Order Statistics Probabilities

[mathmajor23]

Homework Statement

Let X_i ~ 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?

 

[Ray Vickson]

Homework Statement

Let X_i ~ 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?

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

RGV

 

[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

 

[Ray Vickson]

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

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