Conditional distribution of order statistics Let $X_{(1)},...,X_{(n)}$ be the order statistics of a set of $n$ independent uniform $(0,1)$ random variables. Find the conditional distribution of $X_{(n)}$ given that $X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}$
I need to find: $P[X_{(n)}\le x_n| X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}]$ which is equal to: $${P[X_{(n)}\le x_n, X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}]\over P[X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}]}$$
but I have no idea how to compute the above probability and also I don´t know how to use the fact that the variables are uniform and independent. I would really appreciate if you can help me with this problem
 A: Sure, you can work with the CDF as well but there is an easier way. Recall that for a random sample of size $n$ from a uniform (0,1) distribution, the joint density for all the order statistics is given by
$$f_{X_{(1)},X_{(2)},\ldots,X_{(n)}} \left(x_{(1)},x_{(2)},\ldots, x_{(n)} \right)=\begin{cases} n!  & 0<X_{(1)}<X_{(2)}<\ldots<1 \\ 0 & \text{otherwise} \end{cases} $$
This is just all the possible permutations of the random variables. This density will
serve as the numerator of your fraction. We need to integrate out the greatest order statistic so as to get the joint density of the remaining order statistics. Thus
$$f_{X_{(1)},X_{(2)},\ldots,X_{(n-1)}} \left(x_{(1)},\ldots,x_{(n-1)} \right)=\int \limits_{x_{(n-1)}}^{1} n!\ \mathrm{dy} =n!\left(1-x_{(n-1)}\right)$$
Putting these two together, gives the conditional probability you are looking for.
A: Isn't the conditional distribution of $X_{(n)}$ simply uniform on $(s_{n-1},1)$?
A: Assuming that $0< s_1< \cdots < s_{n-1}< s_n < 1$.
Let $S=\{s_1, s_2, \ldots, s_{n-1}\}$, let $X=\{X_1, \ldots X_n\}$, let $X_\ast$ be an arbitrary member of $X$
$\mathsf P[X_{(n)}\le s_n| X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}] = \dfrac{\mathsf P[X_{\ast}\leq  s_n \mid X\setminus \{X_\ast\} = S]}{\mathsf P[X_{\ast}> s_{n-1}\mid X\setminus \{X_\ast\} = S]}$
The conditional probability that the $n$-th smallest number in the list is not greater than $s_n$ when given fixed values for all smaller numbers in the list, is the probability that the unfixed result (which ever one it is) is not greater than $s_n$ given that it is greater than $s_{n-1}$.
Since all members in the list have an identical independent uniform distribution, this means $\mathsf P[X_{(n)}\le s_n| X_{(1)}=s_1,X_{(2)}=s_2....,X_{(n-1)}=s_{n-1}] = \dfrac{\mathsf P[s_{n-1}<X_{\ast}\leq  s_n ]}{\mathsf P[X_{\ast}> s_{n-1}]} = \dfrac{s_{n}-s_{n-1}}{1-s_{n-1}}$
A: Given all $n-1$ values of $\mathbf{x}$, the probability of the largest one being smaller than $x_n$, i.e. $P(X_{(n)}\leq x_{n}|\cdots)$ is equal to $1-P(X_{(n)}> x_{n}|\cdots)$. The key point is that if $X_{(n)}$ is larger than $x_{n}$, due to the order statistic we also know that it is larger than all other $x_i$, $i<n$. This says that $$1-P(X_{(n)}> x_{n}|X_{n-1}=x_{n-1},\cdots)=1-P(X_{(n)}> x_{n},X_{n}>x_{n-1},\cdots)$$ $$=1-P(X_{(n)}> x_{n})P(X_{n}>x_{n-1})P(X_{n}>x_{n-2})\cdots$$ $$=1-(1-U(x_n))(1-U(x_{n-1}))(1-U(x_{n-2}))\cdots$$
where $U$ is the uniform c.d.f. The second equality is due to independency.
A: According to the definition of conditional density function $f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)}$, then
$$f_{X_{(n)}|X_{(1)}\cdots X_{(n-1)}}(x_n|x_1,\cdots ,x_{n-1})=\frac{f(x_1,x_2,\cdots ,x_n)}{f(x_1,x_2,\cdots ,x_{n-1})}=\frac{n!\times 1}{\int_{s_{n-1}}^1 n!f(x_1)\cdot f(x_2)\cdots f(x_n) dx_n }=\frac{1}{1-s_{n-1}}$$. So, the conditional distribution equals to $\int_0^a \frac{1}{1-s_{n-1}} dx_n=\frac{a}{1-s_{n-1}}$ . We can also say this conditional distribution is uniform on $(s_{n-1},1)$ .
