# 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