Order Statistics Probabilities

1. Apr 24, 2012

mathmajor23

1. The problem statement, all variables and given/known data
Let Xi ~ iid f(x) = (2x)I[0,1](x), i = 1,....,n.

Find the distribution of X(1). What is the probability that the smallest one exceeds .2?

2. Apr 24, 2012

Ray Vickson

What work have you done on this so far? You need to show an attempt to solve this yourself before we can help.

RGV

3. Apr 25, 2012

mathmajor23

FX(1)(x) = P(X(1) <= x)
= P(X1,....,Xn <= X)
=1-P(X1,...,Xn > X)
=1-P(X1>X)^n since the xi's are iid.
=1-[1-P(X1 <= X)]^n
=1-[1-F(x)]^n
=1-[1-∫ from 0 to x (2tdt)]^n
=1-(1-X^2)^n

For the probability, P(X1>.2) = 1-Fx(1) (0.2) = (1-(0.2)^2)^n = (0.96)^n

4. Apr 25, 2012

Ray Vickson

I hope you were just being sloppy and don't really believe that
$$P(X_1, X_2, \ldots, X_n \leq x) = 1 - P(X_1, X_2, \ldots, X_n > x) \, \leftarrow \text{FALSE!}$$ because that is not valid. For two events A and B we have P(A) = 1 - P(B) if the events A and B are complementary (that is, have no points in common and together cover the entire sample sample space). Do the events $\{X_1, \ldots,X_n \leq x \}$ and $\{X_1,\dots,X_n > x\}$ look complementary to you? (Try drawing these for n = 2.) Of course, what is true is that $P(X_i > x) = 1-P(X_i \leq x)$ for each $X_i$ separately. You need to decide whether or not we have
$$P(X_1,\ldots,X_n > x) = (1-x^2)^n \text{ or } P(X_1,\ldots,X_n > x) = 1-(1-x^2)^n \,.$$

RGV

Last edited: Apr 25, 2012