Given $U_1,U_2,…,U_n$ identical and independent uniform distributions of the form $U(0,1)$. Let $U_{(1)}<U_{(2)}<…<U_{(n)}$ be their order statistics, then what is their joint distribution $\left( U_{(1)},U_{(2)},…,U_{(n)}\right)$?
Any help is much appreciated.
EDIT: Yes, I forgot to put what I thought it was on here. So the the PDF of $f_U(u)=1$ if $u\in[0,1]$ and $f_U(u)=0$ if $u\notin[0,1]$. Because each $U_i$ is independent, to find the join probability density function, I just multiply all of their PDFs:
\begin{align}
f_{U_1…U_n}(u_1,…,u_n)&=f_{U_1}(u_1)\cdot…\cdot f_{U_n}(u_n)\\
= &1 \quad \text{if} \quad u_1,u_2,…,u_n\in[0,1]\\
&0 \quad \text{if} \quad u_1,u_2,…,u_n\notin[0,1]
\end{align}
Although I am not sure how to say that because they are all multiplied, then if any of $u_i\notin [0,1]$ then $f_{U_1…U_n}(u_1,…,u_n)=0$.
Best Answer
edit: OP has changed the question.
Let the random variables $X_1, X_2, \ldots X_n$ be i.i.d. continuous random variables with common pdf $f(x)$ and cdf $F(x)$. Denote $Y_i = X_{(i)}$, where $X_{(i)}$ represents the $i$th ordered statistic. The joint pdf of $Y_1, \ldots, Y_n$ is given by \begin{equation*} f_{\mathbf{Y}}(y_1, \ldots, y_n) = \begin{cases} n! \prod_{i=1}^{n} f(y_i), & \text{ if } - \infty < y_1 \leq \ldots \leq y_n < \infty\\ 0, & \text{elsewhere} \end{cases} \end{equation*}
The multiplier $n!$ occurs because we can arrange the $y_1, \ldots y_n$ in $n!$ ways and the pdf for any such arrangement is the product $\prod_{i=1}^{n} f(y_i)$ via the iid assumption.
For the uniform distribution, $f(u) = 1$, $0 < u < 1$, hence $\prod_{i=1}^{n} f(y_i) = 1$. The joint pdf is thus
\begin{equation*} f_{\mathbf{Y}}(y_1, \ldots, y_n) = \begin{cases} n!, & \text{ if } 0 < y_1 \leq \ldots \leq y_n < 1\\ 0, & \text{elsewhere} \end{cases} \end{equation*}