# Order statistics

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• Jun 18th 2013, 05:14 PM
Jame
Order statistics
Let $Y_1, ... , Y_n$ be the order statistics of a random sample of size $n$ from a uniform distribution on the interval (0,2).

What is $P[Y_1 < \frac{1}{2} < Y_n]$

If $U_1, ... , U_n$ are the uniform RVs from the sample then

$Y_1 = min(U_1, ... , U_n)$

$P(Y_1 \leq y) = 1 - P(Y > y) = 1 - [P(U_1 > y) * ... * P(U_n > y)] = 1 - [1 - \frac{y}{2}]^n$

Thus $P(Y_1 \leq \frac{1}{2}) = 1 - \frac{3^n}{4^n}$

$Y_n = max(U_1,...,U_n)$

$P(Y_n \leq y) = P(U_1 \leq y)*...*P(U_n \leq y) = (\frac{y}{2} )^n$

$P(Y_n > y) = 1 - (\frac{y}{2} ) ^ n$

Thus $P(Y_n > \frac{1}{2}) = 1 - \frac{1}{4^n}$

However, I am not sure how to use these probabilites to find the interval I'm interested in $[Y_1 < \frac{1}{2} < Y_n]$
(The end points are not real numbers, but RVs. I tried subtraction, but I don't think thats right. Is there a different way to think of this interval?)

Thank you for your assistance.
• Jun 19th 2013, 02:42 AM
chiro
Re: Order statistics
Hey Jame.

You need to get the joint distribution of Y1 and Yn and then get the region of integration and integrate to find the result.

Because these are related it means that it will likely be complicated unless you can find a dependent function of min in terms of max (or max in terms of min).