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Math Help - Uniform Distribution!!!

  1. #1
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    Uniform Distribution!!!

    Suppose that X1, X2 and X3 are independent and uniformly distributed over (0,1). Define Y=max(X1, X2, X3).

    Find EY (expected value of Y) [Hint compute P(Y is less than or equal to y) and deduce this from the density of Y.

    I dont know where to start on this problem... Can someone give me some clues/steps to get me started at least?


    Thanks!
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  2. #2
    Senior Member Peritus's Avatar
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    As the hint suggests we will find the cdf (comulative distribution function) Y:

    F(y) = P(Y <= y) = P(max(X1, X2, X3) <= y)

    Now the important part is to notice that max(X1, X2, X3) <= y implies that each & every one of the r.v.'s (X1, X2, X3) is smaller than y, and thus:

    P(max(X1, X2, X3) <= y) = P(X1 <= y)P(X2 <= y)P(X3 <= y) = Fx1(y)Fx2(y)Fx3(y) = y^3 0 <= y <= 1

    now we can compute the pdf of Y by taking the derivative of it's cdf:

    fy(Y) = 3*y^2

    E(Y) = integ(y*fy(Y))[0,1] = integ(3*y^3)[0,1] = 0.75*y^4 |[0,1] = 0.75
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  3. #3
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    Thanks! One quick question, how did u get from the Fx1(y)Fx2(y)Fx3(y)=y^3? Thanks!!!!
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  4. #4
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    Also, the next problem asks for P(Y is greater than y) and Y=min(x1,x2,x3)... i know this is the same set up but how do we take account on the min.... that would mean that the lowers value of the 3 x values is less than or equal to y and the other two could be either less than or equal to y or greater than y. Is that right?
    Last edited by clipperdude21; November 10th 2007 at 11:23 AM.
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  5. #5
    Senior Member Peritus's Avatar
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    let's look at the following cdf:

    P(min(X1,X2,X3) <= y)

    the complementing probability is:
    P(min(X1,X2,X3) <= y) = 1 - P(min(X1,X2,X3) >= y)

    now you can continue in the same way you did in the previous problem...
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  6. #6
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    I think i understand that part but im not sure what to do with the min part of the problem. For max, it made since since even the biggest of the 3 values was less or equal to y. For min, does that mean that x1 is less than equal to y but the other two x values (x2 and x3) are greater than y?

    Like this: P(min(X1, X2, X3) <= y) = P(X1 <= y)P(X2 >= y)P(X3 >= y) = Fx1(y)(1-Fx2(y))(1-Fx3(y))


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  7. #7
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    Peritus, thanks for bringing more clarity to the problem. Could you explain the P(min(X1,X2,X3) <= y) = 1 - P(min(X1,X2,X3) >= y) more throughly? Does min(X1,X2,X3)>=y imply that since the minimum RV is >=y, all X1, X2, and X3 are >=y?

    If that is the case; would this be correct?
    P(min(X1,X2,X3) <= y) = 1 - P(min(X1,X2,X3) >= y)

    P(min(X1,X2,X3) >= y) = P(X1 >=y)P(X2 >=y)(PX3 >=y) = {1-P(X1 <=y)}{1-P(X2 <=y)}{1-P(X2 <=y)} = (1-y)(1-y)(1-y) = -y^3 + 3y^2 -3y + 1

    P(min(X1,X2,X3) <= y) = y^3 - 3y^2 + 3y

    Then to find the density function, we take the derivative of y^3 - 3y^2 + 3y ?

    (on a side note, clipper dude, you're in math 3C with boivert at ucla, right? same here)
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