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Question in Order Statistics

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  • Question in Order Statistics

    Section 9 of the Actex manual, the distribution for each of the order statistics Y1-Yn is given as a formula, then an example (9-11) is given to illustrate its use, in which three exponentially distributed (with mean 1) variables are used to find the expectation of the second flight arriving, which is calculated at 5/6. I also did the first and third at 1/3 and 11/6 respectively.

    I noted that 1/3 = mean/number of variables
    5/6 = (2*mean-1/3)/2
    11/6= (3*mean-1/3-5/6)/1

    Is this a real pattern that can be generalized?

  • #2
    I haven't given this a great deal of thought, and there may
    be more efficient ways of doing the derivations, but I have
    the following comments.

    For an exponential distribution with mean theta, if we consider
    the order statistics of a random sample X1, X2, ..., Xn , of size n,
    then the first order statistic Y1 (the minimum of the sample)
    will have a pdf of n x [1 - F(t)]^(n-1) x f(t) .
    This becomes (n/theta) x e^(-nt/theta) .
    The expected value is the integral from 0 to infinity of
    (n/theta) x t x e^(-nt/theta) .
    This integral is theta/n , so it is no coincidence that
    when n=3 the mean of the first order statistic is theta/3 ,
    which is 1/3 in your example because theta is 1.
    There is a general rule for exponential integration that was used
    in the integral above:
    integral from 0 to infin of t^k x e^-at is k!/(a^(k+1)) .
    In the integral above, k=1 and a = n/theta .

    For the 2nd order stat Y2, the pdf is
    n(n-1) x F(t) x [1 - F(t)]^(n-2) x f(t) .
    For the exponential dist, this is
    (n(n-1)/theta) x [1 - e^(-t/theta)] x e^(-(n-1)t/theta)
    which becomes
    (n(n-1)/theta) x [e^(-(n-1)t/theta) - e^(-nt/theta) ]
    The mean of Y2 is the integral of
    t x (n(n-1)/theta) x [e^(-(n-1)t/theta) - e^(-nt/theta) ]
    which is (2n-1)theta/(n(n-1)) .
    For n=3 , this is 5theta/6 .
    For general n, the mean of the 2nd order stat, which we
    just saw is (2n-1)theta/(n(n-1)) , can be written as
    (2 theta - theta/n)/(n-1) , which shows that
    E[Y2] = (2 theta - E[Y1])/(n-1) (plug in n=3 for your result).

    I haven't taken the time to work through the general
    form of the mean of Y3 for the exponential.
    However, for a sample of size n=3 , it must be true
    that E[Y1] + E[Y2] + E[Y3] = 3 theta (since each of the
    Y's is on of the original sample X's, it must be true
    that Y1 + Y2 + Y3 = X1 + X2 + X3 , so
    E[Y1 + Y2 + Y3 ] = E[X1 + X2 + X3 ] = 3 theta.
    It follows that E[Y3] = 3theta - E[Y1] - E[Y2] .

    I'm sorry if this is a little complicated. I haven't thought
    it through enough to come up with a simpler solution
    (if there is one).
    Sam Broverman

    [email protected]


    • #3
      Man, this board needs to be LaTex enabled!!


      • #4
        Yes, this seems to generalize, albeit only to order statistics of exponentials. Letting X1, ..., Xn be iid exponential with mean mu, and Y1, ..., Yn their order statistics, then as Dr. Broverman pointed out, it is easy to see that Y1 is an exponential with mean mu/n.

        For the higher order statistics, the memoryless property of exponentials shows that Y{k+1}-Y_k is likewise exponential with mean mu/(n-k). This means that EYj=mu (sum_{i=1}^j 1/(n-i+1). Rearranging terms and doing more algebra than I want to type gives your series.

        I don't know whether or not the memoryless property for exponentials is on the syllabus for P, but I assume that it is for M. Basically it says that if T is a reasonable random variable and you are given that Xi>T, then Xi-T is an exponential with mean mu that is independent of T. The fact that Y{k+1}-Yk is exponential with mean mu/(n-k) then is just the fact that n-k of the exponentials are larger than Yk, and independent of Yk.


        • #5
          Makes total sense that exponentials would possibly be a special case - yeah, we learned about the memory-less exponential property in my probability course.

          I'm getting close to being ready for the test (three weeks!) I just finished the 123 problems, got 103/123, and found out I need a lot of work on some of the joint distribution (specifically conditional distributions and combinations of random variables (such as Y=X1+5*X2, what's the joint distribution? problems)).