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 $f_{U_1...U_n}(u_1,...,u_n)=0$.

  • 1
    $\begingroup$ If $U_1 < U_2$ they can't be independent... $\endgroup$ – Lorenzo Najt Sep 1 '20 at 0:21
  • $\begingroup$ i.i.d. random variables cannot be ordered in that fashion. $\endgroup$ – Kavi Rama Murthy Sep 1 '20 at 0:21
  • $\begingroup$ OP: Is what you mean to ask something about order statistics? Or do you want to know the joint distribution of variables $V_1, V_2, \ldots, V_n$, where $V_k$ is the $k$th lowest entry out of the i.i.d. $\{U_i\}$? $\endgroup$ – Brian Tung Sep 1 '20 at 0:25
  • $\begingroup$ @BrianTung, yes, I am still a novice so I wasn't sure what that meant or if it affected the joint distribution. I can reword the original question to help make it fit. $\endgroup$ – Tsangares Sep 1 '20 at 0:33
  • $\begingroup$ @BrianTung, I changed it does that work? $\endgroup$ – Tsangares Sep 1 '20 at 0:36

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*}

  • $\begingroup$ Do you mean we can rearrange the $y_1,...,y_n$ in $n!$ ways, or am I not understanding something? What if some of the $y_i$'s were equal? $\endgroup$ – Matthew Pilling Sep 1 '20 at 1:13
  • $\begingroup$ $n!$ ways - I have fixed the typo. Since $f(x)$ is continuous, $P(Y_i = Y_j)=0$ for $i \neq j$. $\endgroup$ – Jackson Sep 1 '20 at 1:43
  • $\begingroup$ Normal distribution? $\endgroup$ – Nap D. Lover Sep 1 '20 at 15:45
  • $\begingroup$ @NapD.Lover, edited, thank you. $\endgroup$ – Jackson Sep 1 '20 at 15:52
  • $\begingroup$ @Jackson Why does the $n!$ come into play? If there is an order statistic, then isn't there only one way it can be configured? $\endgroup$ – Tsangares Sep 1 '20 at 19:27

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