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Order Statistics

  1. Jan 28, 2009 #1

    Suppose that we have N independent and identically distributed Random Variables (RVs) say [tex]X_1, X_2, \ldots, X_N[/tex]. Suppose then that we arrange these RVs in magnitude and in an ascending order as [tex]X_{(1)}\leq X_{(2)}\leq \dots \leq X_{(N)}[/tex]. Now, the question is: what is the probability that we select at random any of these RVs? For example, mathematically:


    Thanks in advance
  2. jcsd
  3. May 18, 2009 #2
    How is X defined?
  4. May 18, 2009 #3
    Suppose that the Probability Density Function (PDF) and the Cumulative Distribution Function (CDF) of these Random Variables (RVs) are [tex]f_X(x)[/tex] and [tex]F_X(x)[/tex], respectively.
  5. May 19, 2009 #4
    I don't understand the problem. If I am going to select one variable out of N variables, why do I need to put them in "<" order first?
  6. May 19, 2009 #5
    In some problems, such as in antenna selection problem, we need to select the antenna that maximizes the performance. In order to compute this performance, we need to arrange the the Signal-to-Noise ratios (SNRs) in ascending order, and find its PDF and CDF. This is a special case, in other cases we need to select an antenna Whose SNR is above a certain threshold, but not necessarly the maximum.
  7. May 19, 2009 #6
    Are you looking at an individual order stat. (for example, max{X1, ..., XN}, or X(N)), and asking what is the probability that X(N) = X3?
  8. May 21, 2009 #7
    I am sorry, I might not state the problem correctly. Suppose we have [tex]N[/tex] random variables that we arrange in magnitude and ascending order. Now suppose that, using random selection cretirion, we select the [tex]r^{th}[/tex] order statistic, which is the [tex]r^{th}[/tex] smallest RV from the arranged ones. Referring to Order Statistics, the CDF of this selected RV is given by:
    [tex]\sum_{i=r}^N {N \choose i} \, F_X^i(x)\,[1-F_X(x)]^{N-i}[/tex]​
    I didn't understand the last equation. How did he obtain this?
  9. May 21, 2009 #8
    That's a question I can answer. I'll use F(x) as a shorthand for FX(x) and write F(r)(x) for the CDF of the r'th order stat. X(r).

    F(r)(x) = Prob{at least r of the X's are less than x}
    = Prob{at least r of the X's are < x and the remaining X's are > x}
    = Prob{at least r of the X's are < x} times Prob{remaining X's are > x}.

    Let's say r = 3. One way that the event {at least r of the X's are < x and remaining X's are > x} could happen is, for example, {X1, X2, X3 < x} and {X4, ..., XN > x}. The corresponding probability is then F(x)3[1 - F(x)]N-3.

    But it can also be {X1, X2, X4 < x} and {X3, X5, ..., XN > x}, or {X1, X3, X4 < x} and {X2, X5, ..., XN > x}, etc. Since all X's have the same F, the probability of each of these subcases is F(x)3[1 - F(x)]N-3. In all, there are (N,3) = N!/(3!(N-3)!) subcases such that exactly 3 of the X's are less than x, and the remaining N-3 of the X's are greater than x.

    Now, still other ways that at least 3 of the X's can be less than x are: {X1, X2, X3, X4 < x} and {X5, X6, ..., XN > x}, or {X1, X2, X3, X4, X5 < x} and {X6, X7, ..., XN > x}, ..., or {all X's < x}.

    In each of the above cases, you have (N,i) ways (subcases) to order the X's, where i is a number between r and N. The probability of each subcase is F(x)i[1 - F(x)]N-i, so the probability of each case is (N,i)F(x)i[1 - F(x)]N-i.

    Summing over all possible cases (i > r), we arrive at the stated formula.
    Last edited: May 21, 2009
  10. May 21, 2009 #9
    Yes, that is right.
  11. May 21, 2009 #10
    Good! No problem.
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