Does the Distribution of X + Y mod a Remain Uniform?

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Discussion Overview

The discussion centers on the distribution of the sum of two random variables, X and Y, when taken modulo a constant a. Participants explore whether the resulting distribution remains uniform under certain conditions, particularly focusing on the independence of X and Y and the implications of Y's distribution.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that 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 Y's distribution.
  • One participant suggests that since X and Y are independent, for any fixed value of Y, the distribution of X + Y (mod a) remains uniform, implying the theorem holds for any combination of values from Y.
  • Another participant mentions that the proof is straightforward, indicating that the theorem holds irrespective of Y's distribution.
  • A participant expresses initial skepticism about the result being counter-intuitive, acknowledging differing intuitions among colleagues.
  • One participant references a paper that discusses distribution functions in arithmetic domains modulo a, suggesting that the theorem follows from a specific section of the paper.
  • A later reply raises a question about the applicability of the theorem when Y is defined over [0,a], proposing that the distributions of (X + Y) mod a and (X + (Y mod a)) mod a are equivalent, thus supporting the original theorem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the applicability of the theorem under different conditions for Y. There are competing views regarding the implications of the cited paper and the conditions under which the theorem holds.

Contextual Notes

Some limitations include the dependence on the specific distribution of Y and the conditions under which the theorem is claimed to hold. The discussion does not resolve whether the theorem is universally applicable or under what specific circumstances it may fail.

areslagae
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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,
 
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areslagae said:
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,

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.
 
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.
 
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:
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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