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Modulo sum of random variables

  1. Sep 2, 2010 #1
    If X is uniformly distributed over [0,a), and Y is independent, then X + Y (mod a) is uniformly distributed over [0,a), independent of the distribution of Y.

    Can anyone point me to a statistics text that shows this?

    Thanks,
     
  2. jcsd
  3. Sep 2, 2010 #2

    CRGreathouse

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    Consider any possible value from Y. Since X and Y are independent, X + that value is uniformly distributed mod a. Now since this is true for any value, it is true for any combination of values.
     
  4. Sep 2, 2010 #3

    mathman

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    I don't know of a text, but the proof is simple enough. Let Y=y, then X+y (mod a) is uniformly distributed, since X is independent of Y. Therefore the theorem holds irrespective of the distribution of Y.
     
  5. Sep 3, 2010 #4
    Thanks for both your replies!

    At first, me (and my collegues) found this result somewhat counter-intuitive. It seems that you do not, but you most likely you have a deeper intuition.

    Meanwhile, I also found the following paper which is interesting in this context:
    The Distribution Functions of Random Variables in Arithmetic Domains Modulo a
    P. Scheinok
    http://www.jstor.org/stable/2310973

    It seems that the theorem from my first post follows from 3.3, with g_Y()=1/a.

    Thanks,
     
    Last edited by a moderator: Apr 25, 2017
  6. Sep 3, 2010 #5
    Hi all,

    I have a quick additional question.

    A colleague pointed out to me that the cited paper only proves the theorem from my first post in the case that Y is defined over [0,a].

    However, the random variables (X + Y) mod a and (X + (Y mod a)) mod a have the same distribution, so the original theorem should hold, no?

    Thanks,
     
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