# What is the joint distribution of n identically distributed uniform distributions from $[0,1]$?

Given $$U_1,U_2,...,U_n$$ identical and independent uniform distributions of the form $$U(0,1)$$. Let $$U_{(1)} be their order statistics, then what is their joint distribution $$\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 $$f_{U_1...U_n}(u_1,...,u_n)=0$$.

• If $U_1 < U_2$ they can't be independent... – Lorenzo Najt Sep 1 '20 at 0:21
• i.i.d. random variables cannot be ordered in that fashion. – Kavi Rama Murthy Sep 1 '20 at 0:21
• 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. $\{U_i\}$? – Brian Tung Sep 1 '20 at 0:25
• @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. – Tsangares Sep 1 '20 at 0:33
• @BrianTung, I changed it does that work? – 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

$$\begin{equation*} f_{\mathbf{Y}}(y_1, \ldots, y_n) = \begin{cases} n!, & \text{ if } 0 < y_1 \leq \ldots \leq y_n < 1\\ 0, & \text{elsewhere} \end{cases} \end{equation*}$$

• 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? – Matthew Pilling Sep 1 '20 at 1:13
• $n!$ ways - I have fixed the typo. Since $f(x)$ is continuous, $P(Y_i = Y_j)=0$ for $i \neq j$. – Jackson Sep 1 '20 at 1:43
• Normal distribution? – Nap D. Lover Sep 1 '20 at 15:45
• @NapD.Lover, edited, thank you. – Jackson Sep 1 '20 at 15:52
• @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? – Tsangares Sep 1 '20 at 19:27