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Placing random variables in order

  1. Oct 10, 2012 #1
    Hello
    Let's say we have some continuous i.i.d random variables [itex]X_1, \ldots X_n[/itex] from a known distribution with some parameter [itex]\theta[/itex]
    We then place them in ascending order [itex]X_{(1)}, \ldots X_{(n)}[/itex] such that [itex]X_{(i)}, < X_{(i+1)}[/itex].

    We call this operation [itex]T(\mathbf{X})[/itex] where [itex]\mathbf{X}[/itex] is our vector [itex]X_1, \ldots X_n[/itex].

    Now let's say we are interested in finding out whether [itex]P(\mathbf{X} = X| T(\mathbf{X}) = t)[/itex] where t and x are both vectors (and outcomes), depends on [itex]\theta[/itex].

    By definition of conditional probability, we have:
    [itex]P(\mathbf{X} = X| T(\mathbf{X}) = t) = \frac{P(\mathbf{X} = X, T(\mathbf{X}) = t)}{P(T(\mathbf{X}) = t)}[/itex]

    Trying to find these 2 probabilities:
    [itex]P(T(\mathbf{X}) = t)[/itex], this is the probability that my ascending ordering [itex]X_{(1)}, \ldots X_{(n)}[/itex] is a certain vector. This probability should simply be
    [itex]\prod^{n}_{i=1}f(x_i)[/itex], since we already know that they are ordered.

    [itex]P(\mathbf{X} = X, T(\mathbf{X}) = t)[/itex], this is the probability that my random vector [itex]\mathbf{X}[/itex] attains a certain (vector)value while the ordering attains a certain (vector)value. But this should also be equal to [itex]\prod^{n}_{i=1}f(x_i)[/itex].

    So
    [itex]P(\mathbf{X} = X| T(\mathbf{X}) = t) = \frac{P(\mathbf{X} = X, T(\mathbf{X}) = t)}{P(T(\mathbf{X}) = t)} = 1[/itex]. If this is correct, what does it mean that the probability is 1? Is it wrong? why?
     
  2. jcsd
  3. Oct 10, 2012 #2

    mfb

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    I agree with [itex]P(\mathbf{X} = X, T(\mathbf{X}) = t) = P(\mathbf{X} = X) = \prod^{n}_{i=1}f(x_i)[/itex]

    However, the probability for a specific ordering, [itex]P(T(\mathbf{X}) = t)[/itex], is different, as you have multiple ways (multiple X) to get the same P(X) - assuming the probability that two Xi are the same is 0, you have exactly n! ways.
    Therefore, [itex]P(T(\mathbf{X}) = t)=\frac{1}{n!}P(\mathbf{X} = X)[/itex].

    This is easy to see with an example:

    ##P(T(\mathbf{X}) = (1,2)) = P(\mathbf{X} = (1,2)) + P(\mathbf{X} = (2,1))##

    ##P(\mathbf{X} = X| T(\mathbf{X}) = t)## is equivalent to "one specific permutation out of n! was chosen" - assuming ##T(X)=t##, of course, otherwise it is 0.
     
  4. Oct 10, 2012 #3

    Mute

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    Are the random variables discrete or continuous?

    http://planetmath.org/encyclopedia/OrderStatistics.html [Broken] claims that the pdf should be ##n! \prod_i f_X(x_i)## for continuous variables ##x##, which seems to agree with what mfb has said.

    In case the OP is not already aware, such a problem is one of order statistics.
     
    Last edited by a moderator: May 6, 2017
  5. Oct 10, 2012 #4
    I see what you mean.
    But since ##P(T(\mathbf{X}) = (1,2)) = P(\mathbf{X} = (1,2)) + P(\mathbf{X} = (2,1)) = 2*P(\mathbf{X} = (1,2))##, shouldnt it rather be:
    [itex]P(T(\mathbf{X}) = t)= n!P(\mathbf{X} = X)[/itex]?

    This way, the probability ##P(\mathbf{X} = X|T(\mathbf{x}) = t)## is equal to ##\frac{1}{n!}## which seems very intuitive: Given that the order ##T## has taken a certain value ##t = X_{(1)},\ldots X_{(n)}## this corresponds to (as pointed out by mfb) excactly ##n!## outcomes of ##\mathbf{X} = X_1 \ldots X_n## such that the probability of ##\mathbf{X}## attaining one of them is excactly ##\frac{1}{n!}##
     
    Last edited: Oct 10, 2012
  6. Oct 10, 2012 #5

    mfb

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    Oh, wrong side of the equation. Of course, otherwise your conditional probability would be n! (>1...) instead of 1/n! (correct).
     
  7. Oct 12, 2012 #6
    Same question but we pick another ##T##: ##T = (\text{min}(\mathbf{X}),\text{max}(\mathbf{X}))##. In this case we should get the same probability ##P(T(X)=t)## as before, namely ##P(T(\mathbf{X}) = t)= n!P(\mathbf{X} = X)##, for consider the case: ##P(T = 1,X_2,3)## this is the sum of the probabilites of ##\mathbf{X}## attaining all possible permutations of ##(1,X_2,3)##, which equals ##3!P(\mathbf{X} = (1,X_2,3) )##.

    For ##P(\mathbf{X} = X, T(\mathbf{X}) = t)## , I believe the same reason applies, since ##T## just takes the largest and smallest value of ##\mathbf{X}##. Hence ##P(\mathbf{X} = X, T(\mathbf{X}) = t) = P(\mathbf{X} = X) = \prod^{n}_{i=1}f(x_i)##

    We therefore get the same result as before. I don't know about the intuition behind this, (look at my interpretation of the previous result), Shouldn't the values of ##\mathbf{X}## that are between the smallest and largest have more "liberty" of attaining their outcomes since we, in this case, only impose the restriction that they are between ##\text{min}(\mathbf{X})## and ##\text{max}(\mathbf{X})##?
     
    Last edited: Oct 12, 2012
  8. Oct 12, 2012 #7

    mfb

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    I agree.

    Why?

    A constructive approach: Pick the index with the lowest value (n choices), and the index with the highest value (n-1 choices), require that all other variables are between those values:
    ##P(T(\mathbf{X}) = t)= n(n-1)f(t_1)f(t_2)\left(\int_{t_1}^{t_2}f(x)\right)^{n-2}##
     
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