1. question involving i.i.d.

Let X1, X2 and X3 be three independent, identically distributed random variables each density function f(x) = 3x^2, 0< x < 1, 0 elsewhere.

Let Y = Max {X1, X2, X3}. Find P(Y> 1/2).

I am at a loss here.

I will (hopefully) show that I know what I.I.D. means by this example.

Suppose you have an urn with two balls in it--1 red and 1 blue. You pick two balls, one at a time w/o replacement.

Let X = the color of the 1st ball and Y = the color of the 2nd ball.

The two outcomes are {red, blue} and {blue, red}.

P(X=red) = 1/2 and (coincidentally) P(X= blue) = 1/2

P(Y=red) = 1/2 and (coincidentally) P(Y= blue) = 1/2

So X and Y are identically distributed.

Now if we know the outcome of the 1st ball we then know the outcome of the 2nd ball. That is, for example, P(X=red|Y=B) = 1 P(X=red)=1/2. So X and Y are NOT independent and hence X and Y are not i.i.d.

Since X1, X2 and X3 are i.i.d. I would think there is no max for them--after all they are identical. Am I missing something here????

So I assume that Y= Max {X1, X2, X3} = X1

Then p(X1 > 1/2) 1- p(X1 < 1/2) = 1- (1/2)3 = 7/8. My solution manual disagrees with this result.

However, if I compute p( X11/3> 1/2) = p( X1 > 1/8) I get the solution manual's answer (511/512)

2. Re: question involving i.i.d.

Just proceed from first principles

$f_X(x) = 3x^2,~0\leq x \leq 1$

$F_X(x) = x^3, ~0 \leq x \leq 1$

$Y= \max(X_1, X_2, X_3)$

$F_Y(y)=\\P[Y < y] = \\\\ P[\max(X_1, X_2, X_3) < y] = \\\\P[X_1 < y\wedge X_2 < y\wedge X_3 < y] =\\\\ P[X_1<y]^3 =\\\\ F_X(y)^3 = \\\\(y^3)^3 =\\\\ y^9$

$P[Y > 1/2] =\\ 1 - P[Y < 1/2] =\\ 1-F_Y\left(\dfrac 1 2 \right) =\\ 1 - \left(\dfrac 1 2 \right)^9 = \dfrac{511}{512}$