TITLE: Distribution of $\max(X_i)\mid\min(X_i)$ when $X_i$ are i.i.d uniform random variables
QUESTION [7 upvotes]: If I have $n$ independent, identically distributed uniform $(a,b)$ random variables, why is this true: $$ \max(x_i) \mid \min(x_i) \sim \mathrm{Uniform}(\min(x_i),b) $$
I agree that the probability density function of $\max(x_i) \mid \min(x_i)$ must be non-zero only in the range $[\min(x_i), b]$.
I also find it reasonable that the posterior distribution is uniform, but I cannot see why it has to be so.
Also, I am confused how to derive this algebraically. Let $X$ be the max of $n$ iid uniform random variables. Let $Y$ be the min of $n$ iid uniform random variables. If I do $$ f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}, $$
$f_Y(y)$ is known. But the joint distribution is unknown ($X$ and $Y$ are not independent).
Thanks.

REPLY [5 votes]: Hmm, I wonder if this question is from one of my statistics students, since I just used this in last night's lecture.
As the previous respondent pointed out, what you say is false.  What is true is that, given $Y =$ min $X_i$,  max $X_i$ has as its conditional distribution that of the maximum of $n - 1$ independent $U(Y,b)$ random variables.
Here is an applied probability approach:  Given  $Y$, we know all the $X_i \ge Y$ and  $X_i = Y$  for one $i$.  This leaves $n-1$ random variables that each have the conditional distribution of a $U(a,b)$ restricted to $[Y,b]$.
If you really want to do this algebraically,  let  $Z =$ max $X_i$,  the joint density is the mixed second partial of $P${$Y \le y, Z \le z$}$ = P${$Z \le z$}$ - (1 -P${$Y > y$}$)$.