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Statistics Problem - Uniform Distribution

  1. Dec 9, 2013 #1
    Hi,

    The question is: http://puu.sh/5GX2G.jpg [Broken]

    http://puu.sh/5GX2G.jpg [Broken]

    I am not exactly sure what the question is asking.


    Here is the answer/solution: http://puu.sh/5GX68.png [Broken]
    But I am not sure what is going on.

    Could someone please explain what exactly the question is asking, I can figure out the rest.

    Thanks,
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 9, 2013 #2
    It's asking you to work out the cumulative distribution function (CDF) F(x). This is defined as the probability that a variate X takes on a value less than or equal to a number x, in other words F(x) = P(X ≤ x).

    Now in this problem X = max(X1, X2, X3, X4), so for any x in order for X ≤ x to be true we must have X1≤ x and X2≤ x and X3≤ x and X4≤ x.

    Are you OK from there?
     
  4. Dec 9, 2013 #3
    Assuming that, I can get the rest of the problem. Once you get, F(x), you can take the derivative to get f(x). To get expected you integrate x*f(x) from 0,1 and x^2 * f(x) for the variance.


    However, I am still confused for the first part (i.e. the cumulative function). The max seems to throw me off. There is another problem similar to this one, except it uses min.

    http://puu.sh/5HsfD.jpg [Broken]

    -------------------------

    The max/min seems to throw me off, what exactly does that represent? :S

    Thanks
     
    Last edited by a moderator: May 6, 2017
  5. Dec 9, 2013 #4

    mathman

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    MrAnchovy gave you the answer.
    X = max(X1, X2, X3, X4) ≤ x is equivalent to all Xi ≤ x.
    Since the Xi are independent, P(X≤x) = P(X1≤x)P(X2≤x)P(X3≤x)P(X4≤x) = x4.
     
  6. Dec 9, 2013 #5
    How would you do min?
     
  7. Dec 9, 2013 #6

    Office_Shredder

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    Gold Member

    You need to calculate
    [tex] P(X_1 \geq x, X_2 \geq x, X_3 \geq x \text{ and } X_4 \geq x ) [/tex]
    Can you figure out what this is?
     
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