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Thread: Probability

  1. #1
    Feb 2010

    Smile Probability

    Hey can someone help me figure this out plz

    If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y≤m)=P(Y≥m)=0.5. If Y1, Y2,...,Yn are independent, exponentially distributed random variables with mean β and median m then Y(n)= max(Y1, Y2,...,Yn) does not have an exponential distribution. Use the general form of FY(n)(y) to show that P(Y(n)>m)= 1-(0.5)^n.

    Thanks alot !!
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  2. #2
    MHF Contributor matheagle's Avatar
    Feb 2009
    This is easy..... P(Y(n)>m)= 1-(0.5)^n

    P(Y(n)>m)= 1-P(Y(n)<m)= 1-P(Y1<m)P(Y2<m)....P(Yn<m)=1-(0.5)^n

    For the first part, just get the density or cdf of the largest order stat.
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