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Thread: question involving i.i.d.

  1. #1
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    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)
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  2. #2
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    Re: question involving i.i.d.

    Just proceed from first principles

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

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

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

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

    $P[Y > 1/2] =\\ 1 - P[Y < 1/2] =\\ 1-F_Y\left(\dfrac 1 2 \right) =\\ 1 - \left(\dfrac 1 2 \right)^9 = \dfrac{511}{512}$
    Last edited by romsek; Dec 6th 2016 at 08:03 PM.
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