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Uniform Distribution, Absolute Value

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the CDF of |X|, given that X is a random variable, uniformly distributed over (-1,3).

    Is |X| uniformly distributed? If yes, over what interval?


    2. Relevant equations



    3. The attempt at a solution

    I found so far that:

    Setting Y=|X|

    Then: Y [tex]\in[/tex] (1,3)

    [tex]F_{Y}(y)=P\left\{Y\leq y\right\}=P\left\{-y\leq X\leq y \right\}= F_{X}(y)-F_{X}(-y)[/tex]

    This sums up to:

    [tex]F_{Y}(y)= \frac{1}{2}y[/tex]

    and differentiating gives the PDF of Y:

    [tex]f_{Y}(y)= \frac{1}{2}[/tex]

    So it seems Y IS uniformly distributed on (1,3).

    My question is, since Y is on a different interval than X, and X is 0 for values less than -1, isn't this the case:

    [tex]F_{X}(-y)=0[/tex] for Y [tex]\in[/tex] (1,3)?

    If so, then the term disappears in the calculation above, and Y's PDF would be [tex]\frac{1}{4}[/tex]

    But tehn this is not even a valid PDF, since integrating over (1,3) doesn't equal 1!!!

    Any ideas?

    Thanks!
     
  2. jcsd
  3. May 19, 2010 #2

    tiny-tim

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    Hi IniquiTrance! :smile:
    Noooo :redface:
     
  4. May 19, 2010 #3
    So if X is 0, what is Y?

    You have written that the density of Y at 0 is zero.
     
  5. May 19, 2010 #4
    Ahhh! Big doors always turn on small hinges!! :blushing:

    So, I guess I should approach it piecewise.

    On (1,3), X and Y share the same PDF, viz., [tex]\frac{1}{4}[/tex].

    Ad it's on (0,1) that:

    [tex]f_{Y}(y)= \frac{1}{2}[/tex]

    Is this right?
     
  6. May 19, 2010 #5
    And Y does not have a uniform distribution, since it does not have a PDF = [tex]\frac{1}{b-a}[/tex] on either sub-interval.
     
  7. May 19, 2010 #6
    Well, to be uniform it would have to have a constant density function over the entire interval, not just some subinterval. But the rest is okay.
     
  8. May 20, 2010 #7

    tiny-tim

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    Hi IniquiTrance! :smile:

    (just got up :zzz: …)
    But does it have a uniform distribution on either sub-interval separately?

    (and don't forget that the question asked for the CDF)
     
  9. May 20, 2010 #8
    Hmm, doesn't a uniform distribution have to have a PDF given by:

    [tex]
    \frac{1}{b-a}
    [/tex]

    I know that on the two subintervals, the PDF's are constants: 1/2 on [0,1), and 1/4 on [1,3). But neither integrates to 1 over the subinterval.

    Can they still be uniform distributions notwithstanding the above?
     
  10. May 20, 2010 #9

    tiny-tim

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    Well, it's an odd way of putting it, that I've not come across before, but in view of the wording of the question …
    … I'm confused as to what answer is expected. :redface:
     
  11. May 20, 2010 #10
    Right, I stil can't decide. From researching up to the 30th page of a google search, it seems that the criterion for a uniform distribution is that its PDF = [tex]

    \frac{1}{b-a}

    [/tex]
    over (a,b).

    I still couldn't find anything attacking the question directly, of given the PDF doesn't equal that quotient, but is constant over a subinterval, is it still uniformly distributed there?
     
  12. May 20, 2010 #11

    tiny-tim

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    Well, I'd say it is :smile:, but I've no idea what the official answer would be. :redface:
     
  13. May 20, 2010 #12
    It seems like it should since the idea is that the probability of the variable occurring in some interval is dependent only on the length of the interval, whenever the PDF is constant, which is the unique feature of uniform distributions.

    Thanks very much for your help! :biggrin:
     
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