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

Let Y = Max {X_{1}, X_{2}, X_{3}}. 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 X_{1}, X_{2}and X_{3 }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 {X_{1}, X_{2}, X_{3}} = X_{1 }Then p(X_{1 }> 1/2) 1- p(X_{1 }< 1/2) = 1- (1/2)^{3 }= 7/8. My solution manual disagrees with this result.

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