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

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If X is uniformly distributed over [0,a) and Y is independent, then X + Y (mod a) is also uniformly distributed over [0,a), regardless of Y's distribution. This result, while initially counter-intuitive to some, is supported by a straightforward proof involving the independence of X and Y. A referenced paper, "The Distribution Functions of Random Variables in Arithmetic Domains Modulo a" by P. Scheinok, provides additional context, particularly in section 3.3. The discussion also highlights that the theorem holds even if Y is not restricted to [0,a), as the distributions of (X + Y) mod a and (X + (Y mod a)) mod a are equivalent. The theorem's validity remains intact under these conditions.
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,
 
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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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