average quality memory) so I prefer to use a crude informal heuristic calculus method to find out each time. $\endgroup$ – André Nicolas Mar 22 '16 at 21:54
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First, the density of $X$ should be more precisely written as: $$f_X(x) = \begin{cases} 1/\theta, & 0 \le x \le \theta \\ 0, & \text{otherwise.} \end{cases}$$ Or using indicator function notation, we could write it as $$f_X(x) = \frac{1}{\theta} \mathbb 1(0 \le x \le \theta).$$ The distinction is important.

As for the order statistic, how do you know it is beta? Far better to proceed from first principles: if $X_{(n)} = \max (X_1, \ldots, X_n)$, then for $0 \le x \le \theta$, we have $$\Pr[X_{(n)} \le x] \overset{\text{ind}}{=} \prod_{i=1}^n \Pr[X_{i} \le x] = \prod_{i=1}^n \frac{x}{\theta} = \left(\frac{x}{\theta}\right)^n.$$ This is because $X_{(n)} \le x$ if and only if every observation $X_1, \ldots, X_n$, is at most $x$. Thus the desired probability is the product of the individual probabilities that each observation is at most $x$, since the $X_i$s are IID.

If $x > \theta$, then $\Pr[X_{(n)} \le x] = 1$, and if $x < 0$, then $\Pr[X_{(n)} \le x] = 0$. So the CDF is $$F_{X_{(n)}}(x) = \begin{cases} 0, & x < 0 \\ (x/\theta)^n, & 0 \le x \le \theta \\ 1, & x > \theta. \end{cases}$$ Now differentiation gives $$f_{X_{(n)}}(x) = \frac{dF_{X_{(n)}}}{dx} = \frac{nx^{n-1}}{\theta} \mathbb 1(0 \le x \le \theta),$$ as claimed. No need to use ad hoc procedures that may or may not be justified.

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The density of the $r$-th order statistic is obtained by the count of arrangements, the density of the r-th statistic, the CDF of those less and the complement of the CDF of those greater.

$\begin{align}f_{X_{(r)}}(x) ~=~& \frac{n!}{(r-1)!1!(n-r)!}~f_X(x)~F_X(x)^{r-1}~(1-F_X(x))^{n-r} \\ ~=~& \frac{n!}{(r-1)!(n-r)!}~ \frac 1\theta ~\Big(\frac x\theta\Big)^{r-1}~\Big(\frac{\theta-x}\theta\Big)^{n-r} ~~\mathbf 1_{x\in[0;\theta]} \end{align}$

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    $\begingroup$ @Istbl just use first principles. You are after the density of one statistic being $x$, $r-1$ being less than $x$, and $n-r$ being greater. $\endgroup$ – Graham Kemp Mar 22 '16 at 21:51

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