Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Yes this is homework and no I'm not looking for direct answers, just tips or some clairification on how to do this problem since it doesn't make sense to me and my professor didn't explain it very well. It says

Let $U_{(1)},...,U_{(n)}$ be the values of $n$ independent uniform $(0,1)$ random variables arranged in increasing order. Let $0\le x\lt y\le 1$. Find and justify a simple formula for $P(U_{(1)} \gt x$ and $U_{(n)} \lt y)$.

I just really don't know what to do or how to interpret this. Thanks!

share|cite|improve this question
up vote 2 down vote accepted

Hint: Let our uniformly distributed random variables be $W_1, W_2, \dots, W_n$. One could also call them $U_1, U_2,\dots, U_n$, but that might risk confusion with the $U_{(i)}$.

$U_{(1)}$ is by definition the minimum of the $W_i$, the smallest result we got in our $n$ experiments. And $U_{(n)}$ is the largest result we got in our $n$ experiments.

The minimum is $\gt x$ and the maximum is $\lt y$ precisely if all the $W_i$ lie in the interval $(x,y)$. What is the probability that $x\lt W_1\lt y$? What is the probability that $x\lt W_i\lt y$ for all $i$?

share|cite|improve this answer
Is there a way to do this problem without order statistics? We haven't covered the topic in class quite yet. – TheHopefulActuary Nov 18 '12 at 19:06
My start towards a solution (which actually gets us very close to the end) does not mention order statistics. Of course it uses the definition of $U_{(1)}$ and $U_{(n)}$, as it must, since the problem is about these. The fact that the minimum is greater than $x$ iff all of our random variables $X_1$ to $X_n$ are $\gt x$ is I think clear. I will add some wording that may help. – André Nicolas Nov 18 '12 at 19:12
Since it is uniform on $(0,1)$ wouldn't the formula be something like $(y-x)$? – TheHopefulActuary Nov 18 '12 at 19:42
We want all of them to end up in the interval, so it is $(y-x)^n$. – André Nicolas Nov 18 '12 at 19:43
Yes, the probability that $X_1$ (what I called $W_1$ in the edited post) lies in the interval is $y-x$, same is true of $X_2$, and so on. But the $X_i$ are independent, so we multiply. – André Nicolas Nov 18 '12 at 19:46

What you are dealing with is order statistics:

Here we first have $n$ random variables $X_1,X_2,...,X_n\sim \operatorname{iidUnif(0,1)}$, each is uniformly distributed on the interval $(0,1)$. Now these $U_{(1)},U_{(2)},...,U_{(n)}$ are the order statistics of the random variables $X_1,X_2,...,X_n$. That is, $U_{(1)}$ is the smallest of the $X_i$'s, and $U_{(2)}$ is the second smallest of the $X_i$'s, and so on.

Note that $U_{(1)}$ is the minimum and $U_{(n)}$ is the maximum of all $X_i$'s. Reinterpret the question as a uniform distribution problem in terms of the $X_i$'s.

This is usually the case for order statistics problems: to reinterpret the problem in terms the original random variables that got ordered.

share|cite|improve this answer
Incase you don't see my comment on Andre's answer, I was wondering if there is a way to do this problem without order statistics because in my class we haven't covered the topic just yet. Thanks! – TheHopefulActuary Nov 18 '12 at 19:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.